{"appState":{"pageLoadApiCallsStatus":true},"categoryState":{"relatedCategories":{"headers":{"timestamp":"2023-05-31T12:01:08+00:00"},"categoryId":33723,"data":{"title":"Calculus","slug":"calculus","image":{"src":null,"width":0,"height":0},"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"parentCategory":{"categoryId":33720,"title":"Math","slug":"math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"}},"childCategories":[],"description":"It's time to conquer calc. With your calculator in hand and these articles by your side, you're ready to take on even the scariest differential equations.","relatedArticles":{"self":"https://dummies-api.dummies.com/v2/articles?category=33723&offset=0&size=5"},"hasArticle":true,"hasBook":true,"articleCount":258,"bookCount":8},"_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"relatedCategoriesLoadedStatus":"success"},"listState":{"list":{"count":10,"total":258,"items":[{"headers":{"creationTime":"2016-03-27T16:54:14+00:00","modifiedTime":"2023-04-11T17:37:22+00:00","timestamp":"2023-04-11T18:01:03+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"Calculus II For Dummies Cheat Sheet","strippedTitle":"calculus ii for dummies cheat sheet","slug":"calculus-ii-for-dummies-cheat-sheet","canonicalUrl":"","seo":{"metaDescription":"Take away the fear of studying calculus by understanding its basic principles, such as derivatives, integration, and compound functions.","noIndex":0,"noFollow":0},"content":"By its nature, calculus can be intimidating. But you can take some of the fear of studying calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. Also discover a few basic rules applied to calculus like Cramer's Rule, the Constant Multiple Rule, and a few others, and you'll be on your way to acing the course.","description":"By its nature, calculus can be intimidating. But you can take some of the fear of studying calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. Also discover a few basic rules applied to calculus like Cramer's Rule, the Constant Multiple Rule, and a few others, and you'll be on your way to acing the course.","blurb":"","authors":[],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}},{"articleId":179232,"title":"Finding the Volume of a Solid with Similar Cross Sections","slug":"finding-the-volume-of-a-solid-with-similar-cross-sections","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179232"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781119986614","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/1119986613/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/1119986613/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/1119986613-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/1119986613/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/1119986613/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-3rd-edition-cover-1119986613-203x255.jpg","width":203,"height":255},"title":"Calculus II For Dummies, 3rd Edition","testBankPinActivationLink":"","bookOutOfPrint":true,"authorsInfo":"<p><b><b data-author-id=\"9399\">Mark Zegarelli</b></b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear &#8212; and fun &#8212; for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.</p>","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <b>Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear &#8212; and fun &#8212; for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"_links":{"self":"https://dummies-api.dummies.com/v2/books/"}},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781119986614&quot;]}]\" id=\"du-slot-6435a05f8da02\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781119986614&quot;]}]\" id=\"du-slot-6435a05f8e2b9\"></div></div>"},"articleType":{"articleType":"Cheat Sheet","articleList":[{"articleId":188540,"title":"The Most Important Derivatives and Antiderivatives to Know","slug":"the-most-important-derivatives-and-antiderivatives-to-know","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188540"}},{"articleId":190797,"title":"The Riemann Sum Formula For the Definite Integral","slug":"the-riemann-sum-formula-for-the-definite-integral","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/190797"}},{"articleId":188541,"title":"Integration by Parts with the DI-agonal Method","slug":"integration-by-parts-with-the-di-agonal-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188541"}},{"articleId":188504,"title":"The Sum Rule, the Constant Multiple Rule, and the Power Rule for Integration","slug":"the-sum-rule-the-constant-multiple-rule-and-the-power-rule-for-integration","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188504"}},{"articleId":188543,"title":"How to Solve Integrals with Variable Substitution","slug":"how-to-solve-integrals-with-variable-substitution","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188543"}},{"articleId":188542,"title":"How to Use Integration by Parts","slug":"how-to-use-integration-by-parts","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188542"}},{"articleId":188544,"title":"How to Solve Compound Functions Where the Inner Function Is <i>ax</i> + <i>b</i>","slug":"how-to-solve-compound-functions-where-the-inner-function-is-ax-b","categoryList":[],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188544"}},{"articleId":188529,"title":"Solve Compound Functions Where the Inner Function Is <i>ax</i>","slug":"solve-compound-functions-where-the-inner-function-is-ax","categoryList":[],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/188529"}}],"content":[{"title":"The most important derivatives and antiderivatives to know","thumb":null,"image":null,"content":"<p>The table below shows you how to differentiate and integrate 18 of the most common functions. As you can see, integration reverses differentiation, returning the function to its original state, up to a constant <i>C</i>.</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246703.image0.png\" alt=\"image0.png\" width=\"401\" height=\"967\" /></p>\n"},{"title":"The Riemann Sum Formula for the definite integral","thumb":null,"image":null,"content":"<p>The Riemann Sum formula provides a precise definition of the definite integral as the limit of an infinite series. The Riemann Sum formula is as follows<i>:</i><i> </i></p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223426.image0.png\" alt=\"image0.png\" width=\"191\" height=\"55\" /></p>\n<p>Below are the steps for approximating an integral using six rectangles:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223427.image1.png\" alt=\"image1.png\" width=\"312\" height=\"36\" /></p>\n<ol class=\"level-one\">\n<li>\n<p class=\"first-para\">Increase the number of rectangles (<i>n</i>) to create a better approximation:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223428.image2.png\" alt=\"image2.png\" width=\"259\" height=\"36\" /></li>\n<li>\n<p class=\"first-para\">Simplify this formula by factoring out <i>w</i> from each term:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223429.image3.png\" alt=\"image3.png\" width=\"239\" height=\"36\" /></li>\n<li>\n<p class=\"first-para\">Use the summation symbol to make this formula even more compact:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223430.image4.png\" alt=\"image4.png\" width=\"125\" height=\"52\" /></p>\n<p class=\"child-para\">The value <i>w</i> is the <i>width</i> of each rectangle:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223431.image5.png\" alt=\"image5.png\" width=\"65\" height=\"41\" /></p>\n<p class=\"child-para\">Each <i>h </i>value<i> </i>is the <i>height</i> of a different rectangle:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223432.image6.png\" alt=\"image6.png\" width=\"35\" height=\"29\" /></p>\n<p class=\"child-para\">So here is the Riemann Sum formula for <i>approximating</i> an integral using <i>n</i> rectangles:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223433.image7.png\" alt=\"image7.png\" width=\"168\" height=\"52\" /></li>\n<li>\n<p class=\"first-para\">For a better approximation, use the limit</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223434.image8.png\" alt=\"image8.png\" width=\"51\" height=\"25\" /></li>\n<li>\n<p class=\"first-para\">to allow the number of rectangles to approach <i>infinity</i>:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/223435.image9.png\" alt=\"image9.png\" width=\"216\" height=\"52\" /></li>\n</ol>\n"},{"title":"Integration by parts with the DI-agonal method","thumb":null,"image":null,"content":"<p>The DI-agonal method is basically integration by parts with a chart that helps you organize information. This method is especially useful when you need to integrate by parts more than once to solve a problem. Use the following table for integration by parts using the DI-agonal method:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246729.image0.jpg\" alt=\"image0.jpg\" width=\"300\" height=\"240\" /></p>\n"},{"title":"The Sum Rule, the Constant Multiple Rule, and the Power Rule for integration","thumb":null,"image":null,"content":"<p>When you perform integration, there are three important rules that you need to know: the Sum Rule, the Constant Multiple Rule, and the Power Rule.</p>\n<p>The Sum Rule for Integration tells you that it’s okay to integrate long expressions term by term. Here it is formally:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246706.image0.png\" alt=\"image0.png\" width=\"239\" height=\"29\" /></p>\n<p>The Constant Multiple Rule for Integration tells you that it’s okay to move a constant outside of an integral before you integrate. Here it is expressed in symbols:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246707.image1.png\" alt=\"image1.png\" width=\"137\" height=\"29\" /></p>\n<p>The Power Rule for Integration allows you to integrate any real power of <i>x</i> (except -1). Here’s the Power Rule expressed formally:</p>\n<p>where <i>n</i> ≠ -1</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246708.image2.png\" alt=\"image2.png\" width=\"120\" height=\"45\" /></p>\n"},{"title":"How to solve integrals with variable substitution","thumb":null,"image":null,"content":"<p>In Calculus, you can use variable substitution to evaluate a complex integral. Variable substitution allows you to integrate when the Sum Rule, Constant Multiple Rule, and Power Rule don’t work.</p>\n<ol class=\"level-one\">\n<li>\n<p class=\"first-para\">Declare a variable <i>u</i>,<i> </i>set it equal to an algebraic expression that appears in the integral, and then substitute <i>u</i> for this expression in the integral.</p>\n</li>\n<li>\n<p class=\"first-para\">Differentiate <i>u</i> to find</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246732.image0.png\" alt=\"image0.png\" width=\"24\" height=\"41\" /></p>\n<p class=\"child-para\">and then isolate all <i>x</i> variables on one side of the equal sign.</p>\n</li>\n<li>\n<p class=\"first-para\">Make another substitution to change <i>dx</i> and all other occurrences of <i>x</i> in the integral to an expression that includes <i>du</i>.</p>\n</li>\n<li>\n<p class=\"first-para\">Integrate by using <i>u</i> as your new variable of integration.</p>\n</li>\n<li>\n<p class=\"first-para\">Express this answer in terms of <i>x</i>.</p>\n</li>\n</ol>\n"},{"title":"How to Use integration by parts","thumb":null,"image":null,"content":"<p>When doing Calculus, the formula for integration by parts gives you the option to break down the product of two functions to its factors and integrate it in an altered form. To use integration by parts in Calculus, follow these steps:</p>\n<ol class=\"level-one\">\n<li>\n<p class=\"first-para\">Decompose the entire integral (including <i>dx</i>) into two factors.</p>\n</li>\n<li>\n<p class=\"first-para\">Let the factor without <i>dx </i>equal <i>u</i> and the factor with <i>dx </i>equal <i>dv</i>.</p>\n</li>\n<li>\n<p class=\"first-para\">Differentiate <i>u</i> to find <i>du</i>, and integrate <i>dv</i> to find <i>v</i>.</p>\n</li>\n<li>\n<p class=\"first-para\">Use the formula:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246735.image0.png\" alt=\"image0.png\" width=\"123\" height=\"29\" /></li>\n<li>\n<p class=\"first-para\">Evaluate the right side of this equation to solve the integral.</p>\n</li>\n</ol>\n"},{"title":"How to solve compound functions where the inner function is ax + b","thumb":null,"image":null,"content":"<p>Some integrals of compound functions <i>f</i><i> </i>(<i>g</i>(<i>x</i>)) are easy to do quickly in Calculus. These include compound functions for which you know how to integrate the outer function <i>f</i>, and the inner function <i>g</i>(<i>x</i>) is of the form <i>ax</i> + <i>b</i> — that is, it differentiates to a constant.</p>\n<p>Here are some examples:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246711.image0.png\" alt=\"image0.png\" width=\"231\" height=\"151\" /></p>\n"},{"title":"Solve compound functions where the inner function is ax","thumb":null,"image":null,"content":"<p>When figuring Calculus problems, some integrals of compound functions <i>f</i><i> </i>(<i>g</i>(<i>x</i>)) are easy to do quickly. These include compound functions for which you know how to integrate the outer function <i>f</i>, and the inner function <i>g</i>(<i>x</i>) is of the form <i>ax</i> — that is, it differentiates to a constant.</p>\n<p>Here are some examples:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/246714.image0.png\" alt=\"image0.png\" width=\"175\" height=\"151\" /></p>\n"}],"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2021-07-12T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":208670},{"headers":{"creationTime":"2016-03-26T18:17:29+00:00","modifiedTime":"2022-09-22T18:17:40+00:00","timestamp":"2022-09-22T19:20:50+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"Substituting with Expressions of the Form f(x) Multiplied by h(g(x))","strippedTitle":"substituting with expressions of the form f(x) multiplied by h(g(x))","slug":"substituting-with-expressions-of-the-form-fx-multiplied-by-hgx","canonicalUrl":"","seo":{"metaDescription":"Learn how to do this calculus problem with this explanation, including the equations and step-by-step instructions.","noIndex":0,"noFollow":0},"content":"When <i>g</i>'(<i>x</i>) = <i>f</i>(<i>x</i>), you can use the substitution <i>u</i> = <i>g</i>(<i>x</i>) to integrate expressions of the form <i>f</i>(<i>x</i>) multiplied by <i>h</i>(<i>g</i>(<i>x</i>)), provided that <i>h</i> is a function that you already know how to integrate.\r\n\r\nVariable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration.\r\n\r\nHere’s a hairy-looking integral that actually responds well to substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314497.image0.png\" alt=\"image0.png\" width=\"120\" height=\"71\" />\r\n\r\nThe key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator. Watch how this plays out in this substitution:\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Declare <i>u</i> equal to the inner function in the denominator and make the substitution:</p>\r\n<p class=\"child-para\">Here’s the substitution:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314498.image1.png\" alt=\"image1.png\" width=\"83\" height=\"52\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Differential <i>d</i><i>u</i> = (2<i>x</i> + 1) <i>dx</i><i>:</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314499.image2.png\" alt=\"image2.png\" width=\"100\" height=\"67\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">The second part of the substitution now becomes clear:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314500.image3.png\" alt=\"image3.png\" width=\"68\" height=\"43\" />\r\n<p class=\"child-para\">Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator. (You may think that this is quite a coincidence, but coincidences like these happen all the time on exams!)</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Integration is now quite straightforward:</p>\r\n<p class=\"child-para\">You take an extra step to remove the fraction before you integrate:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314501.image4.png\" alt=\"image4.png\" width=\"85\" height=\"77\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Substitute back <i>x</i><sup>2</sup> + <i>x</i> – 5 for <i>u:</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314502.image5.png\" alt=\"image5.png\" width=\"153\" height=\"47\" /></li>\r\n</ol>\r\nChecking the answer by differentiating with the Chain Rule reveals how this problem was set up in the first place:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314503.image6.png\" alt=\"image6.png\" width=\"268\" height=\"197\" />\r\n\r\nHere’s another example where you make a variable substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314504.image7.png\" alt=\"image7.png\" width=\"96\" height=\"33\" />\r\n\r\nNotice that the derivative of <i>x</i><sup>4</sup> – 1 is <i>x</i><sup>3</sup>, off by a constant factor. So here’s the declaration, followed by the differentiation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314505.image8.png\" alt=\"image8.png\" width=\"91\" height=\"115\" />\r\n\r\nNow you can just do both substitutions at once:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314506.image9.png\" alt=\"image9.png\" width=\"87\" height=\"77\" />\r\n\r\nAt this point, you can solve the integral simply.