Copyright © Cengage Learning. All rights reserved. 13 The Integral

Copyright © Cengage Learning. All rights reserved Substitution

33 The chain rule for derivatives gives us an extremely useful technique for finding antiderivatives. This technique is called change of variables or substitution. Recall that to differentiate a function like (x 2 + 1) 6, we first think of the function as g(u) where u = x and g(u) = u 6. We then compute the derivative, using the chain rule, as

44 Substitution Any rule for derivatives can be turned into a technique for finding antiderivatives by writing it in integral form. The integral form of the formula is But, if we write g(u) + C = ∫ g(u) du, we get the following interesting equation: This equation is the one usually called the change of variables formula.

55 Substitution We can turn it into a more useful integration technique as follows. Let f = g(u)(du/dx). We can rewrite the change of variables formula using f : In essence, we are making the formal substitution

66 Substitution Substitution Rule If u is a function of x, then we can use the following formula to evaluate an integral:

77 Substitution Rather than use the formula directly, we use the following step-by-step procedure: 1. Write u as a function of x. 2. Take the derivative du/dx and solve for the quantity dx in terms of du. 3. Use the expression you obtain in step 2 to substitute for dx in the given integral and substitute u for its defining expression.

88 Example 1 – Substitution Find ∫ 4x(x 2 + 1) 6 dx. Solution: To use substitution we need to choose an expression to be u. There is no hard and fast rule, but here is one hint that often works: Take u to be an expression that is being raised to a power. In this case, let’s set u = x

99 Example 1 – Solution Continuing the procedure above, we place the calculations for step 2 in a box. cont’d Write u as a function of x. Take the derivative of u with respect to x. Solve for dx: dx = du.

10 Now we substitute u for its defining expression and substitute for dx in the original integral: ∫ 4x(x 2 + 1) 6 dx = ∫ 4xu 6 du = ∫ 2u 6 du. We have boiled the given integral down to the much simpler integral ∫ 2u 6 du, and we can now write down the solution: Example 1 – Solution cont’d Substitute for u and dx. Cancel the xs and simplify. Substitute (x 2 + 1) for u in the answer.

11 Shortcuts

12 Shortcuts Shortcuts: Integrals of Expressions Involving (ax + b) Rule Quick Example

13 Shortcuts Rule Quick Example cont’d