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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 2 + 1 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 2 + 1.

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