\r\n\r\nSimilarly, here’s another example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314507.image10.png\" alt=\"image10.png\" width=\"109\" height=\"31\" />\r\n\r\nAt first glance, this integral looks just plain horrible. But on further inspection, notice that the derivative of cot <i>x</i> is –csc<sup>2</sup> <i>x,</i> so this looks like another good candidate:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314508.image11.png\" alt=\"image11.png\" width=\"111\" height=\"95\" />\r\n\r\nThis results in the following substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314509.image12.png\" alt=\"image12.png\" width=\"81\" height=\"67\" />\r\n\r\nAgain, this is another integral that you can solve.","description":"When <i>g</i>'(<i>x</i>) = <i>f</i>(<i>x</i>), you can use the substitution <i>u</i> = <i>g</i>(<i>x</i>) to integrate expressions of the form <i>f</i>(<i>x</i>) multiplied by <i>h</i>(<i>g</i>(<i>x</i>)), provided that <i>h</i> is a function that you already know how to integrate.\r\n\r\nVariable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration.\r\n\r\nHere’s a hairy-looking integral that actually responds well to substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314497.image0.png\" alt=\"image0.png\" width=\"120\" height=\"71\" />\r\n\r\nThe key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator. Watch how this plays out in this substitution:\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Declare <i>u</i> equal to the inner function in the denominator and make the substitution:</p>\r\n<p class=\"child-para\">Here’s the substitution:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314498.image1.png\" alt=\"image1.png\" width=\"83\" height=\"52\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Differential <i>d</i><i>u</i> = (2<i>x</i> + 1) <i>dx</i><i>:</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314499.image2.png\" alt=\"image2.png\" width=\"100\" height=\"67\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">The second part of the substitution now becomes clear:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314500.image3.png\" alt=\"image3.png\" width=\"68\" height=\"43\" />\r\n<p class=\"child-para\">Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator. (You may think that this is quite a coincidence, but coincidences like these happen all the time on exams!)</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Integration is now quite straightforward:</p>\r\n<p class=\"child-para\">You take an extra step to remove the fraction before you integrate:</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314501.image4.png\" alt=\"image4.png\" width=\"85\" height=\"77\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Substitute back <i>x</i><sup>2</sup> + <i>x</i> – 5 for <i>u:</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314502.image5.png\" alt=\"image5.png\" width=\"153\" height=\"47\" /></li>\r\n</ol>\r\nChecking the answer by differentiating with the Chain Rule reveals how this problem was set up in the first place:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314503.image6.png\" alt=\"image6.png\" width=\"268\" height=\"197\" />\r\n\r\nHere’s another example where you make a variable substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314504.image7.png\" alt=\"image7.png\" width=\"96\" height=\"33\" />\r\n\r\nNotice that the derivative of <i>x</i><sup>4</sup> – 1 is <i>x</i><sup>3</sup>, off by a constant factor. So here’s the declaration, followed by the differentiation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314505.image8.png\" alt=\"image8.png\" width=\"91\" height=\"115\" />\r\n\r\nNow you can just do both substitutions at once:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314506.image9.png\" alt=\"image9.png\" width=\"87\" height=\"77\" />\r\n\r\nAt this point, you can solve the integral simply.\r\n\r\nSimilarly, here’s another example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314507.image10.png\" alt=\"image10.png\" width=\"109\" height=\"31\" />\r\n\r\nAt first glance, this integral looks just plain horrible. But on further inspection, notice that the derivative of cot <i>x</i> is –csc<sup>2</sup> <i>x,</i> so this looks like another good candidate:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314508.image11.png\" alt=\"image11.png\" width=\"111\" height=\"95\" />\r\n\r\nThis results in the following substitution:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314509.image12.png\" alt=\"image12.png\" width=\"81\" height=\"67\" />\r\n\r\nAgain, this is another integral that you can solve.","blurb":"","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[{"articleId":208670,"title":"Calculus II For Dummies Cheat Sheet","slug":"calculus-ii-for-dummies-cheat-sheet","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/208670"}},{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781118161708","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/111816170X-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-2nd-edition-cover-9781118161708-202x255.jpg","width":202,"height":255},"title":"Calculus II For Dummies","testBankPinActivationLink":"","bookOutOfPrint":false,"authorsInfo":"<b data-author-id=\"9399\">Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"_links":{"self":"https://dummies-api.dummies.com/v2/books/"}},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-632cb5928a82a\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-632cb5928ad6c\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-22T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":178425},{"headers":{"creationTime":"2016-03-26T18:16:25+00:00","modifiedTime":"2022-09-19T15:34:13+00:00","timestamp":"2022-09-19T18:01:03+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"Setting Up Partial Fractions When You Have Repeated Linear Factors","strippedTitle":"setting up partial fractions when you have repeated linear factors","slug":"setting-up-partial-fractions-when-you-have-repeated-linear-factors","canonicalUrl":"","seo":{"metaDescription":"Learn how to solve a calculus problem involving partial fractions and repeated linear factors. This article walks you through the process.","noIndex":0,"noFollow":0},"content":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated linear factors. These are difficult to work with because each factor requires more than one partial fraction.\r\nFor each squared linear factor in the denominator, add <i>two</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314691.image0.png\" alt=\"image0.png\" width=\"195\" height=\"51\" />\r\n\r\nFor each quadratic factor in the denominator that’s raised to the third power, add <i>three</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314692.image1.png\" alt=\"image1.png\" width=\"307\" height=\"51\" />\r\n\r\nGenerally speaking, when a linear factor is raised to the <i>n</i>th power, add <i>n</i> partial fractions. For example, suppose that you want to integrate the following expression:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314693.image2.png\" alt=\"image2.png\" width=\"100\" height=\"55\" />\r\n\r\nThis expression contains all linear factors, but one of these factors (<i>x</i> + 5) is nonrepeating and the other (<i>x</i> – 1) is raised to the third power. Set up your partial fractions this way:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314694.image3.png\" alt=\"image3.png\" width=\"227\" height=\"51\" />\r\n\r\nWhich will yield:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314695.image4.png\" alt=\"image4.png\" width=\"269\" height=\"93\" />\r\n\r\nAs you can see, this example adds one partial fraction to account for the nonrepeating factor and three to account for the repeating factor.\r\n\r\nWhen you start out with a linear factor, using partial fractions leaves you with an integral in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314696.image5.png\" alt=\"image5.png\" width=\"92\" height=\"51\" />\r\n\r\nIntegrate all these cases by using the variable substitution <i>u</i> = <i>ax</i> + <i>b</i> so that <i>du = a dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314697.image6.png\" alt=\"image6.png\" width=\"57\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314698.image7.png\" alt=\"image7.png\" width=\"76\" height=\"41\" />\r\n\r\nHere are a few examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314699.image8.png\" alt=\"image8.png\" width=\"205\" height=\"136\" />","description":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated linear factors. These are difficult to work with because each factor requires more than one partial fraction.\r\nFor each squared linear factor in the denominator, add <i>two</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314691.image0.png\" alt=\"image0.png\" width=\"195\" height=\"51\" />\r\n\r\nFor each quadratic factor in the denominator that’s raised to the third power, add <i>three</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314692.image1.png\" alt=\"image1.png\" width=\"307\" height=\"51\" />\r\n\r\nGenerally speaking, when a linear factor is raised to the <i>n</i>th power, add <i>n</i> partial fractions. For example, suppose that you want to integrate the following expression:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314693.image2.png\" alt=\"image2.png\" width=\"100\" height=\"55\" />\r\n\r\nThis expression contains all linear factors, but one of these factors (<i>x</i> + 5) is nonrepeating and the other (<i>x</i> – 1) is raised to the third power. Set up your partial fractions this way:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314694.image3.png\" alt=\"image3.png\" width=\"227\" height=\"51\" />\r\n\r\nWhich will yield:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314695.image4.png\" alt=\"image4.png\" width=\"269\" height=\"93\" />\r\n\r\nAs you can see, this example adds one partial fraction to account for the nonrepeating factor and three to account for the repeating factor.\r\n\r\nWhen you start out with a linear factor, using partial fractions leaves you with an integral in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314696.image5.png\" alt=\"image5.png\" width=\"92\" height=\"51\" />\r\n\r\nIntegrate all these cases by using the variable substitution <i>u</i> = <i>ax</i> + <i>b</i> so that <i>du = a dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314697.image6.png\" alt=\"image6.png\" width=\"57\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314698.image7.png\" alt=\"image7.png\" width=\"76\" height=\"41\" />\r\n\r\nHere are a few examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314699.image8.png\" alt=\"image8.png\" width=\"205\" height=\"136\" />","blurb":"","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[{"articleId":208670,"title":"Calculus II For Dummies Cheat Sheet","slug":"calculus-ii-for-dummies-cheat-sheet","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/208670"}},{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781118161708","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/111816170X-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-2nd-edition-cover-9781118161708-202x255.jpg","width":202,"height":255},"title":"Calculus II For Dummies","testBankPinActivationLink":"","bookOutOfPrint":false,"authorsInfo":"<b data-author-id=\"9399\">Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"_links":{"self":"https://dummies-api.dummies.com/v2/books/"}},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f8272b\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f82e3d\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-19T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":178381},{"headers":{"creationTime":"2016-03-26T18:18:04+00:00","modifiedTime":"2022-09-19T15:31:38+00:00","timestamp":"2022-09-19T18:01:03+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"Setting Up Partial Fractions When You Have Distinct Factors","strippedTitle":"setting up partial fractions when you have distinct factors","slug":"setting-up-partial-fractions-when-you-have-distinct-factors","canonicalUrl":"","seo":{"metaDescription":"Here's how to solve a calculus problem involving partial fractions when the denominator is the product of distinct quadratic factors.","noIndex":0,"noFollow":0},"content":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is when the denominator is the product of <i>distinct</i> <i>quadratic factors </i>— that is, quadratic factors that are nonrepeating.\r\nFor each distinct quadratic factor in the denominator, add a partial fraction of the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314660.image0.png\" alt=\"image0.png\" width=\"107\" height=\"40\" />\r\n\r\nFor example, suppose that you want to integrate this function:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314661.image1.png\" alt=\"image1.png\" width=\"103\" height=\"52\" />\r\n\r\nThe first factor in the denominator is linear, but the second is quadratic and can’t be decomposed to linear factors. So set up your partial fractions as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314662.image2.png\" alt=\"image2.png\" width=\"111\" height=\"41\" />\r\n\r\nThe number of distinct quadratic factors in the denominator tells you how many partial fractions you get. So in this example, two factors in the denominator yield two partial fractions.\r\n<h2 id=\"tab1\" >Working systematically with a system of equations</h2>\r\nSetting up a system of equations is an alternative method for finding the value of unknowns when you’re working with partial fractions. It’s not as simple as plugging in the roots of factors, but it’s your only option when the root of a quadratic factor is imaginary.\r\n\r\nHere’s a problem to illustrate this method:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314663.image3.png\" alt=\"image3.png\" width=\"212\" height=\"52\" />\r\n\r\nTo start out, see how far you can get by plugging in the roots of equations. Begin by getting a common denominator on the right side of the equation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314664.image4.png\" alt=\"image4.png\" width=\"308\" height=\"67\" />\r\n\r\nNow multiply the whole equation by the denominator:\r\n\r\n5<i>x</i> – 6 = (<i>A</i>)(<i>x</i><sup>2</sup> + 3) + (<i>Bx</i> + <i>C</i>)(<i>x</i> – 2)\r\n\r\nThe root of <i>x</i> – 2 is 2, so let <i>x</i> = 2 and see what you get:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314665.image5.png\" alt=\"image5.png\" width=\"136\" height=\"72\" />\r\n\r\nNow you can substitute\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314666.image6.png\" alt=\"image6.png\" width=\"236\" height=\"73\" />\r\n\r\nUnfortunately, <i>x</i><sup>2</sup> + 3 has no root in the real numbers, so you need a different approach. First, get rid of the parentheses on the right side of the equation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314667.image7.png\" alt=\"image7.png\" width=\"268\" height=\"37\" />\r\n\r\nNext, combine similar terms (using <i>x</i> as the variable by which you judge similarity). This is just algebra:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314668.image8.png\" alt=\"image8.png\" width=\"313\" height=\"40\" />\r\n\r\nBecause this equation works for <i>all</i> values of <i>x,</i> you now take what appears to be a questionable step, breaking this equation into three separate equations as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314669.image9.png\" alt=\"image9.png\" width=\"104\" height=\"96\" />\r\n\r\nAt this point, a little algebra tells you that\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314670.image10.png\" alt=\"image10.png\" width=\"136\" height=\"37\" />\r\n\r\nSo you can substitute the values of <i>A,</i> <i>B,</i> and <i>C</i> back into the partial fractions:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314671.image11.png\" alt=\"image11.png\" width=\"256\" height=\"69\" />\r\n\r\nYou can simplify the second fraction a bit:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314672.image12.png\" alt=\"image12.png\" width=\"137\" height=\"52\" />\r\n<h2 id=\"tab2\" >Quadratic factors of the form (<i>ax</i><sup>2</sup> + <i>C</i>)</h2>\r\nWhen you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>C</i>), using partial fractions results in the following two integrals:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314673.image13.png\" alt=\"image13.png\" width=\"104\" height=\"117\" />\r\n\r\nIntegrate the first by using the variable substitution <i>u</i> = <i>ax</i><sup>2</sup> + <i>C</i> so that <i>du</i> = 2<i>ax</i><i> </i><i>dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314674.image14.png\" alt=\"image14.png\" width=\"64\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314675.image15.png\" alt=\"image15.png\" width=\"85\" height=\"41\" />\r\n\r\nHere are some examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314676.image16.png\" alt=\"image16.png\" width=\"229\" height=\"171\" />\r\n\r\nTo evaluate the second integral, use the following formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314677.image17.png\" alt=\"image17.png\" width=\"188\" height=\"41\" />\r\n<h2 id=\"tab3\" >Quadratic factors of the form (<i>ax</i><sup>2</sup> + <i>bx</i> + <i>C</i>)</h2>\r\nMost math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>bx </i>+ <i>C</i>), using partial fractions results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314678.image18.png\" alt=\"image18.png\" width=\"133\" height=\"59\" />\r\n\r\nOkay, that’s way too many letters and not nearly enough numbers. Here’s an example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314679.image19.png\" alt=\"image19.png\" width=\"112\" height=\"41\" />\r\n\r\nThis is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution <i>u</i> = <i>x</i><sup>2</sup> + 6<i>x </i>+ 13 so that <i>du</i> = (2<i>x</i> + 6) <i>dx</i><i>.