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Evaluating Basic Integrals for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Integration Practice Problems for AP Calculus

Antiderivatives and Integration Formulas

Definition: A function F is an antiderivative of another function f if F '(x) = f(x) for all x in some open interval. Any two antiderivatives of f differ by an additive constant C. We denote the set of antiderivatives of f by f(x)dx, called the indefinite integral of f.

Integration Rules:

  1. f(x)dx = F(x) + C F '(x) = f(x)
  2. a f(x)dx = a f(x)dx
  3. f(x)dx = – f(x)dx
  4. [f(x) ± g(x)] dx = f(x)dx ± g(x)dx

Antiderivates and Integration Formulas

  1. tan x dx = ln |sec x| + C or – ln |cos x| + C
  2. cot x dx = ln |sin x| + C or – ln |csc x| + C
  3. sec x dx = ln |sec x + tan x| + C
  4. csc x dx = ln |csc x – cot x| + C
  5. ln x dx = x ln |x| – x + C

Note: After evaluating an integral, always check the result by taking the derivative of the answer (i.e., taking the derivative of the antiderivative).

Antiderivates and Integration Formulas

Evaluating Integrals

Example 1

Evaluate (x5 – 6x2 + x – 1) dx.

Applying the formula .

Example 2

Evaluate .

Rewrite

Example 3

If = 3x2 + 2, and the point (0, –1) lies on the graph of y, find y.

Since = 3x2 + 2, then y is an antiderivative of . Thus, y = (3x2 + 2) dx = x3 + 2x + C. The point (0, –1) is on the graph of y.

Thus, y = x3 + 2x + C becomes –1 = 03 + 2(0) + C or C = –1. Therefore, y = x3 + 2x – 1.

Example 4

Example 5

Example 6

Example 7

Example 8

Evaluate (4 cos x – cot x ) dx.

(4 cos x – cot x ) dx = 4 sin x – ln |sin x| + C.

Example 9

Example 10

Example 11

Example 12

Example 13

Reminder: You can always check the result by taking the derivative of the answer.

Practice problems for these concepts can be found at: Integration Practice Problems for AP Calculus

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