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Copyright © Cengage Learning. All rights reserved.
6 The Integral Copyright © Cengage Learning. All rights reserved.
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Copyright © Cengage Learning. All rights reserved.
6.2 Substitution Copyright © Cengage Learning. All rights reserved.
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Substitution The chain rule for derivatives gives us an extremely useful technique for finding antiderivatives. This technique is called change of variables or substitution. We know that to differentiate a function like (x2 + 1)6, we first think of the function as g(u) where u = x2 + 1 and g(u) = u6. We then compute the derivative, using the chain rule, as
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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.
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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
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Substitution Substitution Rule
If u is a function of x, then we can use the following formula to evaluate an integral:
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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.
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Example 1 – Substitution
Find ∫ 4x(x2 + 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 = x2 + 1.
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Example 1 – Solution cont’d Continuing the procedure above, we place the calculations for step 2 in a box. Write u as a function of x. Take the derivative of u with respect to x. Solve for dx: dx = du.
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Example 1 – Solution ∫ 4x(x2 + 1)6 dx = ∫ 4xu6 du
cont’d Now we substitute u for its defining expression and substitute for dx in the original integral: ∫ 4x(x2 + 1)6 dx = ∫ 4xu6 du = ∫ 2u6 du. We have boiled the given integral down to the much simpler integral ∫ 2u6 du, and we can now write down the solution: Substitute for u and dx. Cancel the xs and simplify. Substitute (x2 + 1) for u in the answer.
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Shortcuts
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Shortcuts Shortcuts: Integrals of Expressions Involving (ax + b)
Rule Quick Example
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Shortcuts cont’d Rule Quick Example
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Shortcuts (Optional) Mike’s Shortcut Rule: Integrals of More
General Expressions Rule Quick Example
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Shortcuts cont’d Rule Quick Example
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