</i> If the numerator were 2<i>x</i> + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314680.image20.png\" alt=\"image20.png\" width=\"137\" height=\"41\" />\r\n\r\nBecause you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314681.image21.png\" alt=\"image21.png\" width=\"137\" height=\"41\" />\r\n\r\nYou have added 0 to the integral, which didn’t change its value. At this point, you can split the integral in two:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314682.image22.png\" alt=\"image22.png\" width=\"288\" height=\"47\" />\r\n\r\nAt this point, you can use variable substitution to change the first integral as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314683.image23.png\" alt=\"image23.png\" width=\"245\" height=\"115\" />\r\n\r\nTo solve the second integral, complete the square in the denominator: Divide the <i>b</i> term (6) by 2 and square it, and then represent the <i>C</i> term (13) as the sum of this and whatever’s left:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314684.image24.png\" alt=\"image24.png\" width=\"152\" height=\"41\" />\r\n\r\nNow split the denominator into two squares:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314685.image25.png\" alt=\"image25.png\" width=\"153\" height=\"51\" />\r\n\r\nTo evaluate this integral, use formula shown in the previous section:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314686.image26.png\" alt=\"image26.png\" width=\"187\" height=\"41\" />\r\n\r\nSo here’s the final answer for the second integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314687.image27.png\" alt=\"image27.png\" width=\"123\" height=\"37\" />\r\n\r\nTherefore, piece together the complete answer as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314688.image28.png\" alt=\"image28.png\" width=\"264\" height=\"132\" />","description":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is when the denominator is the product of <i>distinct</i> <i>quadratic factors </i>— that is, quadratic factors that are nonrepeating.\r\nFor each distinct quadratic factor in the denominator, add a partial fraction of the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314660.image0.png\" alt=\"image0.png\" width=\"107\" height=\"40\" />\r\n\r\nFor example, suppose that you want to integrate this function:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314661.image1.png\" alt=\"image1.png\" width=\"103\" height=\"52\" />\r\n\r\nThe first factor in the denominator is linear, but the second is quadratic and can’t be decomposed to linear factors. So set up your partial fractions as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314662.image2.png\" alt=\"image2.png\" width=\"111\" height=\"41\" />\r\n\r\nThe number of distinct quadratic factors in the denominator tells you how many partial fractions you get. So in this example, two factors in the denominator yield two partial fractions.\r\n<h2 id=\"tab1\" >Working systematically with a system of equations</h2>\r\nSetting up a system of equations is an alternative method for finding the value of unknowns when you’re working with partial fractions. It’s not as simple as plugging in the roots of factors, but it’s your only option when the root of a quadratic factor is imaginary.\r\n\r\nHere’s a problem to illustrate this method:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314663.image3.png\" alt=\"image3.png\" width=\"212\" height=\"52\" />\r\n\r\nTo start out, see how far you can get by plugging in the roots of equations. Begin by getting a common denominator on the right side of the equation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314664.image4.png\" alt=\"image4.png\" width=\"308\" height=\"67\" />\r\n\r\nNow multiply the whole equation by the denominator:\r\n\r\n5<i>x</i> – 6 = (<i>A</i>)(<i>x</i><sup>2</sup> + 3) + (<i>Bx</i> + <i>C</i>)(<i>x</i> – 2)\r\n\r\nThe root of <i>x</i> – 2 is 2, so let <i>x</i> = 2 and see what you get:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314665.image5.png\" alt=\"image5.png\" width=\"136\" height=\"72\" />\r\n\r\nNow you can substitute\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314666.image6.png\" alt=\"image6.png\" width=\"236\" height=\"73\" />\r\n\r\nUnfortunately, <i>x</i><sup>2</sup> + 3 has no root in the real numbers, so you need a different approach. First, get rid of the parentheses on the right side of the equation:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314667.image7.png\" alt=\"image7.png\" width=\"268\" height=\"37\" />\r\n\r\nNext, combine similar terms (using <i>x</i> as the variable by which you judge similarity). This is just algebra:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314668.image8.png\" alt=\"image8.png\" width=\"313\" height=\"40\" />\r\n\r\nBecause this equation works for <i>all</i> values of <i>x,</i> you now take what appears to be a questionable step, breaking this equation into three separate equations as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314669.image9.png\" alt=\"image9.png\" width=\"104\" height=\"96\" />\r\n\r\nAt this point, a little algebra tells you that\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314670.image10.png\" alt=\"image10.png\" width=\"136\" height=\"37\" />\r\n\r\nSo you can substitute the values of <i>A,</i> <i>B,</i> and <i>C</i> back into the partial fractions:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314671.image11.png\" alt=\"image11.png\" width=\"256\" height=\"69\" />\r\n\r\nYou can simplify the second fraction a bit:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314672.image12.png\" alt=\"image12.png\" width=\"137\" height=\"52\" />\r\n<h2 id=\"tab2\" >Quadratic factors of the form (<i>ax</i><sup>2</sup> + <i>C</i>)</h2>\r\nWhen you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>C</i>), using partial fractions results in the following two integrals:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314673.image13.png\" alt=\"image13.png\" width=\"104\" height=\"117\" />\r\n\r\nIntegrate the first by using the variable substitution <i>u</i> = <i>ax</i><sup>2</sup> + <i>C</i> so that <i>du</i> = 2<i>ax</i><i> </i><i>dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314674.image14.png\" alt=\"image14.png\" width=\"64\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314675.image15.png\" alt=\"image15.png\" width=\"85\" height=\"41\" />\r\n\r\nHere are some examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314676.image16.png\" alt=\"image16.png\" width=\"229\" height=\"171\" />\r\n\r\nTo evaluate the second integral, use the following formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314677.image17.png\" alt=\"image17.png\" width=\"188\" height=\"41\" />\r\n<h2 id=\"tab3\" >Quadratic factors of the form (<i>ax</i><sup>2</sup> + <i>bx</i> + <i>C</i>)</h2>\r\nMost math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>bx </i>+ <i>C</i>), using partial fractions results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314678.image18.png\" alt=\"image18.png\" width=\"133\" height=\"59\" />\r\n\r\nOkay, that’s way too many letters and not nearly enough numbers. Here’s an example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314679.image19.png\" alt=\"image19.png\" width=\"112\" height=\"41\" />\r\n\r\nThis is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution <i>u</i> = <i>x</i><sup>2</sup> + 6<i>x </i>+ 13 so that <i>du</i> = (2<i>x</i> + 6) <i>dx</i><i>.</i> If the numerator were 2<i>x</i> + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314680.image20.png\" alt=\"image20.png\" width=\"137\" height=\"41\" />\r\n\r\nBecause you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314681.image21.png\" alt=\"image21.png\" width=\"137\" height=\"41\" />\r\n\r\nYou have added 0 to the integral, which didn’t change its value. At this point, you can split the integral in two:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314682.image22.png\" alt=\"image22.png\" width=\"288\" height=\"47\" />\r\n\r\nAt this point, you can use variable substitution to change the first integral as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314683.image23.png\" alt=\"image23.png\" width=\"245\" height=\"115\" />\r\n\r\nTo solve the second integral, complete the square in the denominator: Divide the <i>b</i> term (6) by 2 and square it, and then represent the <i>C</i> term (13) as the sum of this and whatever’s left:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314684.image24.png\" alt=\"image24.png\" width=\"152\" height=\"41\" />\r\n\r\nNow split the denominator into two squares:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314685.image25.png\" alt=\"image25.png\" width=\"153\" height=\"51\" />\r\n\r\nTo evaluate this integral, use formula shown in the previous section:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314686.image26.png\" alt=\"image26.png\" width=\"187\" height=\"41\" />\r\n\r\nSo here’s the final answer for the second integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314687.image27.png\" alt=\"image27.png\" width=\"123\" height=\"37\" />\r\n\r\nTherefore, piece together the complete answer as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314688.image28.png\" alt=\"image28.png\" width=\"264\" height=\"132\" />","blurb":"","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[{"label":"Working systematically with a system of equations","target":"#tab1"},{"label":"Quadratic factors of the form (ax2 + C)","target":"#tab2"},{"label":"Quadratic factors of the form (ax2 + bx + C)","target":"#tab3"}],"relatedArticles":{"fromBook":[{"articleId":208670,"title":"Calculus II For Dummies Cheat Sheet","slug":"calculus-ii-for-dummies-cheat-sheet","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/208670"}},{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781118161708","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/111816170X-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-2nd-edition-cover-9781118161708-202x255.jpg","width":202,"height":255},"title":"Calculus II For Dummies","testBankPinActivationLink":"","bookOutOfPrint":false,"authorsInfo":"<b data-author-id=\"9399\">Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"_links":{"self":"https://dummies-api.dummies.com/v2/books/"}},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f7b066\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f7b85a\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-19T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":178469},{"headers":{"creationTime":"2016-03-26T17:45:20+00:00","modifiedTime":"2022-09-19T15:28:36+00:00","timestamp":"2022-09-19T18:01:03+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"Setting Up Partial Fractions When You Have Repeated Quadratic Factors","strippedTitle":"setting up partial fractions when you have repeated quadratic factors","slug":"setting-up-partial-fractions-when-you-have-repeated-quadratic-factors","canonicalUrl":"","seo":{"metaDescription":"This article walks you through how to solve a calculus problem involving partial fractions and quadractic factors.","noIndex":0,"noFollow":0},"content":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated quadratic factors.\r\n\r\nThis is your worst nightmare when it comes to partial fractions, because the denominator includes repeated quadratic factors.\r\n\r\nFor each squared quadratic factor in the denominator, add <i>two</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320805.image0.png\" alt=\"image0.png\" width=\"244\" height=\"51\" />\r\n\r\nFor each quadratic factor in the denominator that’s raised to the third power, add <i>three</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320806.image1.png\" alt=\"image1.png\" width=\"380\" height=\"51\" />\r\n\r\nGenerally speaking, when a quadratic factor is raised to the <i>n</i>th power, add <i>n</i> partial fractions. For example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320807.image2.png\" alt=\"image2.png\" width=\"185\" height=\"59\" />\r\n\r\nThis denominator has one nonrepeating linear factor (<i>x</i> – 8), one nonrepeating quadratic factor (<i>x</i><sup>2 </sup>+ <i>x</i> + 1), and one quadratic expression that’s squared (<i>x</i><sup>2</sup> + 3). Here’s how you set up the partial fractions:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320808.image3.png\" alt=\"image3.png\" width=\"253\" height=\"59\" />\r\n\r\nThis example adds one partial fraction for each of the nonrepeating factors and two partial fractions for the squared factor.\r\n\r\nWhen you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>C</i>), using partial fractions results in the following two integrals:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320809.image4.png\" alt=\"image4.png\" width=\"104\" height=\"123\" />\r\n\r\nIntegrate the first by using the variable substitution <i>u</i> = <i>ax</i><sup>2</sup> + <i>C</i> so that <i>du</i> = 2<i>ax</i><i> </i><i>dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320810.image5.png\" alt=\"image5.png\" width=\"68\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320811.image6.png\" alt=\"image6.png\" width=\"85\" height=\"41\" />\r\n\r\nHere are some examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320812.image7.png\" alt=\"image7.png\" width=\"229\" height=\"156\" />\r\n\r\nTo evaluate the second integral, use the following formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320813.image8.png\" alt=\"image8.png\" width=\"188\" height=\"41\" />\r\n\r\nMost math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>bx </i>+ <i>C</i>), using partial fractions results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320814.image9.png\" alt=\"image9.png\" width=\"133\" height=\"59\" />\r\n\r\nOkay, that’s way too many letters and not nearly enough numbers. Here’s an example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320815.image10.png\" alt=\"image10.png\" width=\"112\" height=\"41\" />\r\n\r\nThis is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution <i>u</i> = <i>x</i><sup>2</sup> + 6<i>x </i>+ 13 so that <i>du</i> = (2<i>x</i> + 6) <i>dx</i><i>.</i> If the numerator were 2<i>x</i> + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320816.image11.png\" alt=\"image11.png\" width=\"137\" height=\"41\" />\r\n\r\nBecause you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320817.image12.png\" alt=\"image12.png\" width=\"137\" height=\"41\" />\r\n\r\nThis time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320818.image13.png\" alt=\"image13.png\" width=\"288\" height=\"47\" />\r\n\r\nAt this point, you can use variable substitution to change the first integral as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320819.image14.png\" alt=\"image14.png\" width=\"252\" height=\"101\" />\r\n\r\nTo solve the second integral, complete the square in the denominator: Divide the <i>b</i> term (6) by 2 and square it, and then represent the <i>C</i> term (13) as the sum of this and whatever’s left:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320820.image15.png\" alt=\"image15.png\" width=\"152\" height=\"41\" />\r\n\r\nNow split the denominator into two squares:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320821.image16.png\" alt=\"image16.png\" width=\"153\" height=\"51\" />\r\n\r\nTo evaluate this integral, use the same formula from the previous section:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320822.image17.png\" alt=\"image17.png\" width=\"187\" height=\"41\" />\r\n\r\nSo here’s the final answer for the second integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320823.image18.png\" alt=\"image18.png\" width=\"123\" height=\"37\" />\r\n\r\nTherefore, piece together the complete answer as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320824.image19.png\" alt=\"image19.png\" width=\"264\" height=\"116\" />","description":"Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated quadratic factors.\r\n\r\nThis is your worst nightmare when it comes to partial fractions, because the denominator includes repeated quadratic factors.\r\n\r\nFor each squared quadratic factor in the denominator, add <i>two</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320805.image0.png\" alt=\"image0.png\" width=\"244\" height=\"51\" />\r\n\r\nFor each quadratic factor in the denominator that’s raised to the third power, add <i>three</i> partial fractions in the following form:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320806.image1.png\" alt=\"image1.png\" width=\"380\" height=\"51\" />\r\n\r\nGenerally speaking, when a quadratic factor is raised to the <i>n</i>th power, add <i>n</i> partial fractions. For example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320807.image2.png\" alt=\"image2.png\" width=\"185\" height=\"59\" />\r\n\r\nThis denominator has one nonrepeating linear factor (<i>x</i> – 8), one nonrepeating quadratic factor (<i>x</i><sup>2 </sup>+ <i>x</i> + 1), and one quadratic expression that’s squared (<i>x</i><sup>2</sup> + 3). Here’s how you set up the partial fractions:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320808.image3.png\" alt=\"image3.png\" width=\"253\" height=\"59\" />\r\n\r\nThis example adds one partial fraction for each of the nonrepeating factors and two partial fractions for the squared factor.