Copyright © Cengage Learning. All rights reserved. 14 Further Integration Techniques and Applications of the Integral Copyright © Cengage Learning. All rights reserved.

Copyright © Cengage Learning. All rights reserved. Integration by Parts 14.1 Copyright © Cengage Learning. All rights reserved.

Integration by Parts Integration by parts is an integration technique that comes from the product rule for derivatives. We start with a little notation to simplify things while we introduce integration by parts. If u is a function, denote its derivative by D(u) and an antiderivative by I(u). Thus, for example, if u = 2x2, then D(u) = 4x and I(u) = [If we wished, we could instead take I(u) = + 46, but we usually opt to take the simplest antiderivative.]

Integration by Parts Integration by parts If u and v are continuous functions of x, and u has a continuous derivative, then Quick Example

Derivation of Integration by Parts Formula

Derivation of Integration by Parts Formula As we mentioned, the integration-by-parts formula comes from the product rule for derivatives. We apply the product rule to the function uI(v) D[u  I(v)] = D(u)I(v) + uD(I(v)) = D(u)I(v) + uv because D(I(v)) is the derivative of an antiderivative of v, which is v. Integrating both sides gives u  I(v) A simple rearrangement of the terms now gives us the integration-by-parts formula.

Example 1 – Integration by Parts: Tabular Method Calculate Solution: First, the reason we need to use integration by parts to evaluate this integral is that none of the other techniques of integration that we’ve talked about up to now will help us. Furthermore, we cannot simply find antiderivatives of x and ex and multiply them together. This integral can be found by integration by parts. We want to find the integral of the product of x and ex.

Example 1 – Solution cont’d We must make a decision: Which function will play the role of u and which will play the role of v in the integration-by-parts formula? Because the derivative of x is just 1, differentiating makes it simpler, so we try letting x be u and letting ex be v. We need to calculate D(u) and I(v), which we record in the following table. The table is read as +x · ex −∫1 · ex dx

Example 1 – Solution cont’d Below x in the D column, we put D(x) = 1; below ex in the I column, we put I(ex) = ex. The arrow at an angle connecting x and I(ex) reminds us that the product xI(ex) will appear in the answer; the plus sign on the left of the table reminds us that it is +xI(ex) that appears. The integral sign and the horizontal arrow connecting D(x) and I(ex) remind us that the integral of the product D(x)I(ex) also appears in the answer; the minus sign on the left reminds us that we need to subtract this integral.

Example 1 – Solution Combining these two contributions, we get cont’d Combining these two contributions, we get The integral that appears on the right is much easier than the one we began with, so we can complete the problem:

Derivation of Integration by Parts Formula Integrating a Polynomial Times a Function If one of the factors in the integrand is a polynomial and the other factor is a function that can be integrated repeatedly, put the polynomial in the D column and keep differentiating until you get zero. Then complete the I column to the same depth, and read off the answer.