# Calculus II For Dummies

**Published: **01-24-2012

**An easy-to-understand primer on advanced calculus topics**

Calculus II is a prerequisite for many popular college majors, including pre-med, engineering, and physics. *Calculus II For Dummies* offers expert instruction, advice, and tips to help second semester calculus students get a handle on the subject and ace their exams.

It covers intermediate calculus topics in plain English, featuring in-depth coverage of integration, including substitution, integration techniques and when to use them, approximate integration, and improper integrals. This hands-on guide also covers sequences and series, with introductions to multivariable calculus, differential equations, and numerical analysis. Best of all, it includes practical exercises designed to simplify and enhance understanding of this complex subject.

- Introduction to integration
- Indefinite integrals
- Intermediate Integration topics
- Infinite series
- Advanced topics
- Practice exercises

Confounded by curves? Perplexed by polynomials? This plain-English guide to Calculus II will set you straight!

## Articles From Calculus II For Dummies

### Filter Results

Cheat Sheet / Updated 08-31-2021

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.

View Cheat SheetArticle / Updated 07-13-2021

A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. This information is provided by the Taylor remainder term: f(x) = Tn(x) + Rn(x) Notice that the addition of the remainder term Rn(x) turns the approximation into an equation. Here’s the formula for the remainder term: It’s important to be clear that this equation is true for one specific value of c on the interval between a and x. It does not work for just any value of c on that interval. Ideally, the remainder term gives you the precise difference between the value of a function and the approximation Tn(x). However, because the value of c is uncertain, in practice the remainder term really provides a worst-case scenario for your approximation. The following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is this approximation likely to be? To find out, use the remainder term: cos 1 = T6(x) + R6(x) Adding the associated remainder term changes this approximation into an equation. Here’s the formula for the remainder term: So substituting 1 for x gives you: At this point, you’re apparently stuck, because you don’t know the value of sin c. However, you can plug in c = 0 and c = 1 to give you a range of possible values: Keep in mind that this inequality occurs because of the interval involved, and because that sine increases on that interval. You can get a different bound with a different interval. This simplifies to provide a very close approximation: Thus, the remainder term predicts that the approximate value calculated earlier will be within 0.00017 of the actual value. And, in fact, As you can see, the approximation is within the error bounds predicted by the remainder term.

View ArticleArticle / Updated 07-12-2021

Every nonzero vector has a corresponding unit vector, which has the same direction as that vector but a magnitude of 1. To find the unit vector u of the vector you divide that vector by its magnitude as follows: Note that this formula uses scalar multiplication, because the numerator is a vector and the denominator is a scalar. A scalar is just a fancy word for a real number. The name arises because a scalar scales a vector — that is, it changes the scale of a vector. For example, the real number 2 scales the vector v by a factor of 2 so that 2v is twice as long as v. As you may guess from its name, the unit vector is a vector. For example, to find the unit vector u of the vector you first calculate its magnitude |q|: Now use the previous formula to calculate the unit vector: You can check that the magnitude of resulting vector u really is 1 as follows:

View ArticleArticle / Updated 04-21-2017

The Mean Value Theorem for Integrals guarantees that for every definite integral, a rectangle with the same area and width exists. Moreover, if you superimpose this rectangle on the definite integral, the top of the rectangle intersects the function. This rectangle, by the way, is called the mean-value rectangle for that definite integral. Its existence allows you to calculate the average value of the definite integral. Calculus boasts two Mean Value Theorems — one for derivatives and one for integrals. Here, you will look at the Mean Value Theorem for Integrals. You can find out about the Mean Value Theorem for Derivatives in Calculus For Dummies by Mark Ryan (Wiley). The best way to see how this theorem works is with a visual example: A definite integral and its mean-value rectangle have the same width and area. The first graph in the figure shows the region described by the definite integral This region obviously has a width of 1, and you can evaluate it easily to show that its area is The second graph in the figure shows a rectangle with a width of 1 and an area of It should come as no surprise that this rectangle’s height is also so the top of this rectangle intersects the original function. The fact that the top of the mean-value rectangle intersects the function is mostly a matter of common sense. After all, the height of this rectangle represents the average value that the function attains over a given interval. This value must fall someplace between the function’s maximum and minimum values on that interval. Here’s the formal statement of 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 that interval such that: This equation may look complicated, but it’s basically a restatement of this familiar equation for the area of a rectangle: Area = Height · Width In other words, start with a definite integral that expresses an area, and then draw a rectangle of equal area with the same width (b – a). The height of that rectangle — f(c) — is such that its top edge intersects the function where x = c. The value f(c) is the average value of f(x) over the interval [a, b]. You can calculate it by rearranging the equation stated in the theorem: For example, here’s a figure that illustrates the definite integral and its mean-value rectangle. Now, here’s how you calculate the average value of the shaded area: Not surprisingly, the average value of this integral is 30, a value between the function’s minimum of 8 and its maximum of 64.