\r\n\r\nWhen you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>C</i>), using partial fractions results in the following two integrals:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320809.image4.png\" alt=\"image4.png\" width=\"104\" height=\"123\" />\r\n\r\nIntegrate the first by using the variable substitution <i>u</i> = <i>ax</i><sup>2</sup> + <i>C</i> so that <i>du</i> = 2<i>ax</i><i> </i><i>dx</i> and\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320810.image5.png\" alt=\"image5.png\" width=\"68\" height=\"37\" />\r\n\r\nThis substitution results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320811.image6.png\" alt=\"image6.png\" width=\"85\" height=\"41\" />\r\n\r\nHere are some examples:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320812.image7.png\" alt=\"image7.png\" width=\"229\" height=\"156\" />\r\n\r\nTo evaluate the second integral, use the following formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320813.image8.png\" alt=\"image8.png\" width=\"188\" height=\"41\" />\r\n\r\nMost math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (<i>ax</i><sup>2</sup> + <i>bx </i>+ <i>C</i>), using partial fractions results in the following integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320814.image9.png\" alt=\"image9.png\" width=\"133\" height=\"59\" />\r\n\r\nOkay, that’s way too many letters and not nearly enough numbers. Here’s an example:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320815.image10.png\" alt=\"image10.png\" width=\"112\" height=\"41\" />\r\n\r\nThis is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution <i>u</i> = <i>x</i><sup>2</sup> + 6<i>x </i>+ 13 so that <i>du</i> = (2<i>x</i> + 6) <i>dx</i><i>.</i> If the numerator were 2<i>x</i> + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320816.image11.png\" alt=\"image11.png\" width=\"137\" height=\"41\" />\r\n\r\nBecause you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320817.image12.png\" alt=\"image12.png\" width=\"137\" height=\"41\" />\r\n\r\nThis time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320818.image13.png\" alt=\"image13.png\" width=\"288\" height=\"47\" />\r\n\r\nAt this point, you can use variable substitution to change the first integral as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320819.image14.png\" alt=\"image14.png\" width=\"252\" height=\"101\" />\r\n\r\nTo solve the second integral, complete the square in the denominator: Divide the <i>b</i> term (6) by 2 and square it, and then represent the <i>C</i> term (13) as the sum of this and whatever’s left:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320820.image15.png\" alt=\"image15.png\" width=\"152\" height=\"41\" />\r\n\r\nNow split the denominator into two squares:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320821.image16.png\" alt=\"image16.png\" width=\"153\" height=\"51\" />\r\n\r\nTo evaluate this integral, use the same formula from the previous section:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320822.image17.png\" alt=\"image17.png\" width=\"187\" height=\"41\" />\r\n\r\nSo here’s the final answer for the second integral:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320823.image18.png\" alt=\"image18.png\" width=\"123\" height=\"37\" />\r\n\r\nTherefore, piece together the complete answer as follows:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/320824.image19.png\" alt=\"image19.png\" width=\"264\" height=\"116\" />","blurb":"","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[{"articleId":208670,"title":"Calculus II For Dummies Cheat Sheet","slug":"calculus-ii-for-dummies-cheat-sheet","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/208670"}},{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781118161708","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/111816170X-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-2nd-edition-cover-9781118161708-202x255.jpg","width":202,"height":255},"title":"Calculus II For Dummies","testBankPinActivationLink":"","bookOutOfPrint":false,"authorsInfo":"<b data-author-id=\"9399\">Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <p><b>Mark Zegarelli</b> is an instructor and math and test prep tutor in New Jersey. He is the author of <i>Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies,</i> and <i>Calculus II For Dummies</i>. In his spare time, he enjoys traveling and learning foreign languages.</p> ","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"_links":{"self":"https://dummies-api.dummies.com/v2/books/"}},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f73672\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[&quot;9781118161708&quot;]}]\" id=\"du-slot-6328ae5f73dc1\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-19T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":176280},{"headers":{"creationTime":"2016-03-26T21:40:51+00:00","modifiedTime":"2022-09-19T15:18:33+00:00","timestamp":"2022-09-19T18:01:03+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"How to Find the Average Value with the Mean Value Theorem for Integrals","strippedTitle":"how to find the average value with the mean value theorem for integrals","slug":"how-to-find-the-average-value-with-the-mean-value-theorem-for-integrals","canonicalUrl":"","seo":{"metaDescription":"In calculus, you can find the average value of a function by using the mean value theorem for integrals. Here's how to do it.","noIndex":0,"noFollow":0},"content":"You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look at the following figure.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179915.image0.jpg\" alt=\"image0.jpg\" width=\"535\" height=\"177\" />\r\n\r\nThe graph on the left shows a rectangle whose area is clearly <i>less</i> <i>than</i> the area under the curve between 2 and 5. This rectangle has a height equal to the lowest point on the curve in the interval from 2 to 5.\r\n\r\nThe middle graph shows a rectangle whose height equals the highest point on the curve. Its area is clearly <i>greater</i><i> than</i> the area under the curve. By now you’re thinking, “Isn’t there a rectangle taller than the short one and shorter than the tall one whose area is <i>the same as</i> the area under the curve?” Of course. And this rectangle obviously crosses the curve somewhere in the interval. This so-called mean value rectangle, shown on the right, basically sums up the Mean Value Theorem for Integrals.\r\n\r\nIt’s really just common sense. But here’s the mumbo jumbo.\r\n\r\n<b>The mean value theorem for integrals:</b> If <i>f</i> (<i>x</i>) is a continuous function on the closed interval [<i>a, b</i>], then there exists a number <i>c</i> in the closed interval such that\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179916.image1.png\" alt=\"image1.png\" width=\"244\" height=\"78\" />\r\n\r\nThe theorem basically just guarantees the existence of the mean value rectangle.\r\n\r\nThe area of the mean value rectangle — which is the same as the area under the curve — equals <i>length</i> times <i>width</i>, or <i>base</i> times <i>height</i>, right?\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179917.image2.png\" alt=\"image2.png\" width=\"464\" height=\"114\" />\r\n\r\nThis height is the <i>average value</i> of the function over the interval in question.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179918.image3.png\" alt=\"image3.png\" width=\"460\" height=\"180\" />\r\n\r\nHere’s an example. What’s the average speed of a car between <i>t</i> = 9 seconds and <i>t</i> = 16 seconds whose speed in <i>feet per se</i><i>cond</i> is given by the function,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179919.image4.png\" alt=\"image4.png\" width=\"136\" height=\"42\" />\r\n\r\nAccording to the definition of average value, this average speed is given by\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179920.image5.png\" alt=\"image5.png\" width=\"158\" height=\"78\" />\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Determine the area under the curve between 9 and 16.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179921.image6.png\" alt=\"image6.png\" width=\"164\" height=\"270\" />\r\n<p class=\"child-para\">This area, by the way, is the total distance traveled from 9 to 16 seconds. Do you see why? Consider the mean value rectangle for this problem. Its height is a speed (because the function values, or heights, are speeds) and its base is an amount of time, so its area is <i>speed</i> times <i>time</i> which equals <i>distance</i>. Alternatively, recall that the derivative of position is velocity. So, the antiderivative of velocity — what you just did in this step — is position, and the change of position from 9 to 16 seconds gives the total distance traveled.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Divide this area, total distance, by the time interval from 9 to 16, namely 7.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179922.image7.png\" alt=\"image7.png\" width=\"440\" height=\"66\" />\r\n<p class=\"child-para\">≈ 105.7 feet per second</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179923.image8.png\" alt=\"image8.png\" width=\"444\" height=\"264\" />\r\n<p class=\"child-para\">It makes more sense to think about these problems in terms of division: <i>area</i> equals <i>base</i> times <i>height</i>, so the height of the mean value rectangle equals its area <i>divided</i> by its base.</p>\r\n</li>\r\n</ol>","description":"You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look at the following figure.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179915.image0.jpg\" alt=\"image0.jpg\" width=\"535\" height=\"177\" />\r\n\r\nThe graph on the left shows a rectangle whose area is clearly <i>less</i> <i>than</i> the area under the curve between 2 and 5. This rectangle has a height equal to the lowest point on the curve in the interval from 2 to 5.\r\n\r\nThe middle graph shows a rectangle whose height equals the highest point on the curve. Its area is clearly <i>greater</i><i> than</i> the area under the curve. By now you’re thinking, “Isn’t there a rectangle taller than the short one and shorter than the tall one whose area is <i>the same as</i> the area under the curve?” Of course. And this rectangle obviously crosses the curve somewhere in the interval. This so-called mean value rectangle, shown on the right, basically sums up the Mean Value Theorem for Integrals.\r\n\r\nIt’s really just common sense. But here’s the mumbo jumbo.\r\n\r\n<b>The mean value theorem for integrals:</b> If <i>f</i> (<i>x</i>) is a continuous function on the closed interval [<i>a, b</i>], then there exists a number <i>c</i> in the closed interval such that\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179916.image1.png\" alt=\"image1.png\" width=\"244\" height=\"78\" />\r\n\r\nThe theorem basically just guarantees the existence of the mean value rectangle.\r\n\r\nThe area of the mean value rectangle — which is the same as the area under the curve — equals <i>length</i> times <i>width</i>, or <i>base</i> times <i>height</i>, right?\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179917.image2.png\" alt=\"image2.png\" width=\"464\" height=\"114\" />\r\n\r\nThis height is the <i>average value</i> of the function over the interval in question.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179918.image3.png\" alt=\"image3.png\" width=\"460\" height=\"180\" />\r\n\r\nHere’s an example. What’s the average speed of a car between <i>t</i> = 9 seconds and <i>t</i> = 16 seconds whose speed in <i>feet per se</i><i>cond</i> is given by the function,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179919.image4.png\" alt=\"image4.png\" width=\"136\" height=\"42\" />\r\n\r\nAccording to the definition of average value, this average speed is given by\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179920.image5.png\" alt=\"image5.png\" width=\"158\" height=\"78\" />\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Determine the area under the curve between 9 and 16.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179921.image6.png\" alt=\"image6.png\" width=\"164\" height=\"270\" />\r\n<p class=\"child-para\">This area, by the way, is the total distance traveled from 9 to 16 seconds. Do you see why? Consider the mean value rectangle for this problem. Its height is a speed (because the function values, or heights, are speeds) and its base is an amount of time, so its area is <i>speed</i> times <i>time</i> which equals <i>distance</i>. Alternatively, recall that the derivative of position is velocity. So, the antiderivative of velocity — what you just did in this step — is position, and the change of position from 9 to 16 seconds gives the total distance traveled.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Divide this area, total distance, by the time interval from 9 to 16, namely 7.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179922.image7.png\" alt=\"image7.png\" width=\"440\" height=\"66\" />\r\n<p class=\"child-para\">≈ 105.7 feet per second</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/179923.image8.png\" alt=\"image8.png\" width=\"444\" height=\"264\" />\r\n<p class=\"child-para\">It makes more sense to think about these problems in terms of division: <i>area</i> equals <i>base</i> times <i>height</i>, so the height of the mean value rectangle equals its area <i>divided</i> by its base.</p>\r\n</li>\r\n</ol>","blurb":"","authors":[],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":0,"slug":null,"isbn":null,"categoryList":null,"amazon":null,"image":null,"title":null,"testBankPinActivationLink":null,"bookOutOfPrint":false,"authorsInfo":null,"authors":null,"_links":null},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6328ae5f69d18\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6328ae5f6a4e8\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-19T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":193536},{"headers":{"creationTime":"2016-03-26T21:19:07+00:00","modifiedTime":"2022-09-16T13:55:56+00:00","timestamp":"2022-09-16T18:01:02+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"How to Locate Intervals of Concavity and Inflection Points","strippedTitle":"how to locate intervals of concavity and inflection points","slug":"how-to-locate-intervals-of-concavity-and-inflection-points","canonicalUrl":"","seo":{"metaDescription":"You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or ","noIndex":0,"noFollow":0},"content":"You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. The following method shows you how to find the intervals of concavity and the inflection points of\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204584.image0.png\" alt=\"image0.png\" width=\"120\" height=\"27\" />\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Find the second derivative of <i>f.</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204585.image1.png\" alt=\"image1.png\" width=\"211\" height=\"79\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Set the second derivative equal to zero and solve.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204586.image2.png\" alt=\"image2.png\" width=\"128\" height=\"203\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Determine whether the second derivative is undefined for any <i>x-</i>values.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204587.image3.png\" alt=\"image3.png\" width=\"471\" height=\"75\" />\r\n<p class=\"child-para\">Steps 2 and 3 give you what you could call “second derivative critical numbers” of <i>f</i> because they are analogous to the critical numbers of <i>f</i> that you find using the first derivative. But this set of numbers has no special name. In any event, the important thing to know is that this list is made up of the zeros of <i>f</i><i>′′</i> plus any <i>x</i>-values where <i>f</i><i>′′</i> is undefined.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Plot these numbers on a number line and test the regions with the <i>second</i> derivative.</p>\r\n<p class=\"child-para\">Use -2, -1, 1, and 2 as test numbers.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204588.image4.png\" alt=\"image4.png\" width=\"139\" height=\"115\" />\r\n<p class=\"child-para\">Because -2 is in the left-most region on the number line below, and because the second derivative at -2 equals <i>negative</i> 240, that region gets a negative sign in the figure below, and so on for the other three regions.</p>\r\n\r\n<div class=\"imageBlock\" style=\"width: 400px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204589.image5.jpg\" alt=\"A second derivative sign graph.\" width=\"400\" height=\"240\" />\r\n<div class=\"imageCaption\">A second derivative sign graph</div>\r\n</div>\r\n<p class=\"child-para\">A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. The function has an inflection point (usually) at any <i>x-</i>value where the signs switch from positive to negative or vice versa.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204590.image6.png\" alt=\"image6.png\" width=\"460\" height=\"48\" />\r\n<p class=\"article-tips tip\">If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. An inflection point exists at a given <i>x</i>-value only if there is a tangent line to the function at that number. This is the case wherever the first derivative exists or where there’s a vertical tangent.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Plug these three <i>x-</i>values into <i>f</i> to obtain the function values of the three inflection points.</p>\r\n\r\n<div class=\"imageBlock\" style=\"width: 303px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204591.image7.jpg\" alt=\"A graph showing inflection points and intervals of concavity.\" width=\"303\" height=\"400\" />\r\n<div class=\"imageCaption\">A graph showing inflection points and intervals of concavity</div>\r\n</div>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204592.image8.png\" alt=\"image8.png\" width=\"137\" height=\"115\" />\r\n<p class=\"child-para\">The square root of two equals about 1.4, so there are inflection points at about (-1.4, 39.6), (0, 0), and about (1.4, -39.6).</p>\r\n</li>\r\n</ol>","description":"You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. The following method shows you how to find the intervals of concavity and the inflection points of\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204584.image0.png\" alt=\"image0.png\" width=\"120\" height=\"27\" />\r\n<ol class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">Find the second derivative of <i>f.</i></p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204585.image1.png\" alt=\"image1.png\" width=\"211\" height=\"79\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Set the second derivative equal to zero and solve.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204586.image2.png\" alt=\"image2.png\" width=\"128\" height=\"203\" /></li>\r\n \t<li>\r\n<p class=\"first-para\">Determine whether the second derivative is undefined for any <i>x-</i>values.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204587.image3.png\" alt=\"image3.png\" width=\"471\" height=\"75\" />\r\n<p class=\"child-para\">Steps 2 and 3 give you what you could call “second derivative critical numbers” of <i>f</i> because they are analogous to the critical numbers of <i>f</i> that you find using the first derivative. But this set of numbers has no special name. In any event, the important thing to know is that this list is made up of the zeros of <i>f</i><i>′′</i> plus any <i>x</i>-values where <i>f</i><i>′′</i> is undefined.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Plot these numbers on a number line and test the regions with the <i>second</i> derivative.</p>\r\n<p class=\"child-para\">Use -2, -1, 1, and 2 as test numbers.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204588.image4.png\" alt=\"image4.png\" width=\"139\" height=\"115\" />\r\n<p class=\"child-para\">Because -2 is in the left-most region on the number line below, and because the second derivative at -2 equals <i>negative</i> 240, that region gets a negative sign in the figure below, and so on for the other three regions.</p>\r\n\r\n<div class=\"imageBlock\" style=\"width: 400px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204589.image5.jpg\" alt=\"A second derivative sign graph.\" width=\"400\" height=\"240\" />\r\n<div class=\"imageCaption\">A second derivative sign graph</div>\r\n</div>\r\n<p class=\"child-para\">A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. The function has an inflection point (usually) at any <i>x-</i>value where the signs switch from positive to negative or vice versa.</p>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204590.image6.png\" alt=\"image6.png\" width=\"460\" height=\"48\" />\r\n<p class=\"article-tips tip\">If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. An inflection point exists at a given <i>x</i>-value only if there is a tangent line to the function at that number. This is the case wherever the first derivative exists or where there’s a vertical tangent.</p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">Plug these three <i>x-</i>values into <i>f</i> to obtain the function values of the three inflection points.</p>\r\n\r\n<div class=\"imageBlock\" style=\"width: 303px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204591.image7.jpg\" alt=\"A graph showing inflection points and intervals of concavity.\" width=\"303\" height=\"400\" />\r\n<div class=\"imageCaption\">A graph showing inflection points and intervals of concavity</div>\r\n</div>\r\n<img src=\"https://www.dummies.com/wp-content/uploads/204592.image8.png\" alt=\"image8.png\" width=\"137\" height=\"115\" />\r\n<p class=\"child-para\">The square root of two equals about 1.4, so there are inflection points at about (-1.4, 39.6), (0, 0), and about (1.4, -39.6).</p>\r\n</li>\r\n</ol>","blurb":"","authors":[],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":0,"slug":null,"isbn":null,"categoryList":null,"amazon":null,"image":null,"title":null,"testBankPinActivationLink":null,"bookOutOfPrint":false,"authorsInfo":null,"authors":null,"_links":null},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6324b9ded3533\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6324b9ded3e25\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2021-07-12T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":192163},{"headers":{"creationTime":"2016-03-26T21:19:41+00:00","modifiedTime":"2022-09-15T16:18:17+00:00","timestamp":"2022-09-15T18:01:21+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"How to Do Integration by Parts","strippedTitle":"how to do integration by parts","slug":"how-to-do-integration-by-parts","canonicalUrl":"","seo":{"metaDescription":"Integrating by parts is the integration version of the product rule for differentiation. Here's how to do it, step-by-step.","noIndex":0,"noFollow":0},"content":"Integrating by parts is the integration version of the product rule for differentiation. The basic idea of integration by parts is to transform an integral you <i>can’t</i> do into a simple product minus an integral you <i>can</i> do. Here’s the formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202716.image0.png\" alt=\"image0.png\" width=\"264\" height=\"29\" />\r\n\r\nDon’t try to understand this yet. Wait for the examples that follow.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202717.image1.png\" alt=\"image1.png\" width=\"384\" height=\"29\" />\r\n\r\nIf you remember that, you can easily remember that the integral on the right is just like the one on the left, except with the <i>u</i> and <i>v</i> reversed.\r\n\r\nHere’s the method in a nutshell.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202718.image2.png\" alt=\"image2.png\" width=\"139\" height=\"31\" />\r\n\r\nFirst, you’ve got to split up the integrand into a <i>u</i> and a <i>dv</i> so that it fits the formula. For this problem, choose ln(<i>x</i>) to be your <i>u. </i>Then, everything else is the <i>dv,</i> namely\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202719.image3.png\" alt=\"image3.png\" width=\"41\" height=\"24\" />\r\n\r\nNext, you differentiate <i>u</i> to get your <i>du</i>, and you integrate <i>dv</i> to get your <i>v</i>. Finally, you plug everything into the formula and you’re home free.\r\n<div class=\"imageBlock\" style=\"width: 259px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202720.image4.jpg\" alt=\"The integration by parts box.\" width=\"259\" height=\"259\" />\r\n<div class=\"imageCaption\">The integration by parts box.</div>\r\n</div>\r\nTo help keep everything straight, organize integration-by-parts problems with a box like the one in the above figure. Draw an empty 2-by-2 box, then put your <i>u</i>, ln(<i>x</i>), in the upper-left corner and your <i>dv</i>,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202721.image5.png\" alt=\"image5.png\" width=\"41\" height=\"25\" />\r\n\r\nin the lower-right corner, as in the following figure.\r\n<div class=\"imageBlock\" style=\"width: 421px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202722.image6.jpg\" alt=\"Filling in the box.\" width=\"421\" height=\"259\" />\r\n<div class=\"imageCaption\">Filling in the box.</div>\r\n</div>\r\nThe arrows in this figure remind you to differentiate on the left and to integrate on the right. Think of differentiation — the easier thing — as going down (like going downhill), and integration — the harder thing — as going up (like going uphill).\r\n\r\nNow complete the box:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202723.image7.png\" alt=\"image7.png\" width=\"188\" height=\"225\" />\r\n\r\nThe completed box for\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202724.image8.png\" alt=\"image8.png\" width=\"80\" height=\"31\" />\r\n\r\nis shown in the next figure.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202725.image9.jpg\" alt=\"image9.jpg\" width=\"421\" height=\"259\" />\r\n\r\nA good way to remember the integration-by-parts formula is to start at the upper-left square and draw an imaginary number 7 — across, then down to the left, as shown in the following figure. This is an oh-so-sevenly mnemonic device (get it?—“sevenly” like “heavenly”—ha, ha, ha, ha.)\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202726.image10.jpg\" alt=\"image10.jpg\" width=\"259\" height=\"259\" />\r\n\r\nRemembering how you draw the 7, look back to the figure with the completed box. The integration-by-parts formula tells you to do the top part of the 7, namely\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202727.image11.png\" alt=\"image11.png\" width=\"71\" height=\"48\" />\r\n\r\nminus the integral of the diagonal part of the 7,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202728.image12.png\" alt=\"image12.png\" width=\"80\" height=\"48\" />\r\n\r\n(By the way, this method is much easier to do than to explain. Try the box technique with the 7 mnemonic. You’ll see how this scheme helps you learn the formula and organize these problems.)\r\n\r\nReady to finish? Plug everything into the formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202729.image13.png\" alt=\"image13.png\" width=\"308\" height=\"321\" />\r\n\r\nThis is because –2/3 times any number is still just any number.\r\n<p class=\"Tip\">Here’s a great mnemonic device for how to choose the <i>u</i> (again, once you’ve selected your <i>u</i>, everything else is automatically the <i>dv</i>.</p>\r\nHerbert E. Kasube came up with the acronym <i>LIATE</i> to help you choose your <i>u</i> (calculus nerds can check out Herb’s article in the <i>American Mathematical Monthly</i> 90, 1983 issue):\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><i>L</i></td>\r\n<td>Logarithmic</td>\r\n<td>like log(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>I</i></td>\r\n<td>Inverse trigonometric</td>\r\n<td>like arctan(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>A</i></td>\r\n<td>Algebraic</td>\r\n<td>like 5<i>x</i> + 3 (<i>x</i> can be raised to any power)</td>\r\n</tr>\r\n<tr>\r\n<td><i>T</i></td>\r\n<td>Trigonometric</td>\r\n<td>like cos(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>E</i></td>\r\n<td>Exponential</td>\r\n<td>like 10 raised to the <i>x</i> power</td>\r\n</tr>\r\n</tbody>\r\n</table>\r\n(note that to qualify as an exponential function, the power must contain an <i>x</i>, not just a number.)","description":"Integrating by parts is the integration version of the product rule for differentiation. The basic idea of integration by parts is to transform an integral you <i>can’t</i> do into a simple product minus an integral you <i>can</i> do. Here’s the formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202716.image0.png\" alt=\"image0.png\" width=\"264\" height=\"29\" />\r\n\r\nDon’t try to understand this yet. Wait for the examples that follow.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202717.image1.png\" alt=\"image1.png\" width=\"384\" height=\"29\" />\r\n\r\nIf you remember that, you can easily remember that the integral on the right is just like the one on the left, except with the <i>u</i> and <i>v</i> reversed.\r\n\r\nHere’s the method in a nutshell.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202718.image2.png\" alt=\"image2.png\" width=\"139\" height=\"31\" />\r\n\r\nFirst, you’ve got to split up the integrand into a <i>u</i> and a <i>dv</i> so that it fits the formula. For this problem, choose ln(<i>x</i>) to be your <i>u. </i>Then, everything else is the <i>dv,</i> namely\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202719.image3.png\" alt=\"image3.png\" width=\"41\" height=\"24\" />\r\n\r\nNext, you differentiate <i>u</i> to get your <i>du</i>, and you integrate <i>dv</i> to get your <i>v</i>. Finally, you plug everything into the formula and you’re home free.\r\n<div class=\"imageBlock\" style=\"width: 259px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202720.image4.jpg\" alt=\"The integration by parts box.\" width=\"259\" height=\"259\" />\r\n<div class=\"imageCaption\">The integration by parts box.</div>\r\n</div>\r\nTo help keep everything straight, organize integration-by-parts problems with a box like the one in the above figure. Draw an empty 2-by-2 box, then put your <i>u</i>, ln(<i>x</i>), in the upper-left corner and your <i>dv</i>,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202721.image5.png\" alt=\"image5.png\" width=\"41\" height=\"25\" />\r\n\r\nin the lower-right corner, as in the following figure.\r\n<div class=\"imageBlock\" style=\"width: 421px;\">\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202722.image6.jpg\" alt=\"Filling in the box.\" width=\"421\" height=\"259\" />\r\n<div class=\"imageCaption\">Filling in the box.</div>\r\n</div>\r\nThe arrows in this figure remind you to differentiate on the left and to integrate on the right. Think of differentiation — the easier thing — as going down (like going downhill), and integration — the harder thing — as going up (like going uphill).\r\n\r\nNow complete the box:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202723.image7.png\" alt=\"image7.png\" width=\"188\" height=\"225\" />\r\n\r\nThe completed box for\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202724.image8.png\" alt=\"image8.png\" width=\"80\" height=\"31\" />\r\n\r\nis shown in the next figure.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202725.image9.jpg\" alt=\"image9.jpg\" width=\"421\" height=\"259\" />\r\n\r\nA good way to remember the integration-by-parts formula is to start at the upper-left square and draw an imaginary number 7 — across, then down to the left, as shown in the following figure. This is an oh-so-sevenly mnemonic device (get it?—“sevenly” like “heavenly”—ha, ha, ha, ha.)\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202726.image10.jpg\" alt=\"image10.jpg\" width=\"259\" height=\"259\" />\r\n\r\nRemembering how you draw the 7, look back to the figure with the completed box. The integration-by-parts formula tells you to do the top part of the 7, namely\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202727.image11.png\" alt=\"image11.png\" width=\"71\" height=\"48\" />\r\n\r\nminus the integral of the diagonal part of the 7,\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202728.image12.png\" alt=\"image12.png\" width=\"80\" height=\"48\" />\r\n\r\n(By the way, this method is much easier to do than to explain. Try the box technique with the 7 mnemonic. You’ll see how this scheme helps you learn the formula and organize these problems.)\r\n\r\nReady to finish? Plug everything into the formula:\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/202729.image13.png\" alt=\"image13.png\" width=\"308\" height=\"321\" />\r\n\r\nThis is because –2/3 times any number is still just any number.\r\n<p class=\"Tip\">Here’s a great mnemonic device for how to choose the <i>u</i> (again, once you’ve selected your <i>u</i>, everything else is automatically the <i>dv</i>.</p>\r\nHerbert E. Kasube came up with the acronym <i>LIATE</i> to help you choose your <i>u</i> (calculus nerds can check out Herb’s article in the <i>American Mathematical Monthly</i> 90, 1983 issue):\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><i>L</i></td>\r\n<td>Logarithmic</td>\r\n<td>like log(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>I</i></td>\r\n<td>Inverse trigonometric</td>\r\n<td>like arctan(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>A</i></td>\r\n<td>Algebraic</td>\r\n<td>like 5<i>x</i> + 3 (<i>x</i> can be raised to any power)</td>\r\n</tr>\r\n<tr>\r\n<td><i>T</i></td>\r\n<td>Trigonometric</td>\r\n<td>like cos(<i>x</i>)</td>\r\n</tr>\r\n<tr>\r\n<td><i>E</i></td>\r\n<td>Exponential</td>\r\n<td>like 10 raised to the <i>x</i> power</td>\r\n</tr>\r\n</tbody>\r\n</table>\r\n(note that to qualify as an exponential function, the power must contain an <i>x</i>, not just a number.)","blurb":"","authors":[],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":0,"slug":null,"isbn":null,"categoryList":null,"amazon":null,"image":null,"title":null,"testBankPinActivationLink":null,"bookOutOfPrint":false,"authorsInfo":null,"authors":null,"_links":null},"collections":[],"articleAds":{"footerAd":"<div class=\"du-ad-region row\" id=\"article_page_adhesion_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_adhesion_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6323687155654\"></div></div>","rightAd":"<div class=\"du-ad-region row\" id=\"article_page_right_ad\"><div class=\"du-ad-unit col-md-12\" data-slot-id=\"article_page_right_ad\" data-refreshed=\"false\" \r\n data-target = \"[{&quot;key&quot;:&quot;cat&quot;,&quot;values&quot;:[&quot;academics-the-arts&quot;,&quot;math&quot;,&quot;calculus&quot;]},{&quot;key&quot;:&quot;isbn&quot;,&quot;values&quot;:[null]}]\" id=\"du-slot-6323687155ef6\"></div></div>"},"articleType":{"articleType":"Articles","articleList":null,"content":null,"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five years","lifeExpectancySetFrom":"2022-09-15T00:00:00+00:00","dummiesForKids":"no","sponsoredContent":"no","adInfo":"","adPairKey":[]},"status":"publish","visibility":"public","articleId":192235},{"headers":{"creationTime":"2016-03-26T18:16:29+00:00","modifiedTime":"2022-08-10T15:41:39+00:00","timestamp":"2022-09-14T18:19:52+00:00"},"data":{"breadcrumbs":[{"name":"Academics & The Arts","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33662"},"slug":"academics-the-arts","categoryId":33662},{"name":"Math","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33720"},"slug":"math","categoryId":33720},{"name":"Calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"},"slug":"calculus","categoryId":33723}],"title":"When to Use Integration by Parts in Calculus","strippedTitle":"when to use integration by parts in calculus","slug":"knowing-when-to-integrate-by-parts","canonicalUrl":"","seo":{"metaDescription":"Learn when to use integration by parts in calculus, including for the logarithmic function ln x and the first four inverse trig functions.","noIndex":0,"noFollow":0},"content":"In calculus, it’s important to recognize when integrating by parts is useful. To start off, here are two important cases when integration by parts is definitely the way to go:\r\n<ul class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">The logarithmic function ln <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">The first four inverse trig functions (arcsin <i>x</i><i>,</i> arccos <i>x</i><i>,</i> arctan <i>x</i><i>,</i> and arccot <i>x</i>)</p>\r\n</li>\r\n</ul>\r\nBeyond these cases, integration by parts is useful for integrating the product of more than one type of function or class of function. For example:\r\n<ul class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i> ln <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i> arcsec <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i><sup>2</sup> sin <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>e</i><i><sup>x</sup></i><i> </i>cos <i>x</i></p>\r\n</li>\r\n</ul>\r\nNotice that in each case, you can recognize the product of functions because the variable <i>x</i> appears more than once in the function.