Presentation on theme: "Chapter 7 Additional Integration Topics"— Presentation transcript:
1 Chapter 7 Additional Integration Topics Section 3Integration by Parts
2 Learning Objectives for Section 7.3 Integration by Parts The student will be able to integrate by parts.
3 An Unsolved ProblemIn a previous section we said we would return later to the indefinite integralsince none of the integration techniques considered up to that time could be used to find an antiderivative for ln x. We now develop a very useful technique, called integration by parts that will enable us to find not only the preceding integral, but also many others.
4 Integration by PartsIntegration by parts is based on the product formula for derivatives:We rearrange the above equation to get:Integrating both sides yieldswhich reduces to
5 Integration by Parts Formula If we let u = f (x) and v = g(x) in the preceding formula, the equation transforms into a more convenient form:Integration by Parts Formula
6 Integration by Parts Formula This method is useful when the integral on the left side is difficult and changing it into the integral on the right side makes it easier.Remember, we can easily check our results by differentiating our answers to get the original integrals.Let’s look at an example.
7 ExampleOur previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities.Option 1u dvOption 2u dvLet’s try option 1.
8 Example (continued) We have decided to use u = x and dv = e x dx. This means du = dx and v = e x, yieldingThis is easy to check by differentiating.Note: Option 2 does not work in this example.
9 Selecting u and dv The product u dv must equal the original integrand. It must be possible to integrate dv by one of our known methods.The new integral should not be more involved than the original integralFor integrals involving x p e ax , try u = x p and dv = e axdxFor integrals involving x p (ln x) q, try u = (ln x) q and dv = x p dx
11 Example 2FindusingLet u = ln x and dv = x3 dx, yielding du = 1/x dx and v = x4/4Check by differentiating:
12 Solving the Unsolved Problem Recall that our original problem was to findWe follow the suggestionFor integrals involving x p (ln x)q, try u = (ln x)q and dv = x p dxSet p = 0 and q = 1 and choose u = ln x and dv = dx.Then du = 1/x dx and v = x, hence
13 SummaryWe have developed integration by parts as a useful technique for integrating more complicated functions.Integration by Parts Formula: