Techniques of Integration Chapter 9 Techniques of Integration Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Chapter Outline Integration by Substitution Integration by Parts Evaluation of Definite Integrals Approximation of Definite Integrals Some Applications of the Integral Improper Integrals Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Evaluation of Definite Integrals Section 9.3 Evaluation of Definite Integrals Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section Outline The Definite Integral Evaluating Definite Integrals Change of Limits Rule Finding the Area Under a Curve Integration by Parts and Definite Integrals Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. The Definite Integral where F΄(x) = f (x). Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Evaluating Definite Integrals EXAMPLE Evaluate. SOLUTION First let u = 1 + 2x and therefore du = 2dx. So, we have Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Evaluating Definite Integrals CONTINUED Consequently, Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Change of Limits Rule Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Using the Change of Limits Rule EXAMPLE Evaluate using the Change of Limits Rule. SOLUTION First let u = 1 + 2x and therefore du = 2dx. When x = 0 we have u = 1 + 2(0) = 1. And when x = 1, u = 1 + 2(1) = 3. Thus Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Finding the Area Under a Curve EXAMPLE Find the area of the shaded region. SOLUTION To find the area of the shaded region, we will integrate the given function. But we must know what our limits of integration will be. Therefore, we must determine the three x-intercepts of the function. This is the given function. Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Finding the Area Under a Curve CONTINUED Replace y with 0 to find the x-intercepts. Set each factor equal to 0. Solve for x. Therefore, the left-most region (above the x-axis) starts at x = -3 and ends at x = 0. The right-most region (below the x-axis) starts at x = 0 and ends at x = 3. So, to find the area in the shaded regions, we will use the following. Now let’s find an antiderivative for both integrals. We will use u = 9 – x2 and du = -2xdx. Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Finding the Area Under a Curve CONTINUED Now we solve for the area. Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Finding the Area Under a Curve CONTINUED Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Integration by Parts & Definite Integrals EXAMPLE Evaluate. SOLUTION To solve this integral, we will need integration by parts. Our calculations can be set up as follows: Then Copyright © 2014, 2010, 2007 Pearson Education, Inc.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. Integration by Parts & Definite Integrals CONTINUED Therefore, we have = 1 Copyright © 2014, 2010, 2007 Pearson Education, Inc.