\r\n<p class=\"Tip\">Whenever you’re faced with integrating the product of functions, consider variable substitution before you think about integration by parts. For example, <i>x</i> cos (<i>x</i><sup>2</sup>) is a job for variable substitution, not integration by parts.</p>\r\nWhen you decide to use integration by parts, your next question is how to split up the function and assign the variables <i>u</i> and <i>dv.</i> Fortunately, a helpful mnemonic exists to make this decision: <b><u>L</u></b>ovely <b><u>I</u></b>ntegrals <b><u>A</u></b>re <b><u>T</u></b>errific, which stands for <b><u>L</u></b>ogarithmic, <b><u>I</u></b>nverse trig, <b><u>A</u></b>lgebraic, <b><u>T</u></b>rig. (If you prefer, you can also use the mnemonic <b><u>L</u></b>ousy <b><u>I</u></b>ntegrals <b><u>A</u></b>re <b><u>T</u></b>errible.) Always choose the <i>first</i> function in this list as the factor to set equal to <i>u,</i> and then set the rest of the product (including <i>dx</i>) equal to <i>dv.</i>\r\n\r\nYou can use integration by parts to integrate any of the functions listed in the table.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314527.image0.png\" alt=\"image0.png\" width=\"508\" height=\"429\" />\r\n\r\nWhen you’re integrating by parts, here’s the most basic rule when deciding which term to integrate and which to differentiate: If you only know how to integrate only one of the two, that’s the one you integrate!","description":"In calculus, it’s important to recognize when integrating by parts is useful. To start off, here are two important cases when integration by parts is definitely the way to go:\r\n<ul class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\">The logarithmic function ln <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\">The first four inverse trig functions (arcsin <i>x</i><i>,</i> arccos <i>x</i><i>,</i> arctan <i>x</i><i>,</i> and arccot <i>x</i>)</p>\r\n</li>\r\n</ul>\r\nBeyond these cases, integration by parts is useful for integrating the product of more than one type of function or class of function. For example:\r\n<ul class=\"level-one\">\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i> ln <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i> arcsec <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>x</i><sup>2</sup> sin <i>x</i></p>\r\n</li>\r\n \t<li>\r\n<p class=\"first-para\"><i>e</i><i><sup>x</sup></i><i> </i>cos <i>x</i></p>\r\n</li>\r\n</ul>\r\nNotice that in each case, you can recognize the product of functions because the variable <i>x</i> appears more than once in the function.\r\n<p class=\"Tip\">Whenever you’re faced with integrating the product of functions, consider variable substitution before you think about integration by parts. For example, <i>x</i> cos (<i>x</i><sup>2</sup>) is a job for variable substitution, not integration by parts.</p>\r\nWhen you decide to use integration by parts, your next question is how to split up the function and assign the variables <i>u</i> and <i>dv.</i> Fortunately, a helpful mnemonic exists to make this decision: <b><u>L</u></b>ovely <b><u>I</u></b>ntegrals <b><u>A</u></b>re <b><u>T</u></b>errific, which stands for <b><u>L</u></b>ogarithmic, <b><u>I</u></b>nverse trig, <b><u>A</u></b>lgebraic, <b><u>T</u></b>rig. (If you prefer, you can also use the mnemonic <b><u>L</u></b>ousy <b><u>I</u></b>ntegrals <b><u>A</u></b>re <b><u>T</u></b>errible.) Always choose the <i>first</i> function in this list as the factor to set equal to <i>u,</i> and then set the rest of the product (including <i>dx</i>) equal to <i>dv.</i>\r\n\r\nYou can use integration by parts to integrate any of the functions listed in the table.\r\n\r\n<img src=\"https://www.dummies.com/wp-content/uploads/314527.image0.png\" alt=\"image0.png\" width=\"508\" height=\"429\" />\r\n\r\nWhen you’re integrating by parts, here’s the most basic rule when deciding which term to integrate and which to differentiate: If you only know how to integrate only one of the two, that’s the one you integrate!","blurb":"","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <b>Mark Zegarelli</b> is a professional writer with degrees in both English and Math from Rutgers University. He has earned his living for many years writing vast quantities of logic puzzles, a hefty chunk of software documentation, and the occasional book or film review. Along the way, he&#8217;s also paid a few bills doing housecleaning, decorative painting, and (for ten hours) retail sales. He likes writing best, though.","hasArticle":false,"_links":{"self":"https://dummies-api.dummies.com/v2/authors/9399"}}],"primaryCategoryTaxonomy":{"categoryId":33723,"title":"Calculus","slug":"calculus","_links":{"self":"https://dummies-api.dummies.com/v2/categories/33723"}},"secondaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"tertiaryCategoryTaxonomy":{"categoryId":0,"title":null,"slug":null,"_links":null},"trendingArticles":null,"inThisArticle":[],"relatedArticles":{"fromBook":[{"articleId":208670,"title":"Calculus II For Dummies Cheat Sheet","slug":"calculus-ii-for-dummies-cheat-sheet","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/208670"}},{"articleId":179236,"title":"Computing Integrals and Representing Integrals as Functions","slug":"computing-integrals-and-representing-integrals-as-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179236"}},{"articleId":179235,"title":"Drawing with 3-D Cartesian Coordinates","slug":"drawing-with-3-d-cartesian-coordinates","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179235"}},{"articleId":179234,"title":"Evaluating Triple Integrals","slug":"evaluating-triple-integrals","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179234"}},{"articleId":179233,"title":"Find the Area Between Two Functions","slug":"find-the-area-between-two-functions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/179233"}}],"fromCategory":[{"articleId":256336,"title":"Solve a Difficult Limit Problem Using the Sandwich Method","slug":"solve-a-difficult-limit-problem-using-the-sandwich-method","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/256336"}},{"articleId":255765,"title":"Solve Limit Problems on a Calculator Using Graphing Mode","slug":"solve-limit-problems-on-a-calculator-using-graphing-mode","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255765"}},{"articleId":255755,"title":"Solve Limit Problems on a Calculator Using the Arrow-Number","slug":"solve-limit-problems-on-a-calculator-using-the-arrow-number","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255755"}},{"articleId":255261,"title":"Limit and Continuity Graphs: Practice Questions","slug":"limit-and-continuity-graphs-practice-questions","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255261"}},{"articleId":255255,"title":"Use the Vertical Line Test to Identify a Function","slug":"use-the-vertical-line-test-to-identify-a-function","categoryList":["academics-the-arts","math","calculus"],"_links":{"self":"https://dummies-api.dummies.com/v2/articles/255255"}}]},"hasRelatedBookFromSearch":false,"relatedBook":{"bookId":282046,"slug":"calculus-ii-for-dummies-2nd-edition","isbn":"9781118161708","categoryList":["academics-the-arts","math","calculus"],"amazon":{"default":"https://www.amazon.com/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","ca":"https://www.amazon.ca/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","indigo_ca":"http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/111816170X-item.html&cjsku=978111945484","gb":"https://www.amazon.co.uk/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20","de":"https://www.amazon.de/gp/product/111816170X/ref=as_li_tl?ie=UTF8&tag=wiley01-20"},"image":{"src":"https://www.dummies.com/wp-content/uploads/calculus-ii-for-dummies-2nd-edition-cover-9781118161708-202x255.jpg","width":202,"height":255},"title":"Calculus II For Dummies","testBankPinActivationLink":"","bookOutOfPrint":false,"authorsInfo":"<b data-author-id=\"9399\">Mark Zegarelli</b>, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of <i>Logic For Dummies</i> and <i>Basic Math &amp; Pre-Algebra For Dummies</i>.","authors":[{"authorId":9399,"name":"Mark Zegarelli","slug":"mark-zegarelli","description":" <b>Mark Zegarelli</b> is a professional writer with degrees in both English and Math from Rutgers University. He has earned his living for many years writing vast quantities of logic puzzles, a hefty chunk of software documentation, and the occasional book or film review. Along the way, he&#8217;s also paid a few bills doing housecleaning, decorative painting, and (for ten hours) retail sales. 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This Cheat Sheet provides some basic formulas you can refer to regularly to make solving calculus problems a breeze (well, maybe not a breeze, but definitely easier).","description":"Solving calculus problems is a great way to master the various rules, theorems, and calculations you encounter in a typical Calculus class. This Cheat Sheet provides some basic formulas you can refer to regularly to make solving calculus problems a breeze (well, maybe not a breeze, but definitely easier).","blurb":"","authors":[{"authorId":9453,"name":"Patrick Jones","slug":"patrick-jones","description":" <p><b>Patrick Jones</b> has a master&#8217;s degree in Mathematics from the University of Louisville. He has taught at University of Louisville, Vanderbilt University, and Austin Community College. 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The list isn’t comprehensive, but it should cover the items you’ll use most often.</p>\n<h2>Limit Definition of a Derivative</h2>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421675.image0.png\" alt=\"image0.png\" width=\"148\" height=\"45\" /></p>\n<h2>Definition: Continuous at a number <i>a</i></h2>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421676.image1.png\" alt=\"image1.png\" width=\"349\" height=\"32\" /></p>\n<h2>The Intermediate Value Theorem</h2>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421677.image2.png\" alt=\"image2.png\" width=\"538\" height=\"49\" /></p>\n<h2>Definition of a <b>Critical Number</b></h2>\n<p>A <i>critical number</i> of a function <i>f</i> is a number <i>c</i> in the domain of <i>f</i> such that either <i>f</i><i><sup> </sup></i><i>&#8216;</i>(<i>c</i>)<i> </i>= 0 or <i>f</i><i><sup> </sup></i><i>&#8216;</i>(<i>c</i>) does not exist.</p>\n<h2><b>Rolle’s Theorem</b></h2>\n<p>Let <i>f</i> be a function that satisfies the following three hypotheses:</p>\n<ul class=\"level-one\">\n<li>\n<p class=\"first-para\"><i>f</i> is continuous on the closed interval [<i>a, b</i>].</p>\n</li>\n<li>\n<p class=\"first-para\"><i>f</i> is differentiable on the open interval (<i>a, b</i>).</p>\n</li>\n<li>\n<p class=\"first-para\"><i>f</i><i></i>(<i>a</i>)<i> </i>= <i>f</i><i></i>(<i>b</i>).</p>\n</li>\n</ul>\n<p>Then there is a number <i>c</i> in (<i>a, b</i>) such that <i>f</i><i><sup> </sup></i><i>&#8216;</i>(<i>c</i>)<i> </i>= 0.</p>\n<h2><b>The Mean Value Theorem</b></h2>\n<p>Let <i>f</i> be a function that satisfies the following hypotheses:</p>\n<ul class=\"level-one\">\n<li>\n<p class=\"first-para\"><i>f</i> is continuous on the closed interval [<i>a, b</i>].</p>\n</li>\n<li>\n<p class=\"first-para\"><i>f</i> is differentiable on the open interval (<i>a, b</i>).</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421678.image3.png\" alt=\"image3.png\" width=\"372\" height=\"41\" /></li>\n</ul>\n<h2><b>Newton’s Method Approximation Formula</b></h2>\n<p>Newton’s method is a technique that tries to find a root of an equation. To begin, you try to pick a number that’s “close” to the value of a root and call this value <i>x</i><sub>1</sub>. Picking <i>x</i><sub>1</sub> may involve some trial and error; if you’re dealing with a continuous function on some interval (or possibly the entire real line), the intermediate value theorem may narrow down the interval under consideration. After picking <i>x</i><sub>1</sub>, you use the recursive formula given here to find successive approximations:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421679.image4.png\" alt=\"image4.png\" width=\"138\" height=\"45\" /></p>\n<p>A word of caution: Always verify that your final approximation is correct (or close to the value of the root). Newton’s method can fail in some instances, based on the value picked for <i>x</i><sub>1</sub>. Any calculus text that covers Newton’s method should point out these shortcomings.</p>\n<h2><b>The Fundamental Theorem of Calculus </b></h2>\n<p>Suppose <i>f</i> is continuous on [<i>a, b</i>]. Then the following statements are true:</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421680.image5.png\" alt=\"image5.png\" width=\"225\" height=\"38\" /><br />\n<img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421681.image6.png\" alt=\"image6.png\" width=\"432\" height=\"38\" /></p>\n<h2><b>The Trapezoid Rule</b></h2>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421682.image7.png\" alt=\"image7.png\" width=\"477\" height=\"40\" /></p>\n<p>where</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421683.image8.png\" alt=\"image8.png\" width=\"197\" height=\"38\" /></p>\n<h2><b>Simpson’s Rule</b></h2>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421684.image9.png\" alt=\"image9.png\" width=\"400\" height=\"40\" /></p>\n<p>&nbsp;</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421685.image10.png\" alt=\"image10.png\" width=\"234\" height=\"40\" /></p>\n<p>where <i>n</i> is even and</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421686.image11.png\" alt=\"image11.png\" width=\"82\" height=\"38\" /></p>\n"},{"title":"Special limit formulas","thumb":null,"image":null,"content":"<p>Many people first encounter the following limits in a Calculus textbook when trying to prove the derivative formulas for the sine function and the cosine function. These results aren’t immediately obvious and actually take a bit of work to justify. Any calculus text should provide more explanation if you’re interested in seeing it!</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421669.image0.png\" alt=\"image0.png\" width=\"181\" height=\"38\" /><br />\n<img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421670.image1.png\" alt=\"image1.png\" width=\"126\" height=\"38\" /></p>\n"},{"title":"Derivative and integration formulas for hyperbolic functions","thumb":null,"image":null,"content":"<p>The hyperbolic functions are certain combinations of the exponential functions <i>e</i><i><sup>x</sup></i> and <i>e</i><i><sup>–</sup></i><i><sup>x</sup></i>. These functions occur often enough in differential equations and engineering that they’re typically introduced in a Calculus course. Some of the real-life applications of these functions relate to the study of electric transmission and suspension cables.</p>\n<p><img loading=\"lazy\" src=\"https://www.dummies.com/wp-content/uploads/421672.image0.png\" alt=\"A chart of derivative and integration formulas for hyperbolic functions.\" width=\"535\" height=\"514\" /></p>\n"}],"videoInfo":{"videoId":null,"name":null,"accountId":null,"playerId":null,"thumbnailUrl":null,"description":null,"uploadDate":null}},"sponsorship":{"sponsorshipPage":false,"backgroundImage":{"src":null,"width":0,"height":0},"brandingLine":"","brandingLink":"","brandingLogo":{"src":null,"width":0,"height":0},"sponsorAd":"","sponsorEbookTitle":"","sponsorEbookLink":"","sponsorEbookImage":{"src":null,"width":0,"height":0}},"primaryLearningPath":"Advance","lifeExpectancy":"Five 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Calculus Articles

It's time to conquer calc. With your calculator in hand and these articles by your side, you're ready to take on even the scariest differential equations.

Articles From Calculus

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Calculus Calculus II For Dummies Cheat Sheet

Cheat Sheet / Updated 04-11-2023

By its nature, calculus can be intimidating. But you can take some of the fear of studying calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. Also discover a few basic rules applied to calculus like Cramer's Rule, the Constant Multiple Rule, and a few others, and you'll be on your way to acing the course.

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Calculus Substituting with Expressions of the Form f(x) Multiplied by h(g(x))

Article / Updated 09-22-2022

When g'(x) = f(x), you can use the substitution u = g(x) to integrate expressions of the form f(x) multiplied by h(g(x)), provided that h is a function that you already know how to integrate. Variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration. Here’s a hairy-looking integral that actually responds well to substitution: The key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator. Watch how this plays out in this substitution: Declare u equal to the inner function in the denominator and make the substitution: Here’s the substitution: Differential du = (2x + 1) dx: The second part of the substitution now becomes clear: Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator. (You may think that this is quite a coincidence, but coincidences like these happen all the time on exams!) Integration is now quite straightforward: You take an extra step to remove the fraction before you integrate: Substitute back x2 + x – 5 for u: Checking the answer by differentiating with the Chain Rule reveals how this problem was set up in the first place: Here’s another example where you make a variable substitution: Notice that the derivative of x4 – 1 is x3, off by a constant factor. So here’s the declaration, followed by the differentiation: Now you can just do both substitutions at once: At this point, you can solve the integral simply. Similarly, here’s another example: At first glance, this integral looks just plain horrible. But on further inspection, notice that the derivative of cot x is –csc2 x, so this looks like another good candidate: This results in the following substitution: Again, this is another integral that you can solve.

