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:
- f(x)dx = F(x) + C F '(x) = f(x)
- a f(x)dx = a f(x)dx
- – f(x)dx = – f(x)dx
- [f(x) ± g(x)] dx = f(x)dx ± g(x)dx
- tan x dx = ln |sec x| + C or – ln |cos x| + C
- cot x dx = ln |sin x| + C or – ln |csc x| + C
- sec x dx = ln |sec x + tan x| + C
- csc x dx = ln |csc x – cot x| + C
- 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).
Evaluating Integrals
Example 1
Evaluate (x^{5} – 6x^{2} + x – 1) dx.
Applying the formula .
Example 2
Evaluate .
Rewrite
Example 3
If = 3x^{2} + 2, and the point (0, –1) lies on the graph of y, find y.
Since = 3x^{2} + 2, then y is an antiderivative of . Thus, y = (3x^{2} + 2) dx = x^{3} + 2x + C. The point (0, –1) is on the graph of y.
Thus, y = x3 + 2x + C becomes –1 = 0^{3} + 2(0) + C or C = –1. Therefore, y = x^{3} + 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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