Calculus Articles

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.

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:

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:

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:

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.

Cheat Sheet / Updated 09-16-2022

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.

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).

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.)

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!

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).