MATH 12002 - CALCULUS I 4.5: Integration by Substitution Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.5: Integration by Substitution

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Reversing the Chain Rule

Recall the Chain Rule for derivatives: d dx F(g(x)) = F ′(g(x)) · g′(x). It follows that if F ′(x) = f (x), then

• f (g(x)) · g′(x) dx = F(g(x)) + C.

If we let u = g(x) be a new variable, we get du

dx = g′(x).

Abusing the notation and treating du

dx like a fraction yields du = g′(x)dx.

The integral becomes

• f (g(x)) · g′(x) dx =
• f (u) du,

and since F(x) is an anitiderivative for f (x), we have

• f (u) du = F(u) + C = F(g(x)) + C.

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For example, consider

• (x2 + 3)8 · 2x dx.

Observe that (x2 + 3)8 can be written as f (g(x)), where f (x) = x8 and g(x) = x2 + 3. The inside of the composite function is g(x) = x2 + 3 and its derivative is g′(x) = 2x. We can turn our integral into a much simpler form by letting u = g(x) = x2 + 3, so that du

dx = g′(x) = 2x and du = 2x dx.

We have

• (x2 + 3)8 · 2x dx =
• u8 du = 1

9u9 + C = 1 9(x2 + 3)9 + C.

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Notes: Usually, we are looking for the inside of a composite function whose derivative is also a factor in the integral. The only thing we are allowed to choose is the function u = g(x). Once u = g(x) is chosen, du = g′(x) dx is completely determined. We do not get to choose du. After choosing u and computing du, re-write everything in the integral in terms of u. Nothing in the integral can be left over and nothing can be added. The substitution must “fit” correctly. The variables x and u should never appear in the integral together. The order of multiplication within the integral does not matter, but the differential (du) must come last: start with

• , end with du.

We will see shortly that the one exception to the “perfect fit” is that the substitution can be off by a constant multiple. This is due to the fact that

• kf (x) dx = k
• f (x) dx for a constant k.

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