Calculus II For Dummies book cover

Calculus II For Dummies

Author:
Mark Zegarelli
Published: January 24, 2012
Buy on Amazon

Overview

The easy (okay, easier) way to master advanced calculus topics and theories

Calculus II For Dummies will help you get through your (notoriously difficult) calc class—or pass a standardized test like the MCAT with flying colors. Calculus is required for many majors, but not everyone’s a natural at it. This friendly book breaks down tricky concepts in plain English, in a way that you can understand. Practical examples and detailed walkthroughs help you manage differentiation, integration, and everything in between. You’ll refresh your knowledge of algebra, pre-calc and Calculus I topics, then move on to the more advanced stuff, with plenty of problem-solving tips along the way.

  • Review Algebra, Pre-Calculus, and Calculus I concepts
  • Make sense of complicated processes and equations
  • Get clear explanations of how to use trigonometry functions
  • Walk through practice examples to master Calc II

Use this essential resource as a supplement to your textbook or as refresher before taking a test—it’s packed with all the helpful knowledge you need to succeed in Calculus II.

The easy (okay, easier) way to master advanced calculus topics and theories

Calculus II For Dummies will help you get through your (notoriously difficult) calc class—or pass a standardized test like the MCAT with flying colors. Calculus is required for many majors, but not everyone’s a natural at it. This friendly book breaks down tricky concepts in plain English, in a way that you can understand. Practical examples and detailed walkthroughs help you manage differentiation, integration, and everything in between. You’ll refresh your knowledge of

algebra, pre-calc and Calculus I topics, then move on to the more advanced stuff, with plenty of problem-solving tips along the way.
  • Review Algebra, Pre-Calculus, and Calculus I concepts
  • Make sense of complicated processes and equations
  • Get clear explanations of how to use trigonometry functions
  • Walk through practice examples to master Calc II

Use this essential resource as a supplement to your textbook or as refresher before taking a test—it’s packed with all the helpful knowledge you need to succeed in Calculus II.

Calculus II For Dummies Cheat Sheet

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.

About The Author

Mark Zegarelli, 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 Logic For Dummies and Basic Math & Pre-Algebra For Dummies.

Articles From The Book

82 results

Calculus Articles

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

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:

  1. Declare u equal to the inner function in the denominator and make the substitution:

    Here’s the substitution:

  2. Differential du = (2x + 1) dx:

  3. 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!)

  4. Integration is now quite straightforward:

    You take an extra step to remove the fraction before you integrate:

  5. 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.

Calculus Articles

Setting Up Partial Fractions When You Have Repeated Linear Factors

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:

Calculus Articles

Setting Up Partial Fractions When You Have Distinct Factors

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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