View ArticleArticle / Updated 04-21-2017

You can express every product of powers of trig functions, no matter how weird, as the product of any pair of trig functions. The three most useful pairings are sine and cosine, tangent and secant, and cotangent and cosecant. The table shows you how to express all six trig functions as each of these pairings. For example, look at the following function: As it stands, you can’t do much to integrate this monster. But try expressing it in terms of each possible pairing of trig functions: As it turns out, the most useful pairing for integration in this case is cot6 x csc2 x. No fraction is present — that is, both terms are raised to positive powers — and the cosecant term is raised to an even power.

View ArticleArticle / Updated 04-21-2017

The geometric series is a simplified form of a larger set of series called the power series. A power series is any series of the following form: Notice how the power series differs from the geometric series: In a geometric series, every term has the same coefficient. In a power series, the coefficients may be different — usually according to a rule that’s specified in the sigma notation. Here are a few examples of power series: You can think of a power series as a polynomial with an infinite number of terms. For this reason, many useful features of polynomials carry over to power series. The most general form of the power series is as follows: This form is for a power series that’s centered at a. Notice that when a = 0, this form collapses to the simpler version: So a power series in this form is centered at 0.

View ArticleArticle / Updated 04-21-2017

Understanding sequences is an important first step toward understanding series. The simplest notation for defining a sequence is a variable with the subscript n surrounded by braces. For example: You can reference a specific term in the sequence by using the subscript: Make sure you understand the difference between notation with and without braces: The notation {an} with braces refers to the entire sequence. The notation an without braces refers to the nth term of the sequence. When defining a sequence, instead of listing the first few terms, you can state a rule based on n. (This is similar to how a function is typically defined.) For example: Sometimes, for increased clarity, the notation includes the first few terms plus a rule for finding the nth term of the sequence. For example: This notation can be made more concise by appending starting and ending values for n: This last example points out the fact that the initial value of n doesn’t have to be 1, which gives you greater flexibility to define a number series by using a rule. Don’t let the fancy notation for number sequences get to you. When you’re faced with a new sequence that’s defined by a rule, jot down the first four or five numbers in that sequence. After you see the pattern, you’ll likely find that a problem is much easier.

View ArticleArticle / Updated 04-21-2017

When the function you’re integrating includes a term of the form (a2 + x2)n, draw your trigonometry substitution triangle for the tangent case. For example, suppose that you want to evaluate the following integral: This is a tangent case, because a constant plus a multiple of x2 is being raised to a power (–2). Here’s how you use trig substitution to integrate: Draw the trig substitution triangle for the tangent case. The figure shows you how to fill in the triangle for the tangent case. Notice that the radical of what’s inside the parentheses goes on the hypotenuse of the triangle. Then, to fill in the other two sides of the triangle, use the square roots of the two terms inside the radical — that is, 2 and 3x. Place the constant term 2 on the adjacent side and the variable term 3x on the opposite side. With the tangent case, make sure not to mix up your placement of the variable and the constant. Identify the separate pieces of the integral (including dx) that you need to express in terms of theta. In this case, the function contains two separate pieces that contain x: Express these pieces in terms of trig functions of theta. In the tangent case, all trig functions should be initially expressed as tangents and secants. To represent the rational portion as a trig function of theta, build a fraction using the radical as the numerator and the constant 2 as the denominator. Then set this fraction equal to the appropriate trig function: Because this fraction is the hypotenuse of the triangle over the adjacent side it’s equal to Now use algebra and trig identities to tweak this equation into shape: Next, express dx as a trig function of theta. To do so, build another fraction with the variable 3x in the numerator and the constant 2 in the denominator: This time, the fraction is the opposite side of the triangle over the adjacent side so it equals Now solve for x and then differentiate: Express the integral in terms of theta and evaluate it: Now some cancellation and reorganization turns this nasty-looking integral into something manageable: At this point, you can evaluate this integral: So here’s the substitution: And here is the antiderivative: Change the two theta terms back into x terms: You need to find a way to express theta in terms of x. Here’s the simplest way: So here’s a substitution that gives you an answer: This answer is valid, but most professors won’t be crazy about that ugly second term, with the sine of an arctangent. To simplify it, apply the double-angle sine formula to Now use your trig substitution triangle to substitute values for in terms of x: Finally, use this result to express the answer in terms of x:

View ArticleArticle / Updated 04-21-2017

When the function you’re integrating includes a term of the form (a2 – bx2)n, draw your trig substitution triangle for the sine case. For example, suppose that you want to evaluate the following integral: This is a sine case, because a constant minus a multiple of x2 is being raised to a power Here’s how you use trig substitution to handle the job: Draw the trig substitution triangle for the correct case. This figure shows you how to fill in the triangle for the sine case. Notice that the radical goes on the adjacent side of the triangle. Then, to fill in the other two sides of the triangle, you use the square roots of the two terms inside the radical — that is, 2 and x. Place 2 on the hypotenuse and x on the opposite side. You can check to make sure that this placement is correct by using the Pythagorean theorem: Identify the separate pieces of the integral (including dx) that you need to express in terms of theta. In this case, the function contains two separate pieces that contain x: Express these pieces in terms of trig functions of theta. This is the real work of trig substitution, but when your triangle is set up properly, this work becomes a lot easier. In the sine case, all trig functions should be sines and cosines. To represent the radical portion as a trig function of theta, first build a fraction using the radical as the numerator and the constant 2 as the denominator. Then set this fraction equal to the appropriate trig function: Because the numerator is the adjacent side of the triangle and the denominator is the hypotenuse this fraction is equal to Now a little algebra gets the radical alone on one side of the equation: Next, you want to express dx as a trig function of theta. To do so, build another fraction with the variable x in the numerator and the constant 2 in the denominator. Then set this fraction equal to the correct trig function: This time, the numerator is the opposite side of the triangle and the denominator is the hypotenuse so this fraction is equal to Now solve for x and then differentiate: Rewrite the integral in terms of theta and evaluate it: To change those two theta terms into x terms, reuse the following equation: So here’s a substitution that gives you an answer: This answer is perfectly valid so, technically speaking, you can stop here. However, some professors frown upon the nesting of trig and inverse trig functions, so they’ll prefer a simplified version of To find this, start by applying the double-angle sine formula to Now use your trig substitution triangle to substitute values for in terms of x: To finish, substitute this expression for that problematic second term to get your final answer in a simplified form:

View ArticleArticle / Updated 04-21-2017

Differential equations (DEs) come in many varieties. And different varieties of DEs can be solved using different methods. You can classify DEs as ordinary and partial Des. In addition to this distinction they can be further distinguished by their order. Here are some examples: Solving a differential equation means finding the value of the dependent variable in terms of the independent variable. The following examples use y as the dependent variable, so the goal in each problem is to solve for y in terms of x. An ordinary differential equation (ODE) has only derivatives of one variable — that is, it has no partial derivatives. Here are a few examples of ODEs: In contrast, a partial differential equation (PDE) has at least one partial derivative. Here are a few examples of PDEs: DEs are further classified according to their order. This classification is similar to the classification of polynomial equations by degree. First-order ODEs contain only first derivatives. For example: Higher-order ODEs are classified, as polynomials are, by the greatest order of their derivatives. Here are examples of second-, third-, and fourth-order ODEs: As with polynomials, generally speaking, a higher-order DE is more difficult to solve than one of lower order. What constitutes a linear differential equation depends slightly on who you ask. For practical purposes, a linear first-order DE fits into the following form: where a(x) and b(x) are functions of x. Here are a few examples of linear first-order DEs: Linear DEs can often be solved, or at least simplified, using an integrating factor. A linear second-degree DE fits into the following form: where a, b, and c are all constants. Here are some examples: Note that the constant a can always be reduced to 1, resulting in adjustments to the other two coefficients.

View Article