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Calculus Setting Up Partial Fractions When You Have Repeated Linear Factors

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated linear factors. These are difficult to work with because each factor requires more than one partial fraction. For each squared linear factor in the denominator, add two partial fractions in the following form: For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form: Generally speaking, when a linear factor is raised to the nth power, add n partial fractions. For example, suppose that you want to integrate the following expression: This expression contains all linear factors, but one of these factors (x + 5) is nonrepeating and the other (x – 1) is raised to the third power. Set up your partial fractions this way: Which will yield: As you can see, this example adds one partial fraction to account for the nonrepeating factor and three to account for the repeating factor. When you start out with a linear factor, using partial fractions leaves you with an integral in the following form: Integrate all these cases by using the variable substitution u = ax + b so that du = a dx and This substitution results in the following integral: Here are a few examples:

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Calculus Setting Up Partial Fractions When You Have Distinct Factors

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is when the denominator is the product of distinct quadratic factors — that is, quadratic factors that are nonrepeating. For each distinct quadratic factor in the denominator, add a partial fraction of the following form: For example, suppose that you want to integrate this function: The first factor in the denominator is linear, but the second is quadratic and can’t be decomposed to linear factors. So set up your partial fractions as follows: The number of distinct quadratic factors in the denominator tells you how many partial fractions you get. So in this example, two factors in the denominator yield two partial fractions. Working systematically with a system of equations Setting up a system of equations is an alternative method for finding the value of unknowns when you’re working with partial fractions. It’s not as simple as plugging in the roots of factors, but it’s your only option when the root of a quadratic factor is imaginary. Here’s a problem to illustrate this method: To start out, see how far you can get by plugging in the roots of equations. Begin by getting a common denominator on the right side of the equation: Now multiply the whole equation by the denominator: 5x – 6 = (A)(x2 + 3) + (Bx + C)(x – 2) The root of x – 2 is 2, so let x = 2 and see what you get: Now you can substitute Unfortunately, x2 + 3 has no root in the real numbers, so you need a different approach. First, get rid of the parentheses on the right side of the equation: Next, combine similar terms (using x as the variable by which you judge similarity). This is just algebra: Because this equation works for all values of x, you now take what appears to be a questionable step, breaking this equation into three separate equations as follows: At this point, a little algebra tells you that So you can substitute the values of A, B, and C back into the partial fractions: You can simplify the second fraction a bit: Quadratic factors of the form (ax2 + C) When you start out with a quadratic factor of the form (ax2 + C), using partial fractions results in the following two integrals: Integrate the first by using the variable substitution u = ax2 + C so that du = 2ax dx and This substitution results in the following integral: Here are some examples: To evaluate the second integral, use the following formula: Quadratic factors of the form (ax2 + bx + C) Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral: Okay, that’s way too many letters and not nearly enough numbers. Here’s an example: This is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution u = x2 + 6x + 13 so that du = (2x + 6) dx. If the numerator were 2x + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2: Because you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator: You have added 0 to the integral, which didn’t change its value. At this point, you can split the integral in two: At this point, you can use variable substitution to change the first integral as follows: To solve the second integral, complete the square in the denominator: Divide the b term (6) by 2 and square it, and then represent the C term (13) as the sum of this and whatever’s left: Now split the denominator into two squares: To evaluate this integral, use formula shown in the previous section: So here’s the final answer for the second integral: Therefore, piece together the complete answer as follows:

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Calculus Setting Up Partial Fractions When You Have Repeated Quadratic Factors

Article / Updated 09-19-2022

Your first step in any problem that involves partial fractions is to recognize which case you’re dealing with so that you can solve the problem. One case where you can use partial fractions is with repeated quadratic factors. This is your worst nightmare when it comes to partial fractions, because the denominator includes repeated quadratic factors. For each squared quadratic factor in the denominator, add two partial fractions in the following form: For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form: Generally speaking, when a quadratic factor is raised to the nth power, add n partial fractions. For example: This denominator has one nonrepeating linear factor (x – 8), one nonrepeating quadratic factor (x2 + x + 1), and one quadratic expression that’s squared (x2 + 3). Here’s how you set up the partial fractions: This example adds one partial fraction for each of the nonrepeating factors and two partial fractions for the squared factor. When you start out with a quadratic factor of the form (ax2 + C), using partial fractions results in the following two integrals: Integrate the first by using the variable substitution u = ax2 + C so that du = 2ax dx and This substitution results in the following integral: Here are some examples: To evaluate the second integral, use the following formula: Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral: Okay, that’s way too many letters and not nearly enough numbers. Here’s an example: This is about the hairiest integral you’re ever going to see at the far end of a partial fraction. To evaluate it, you want to use the variable substitution u = x2 + 6x + 13 so that du = (2x + 6) dx. If the numerator were 2x + 6, you’d be in great shape. So you need to tweak the numerator a bit. First multiply it by 2 and divide the whole integral by 2: Because you multiplied the entire integral by 1, no net change has occurred. Now add 6 and –6 to the numerator: This time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two: At this point, you can use variable substitution to change the first integral as follows: To solve the second integral, complete the square in the denominator: Divide the b term (6) by 2 and square it, and then represent the C term (13) as the sum of this and whatever’s left: Now split the denominator into two squares: To evaluate this integral, use the same formula from the previous section: So here’s the final answer for the second integral: Therefore, piece together the complete answer as follows:

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Calculus How to Find the Average Value with the Mean Value Theorem for Integrals

Article / Updated 09-19-2022

You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look at the following figure. The graph on the left shows a rectangle whose area is clearly less than the area under the curve between 2 and 5. This rectangle has a height equal to the lowest point on the curve in the interval from 2 to 5. The middle graph shows a rectangle whose height equals the highest point on the curve. Its area is clearly greater than the area under the curve. By now you’re thinking, “Isn’t there a rectangle taller than the short one and shorter than the tall one whose area is the same as the area under the curve?” Of course. And this rectangle obviously crosses the curve somewhere in the interval. This so-called mean value rectangle, shown on the right, basically sums up the Mean Value Theorem for Integrals. It’s really just common sense. But here’s the mumbo jumbo. The mean value theorem for integrals: If f (x) is a continuous function on the closed interval [a, b], then there exists a number c in the closed interval such that The theorem basically just guarantees the existence of the mean value rectangle. The area of the mean value rectangle — which is the same as the area under the curve — equals length times width, or base times height, right? This height is the average value of the function over the interval in question. Here’s an example. What’s the average speed of a car between t = 9 seconds and t = 16 seconds whose speed in feet per second is given by the function, According to the definition of average value, this average speed is given by Determine the area under the curve between 9 and 16. This area, by the way, is the total distance traveled from 9 to 16 seconds. Do you see why? Consider the mean value rectangle for this problem. Its height is a speed (because the function values, or heights, are speeds) and its base is an amount of time, so its area is speed times time which equals distance. Alternatively, recall that the derivative of position is velocity. So, the antiderivative of velocity — what you just did in this step — is position, and the change of position from 9 to 16 seconds gives the total distance traveled. Divide this area, total distance, by the time interval from 9 to 16, namely 7. ≈ 105.7 feet per second It makes more sense to think about these problems in terms of division: area equals base times height, so the height of the mean value rectangle equals its area divided by its base.

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Calculus How to Locate Intervals of Concavity and Inflection Points

Article / Updated 09-16-2022

You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. The following method shows you how to find the intervals of concavity and the inflection points of Find the second derivative of f. Set the second derivative equal to zero and solve. Determine whether the second derivative is undefined for any x-values. Steps 2 and 3 give you what you could call “second derivative critical numbers” of f because they are analogous to the critical numbers of f that you find using the first derivative. But this set of numbers has no special name. In any event, the important thing to know is that this list is made up of the zeros of f′′ plus any x-values where f′′ is undefined. Plot these numbers on a number line and test the regions with the second derivative. Use -2, -1, 1, and 2 as test numbers. Because -2 is in the left-most region on the number line below, and because the second derivative at -2 equals negative 240, that region gets a negative sign in the figure below, and so on for the other three regions. A second derivative sign graph A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. The function has an inflection point (usually) at any x-value where the signs switch from positive to negative or vice versa. If you get a problem in which the signs switch at a number where the second derivative is undefined, you have to check one more thing before concluding that there’s an inflection point there. An inflection point exists at a given x-value only if there is a tangent line to the function at that number. This is the case wherever the first derivative exists or where there’s a vertical tangent. Plug these three x-values into f to obtain the function values of the three inflection points. A graph showing inflection points and intervals of concavity The square root of two equals about 1.4, so there are inflection points at about (-1.4, 39.6), (0, 0), and about (1.4, -39.6).

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Calculus How to Do Integration by Parts

Article / Updated 09-15-2022

Integrating by parts is the integration version of the product rule for differentiation. The basic idea of integration by parts is to transform an integral you can’t do into a simple product minus an integral you can do. Here’s the formula: Don’t try to understand this yet. Wait for the examples that follow. If you remember that, you can easily remember that the integral on the right is just like the one on the left, except with the u and v reversed. Here’s the method in a nutshell. First, you’ve got to split up the integrand into a u and a dv so that it fits the formula. For this problem, choose ln(x) to be your u. Then, everything else is the dv, namely Next, you differentiate u to get your du, and you integrate dv to get your v. Finally, you plug everything into the formula and you’re home free. The integration by parts box. To help keep everything straight, organize integration-by-parts problems with a box like the one in the above figure. Draw an empty 2-by-2 box, then put your u, ln(x), in the upper-left corner and your dv, in the lower-right corner, as in the following figure. Filling in the box. The arrows in this figure remind you to differentiate on the left and to integrate on the right. Think of differentiation — the easier thing — as going down (like going downhill), and integration — the harder thing — as going up (like going uphill). Now complete the box: The completed box for is shown in the next figure. A good way to remember the integration-by-parts formula is to start at the upper-left square and draw an imaginary number 7 — across, then down to the left, as shown in the following figure. This is an oh-so-sevenly mnemonic device (get it?—“sevenly” like “heavenly”—ha, ha, ha, ha.) Remembering how you draw the 7, look back to the figure with the completed box. The integration-by-parts formula tells you to do the top part of the 7, namely minus the integral of the diagonal part of the 7, (By the way, this method is much easier to do than to explain. Try the box technique with the 7 mnemonic. You’ll see how this scheme helps you learn the formula and organize these problems.) Ready to finish? Plug everything into the formula: This is because –2/3 times any number is still just any number. Here’s a great mnemonic device for how to choose the u (again, once you’ve selected your u, everything else is automatically the dv. Herbert E. Kasube came up with the acronym LIATE to help you choose your u (calculus nerds can check out Herb’s article in the American Mathematical Monthly 90, 1983 issue): L Logarithmic like log(x) I Inverse trigonometric like arctan(x) A Algebraic like 5x + 3 (x can be raised to any power) T Trigonometric like cos(x) E Exponential like 10 raised to the x power (note that to qualify as an exponential function, the power must contain an x, not just a number.)

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Calculus When to Use Integration by Parts in Calculus

Article / Updated 08-10-2022

In calculus, it’s important to recognize when integrating by parts is useful. To start off, here are two important cases when integration by parts is definitely the way to go: The logarithmic function ln x The first four inverse trig functions (arcsin x, arccos x, arctan x, and arccot x) Beyond these cases, integration by parts is useful for integrating the product of more than one type of function or class of function. For example: x ln x x arcsec x x2 sin x ex cos x Notice that in each case, you can recognize the product of functions because the variable x appears more than once in the function. Whenever you’re faced with integrating the product of functions, consider variable substitution before you think about integration by parts. For example, x cos (x2) is a job for variable substitution, not integration by parts. When you decide to use integration by parts, your next question is how to split up the function and assign the variables u and dv. Fortunately, a helpful mnemonic exists to make this decision: Lovely Integrals Are Terrific, which stands for Logarithmic, Inverse trig, Algebraic, Trig. (If you prefer, you can also use the mnemonic Lousy Integrals Are Terrible.) Always choose the first function in this list as the factor to set equal to u, and then set the rest of the product (including dx) equal to dv. You can use integration by parts to integrate any of the functions listed in the table. When you’re integrating by parts, here’s the most basic rule when deciding which term to integrate and which to differentiate: If you only know how to integrate only one of the two, that’s the one you integrate!

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Calculus Calculus: 1001 Practice Problems For Dummies Cheat Sheet

Cheat Sheet / Updated 04-25-2022

Solving calculus problems is a great way to master the various rules, theorems, and calculations you encounter in a typical Calculus class. This Cheat Sheet provides some basic formulas you can refer to regularly to make solving calculus problems a breeze (well, maybe not a breeze, but definitely easier).

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