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## Calculus 2

by: Vernie McCullough

51

0

15

# Calculus 2 MATH 211

Vernie McCullough

GPA 3.58

Staff

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COURSE
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15
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KARMA
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## Popular in Mathematics (M)

This 15 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 211 at Millersville University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 51 views. For similar materials see /class/223535/math-211-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.

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Date Created: 10/15/15
Integration by Parts MATH 211 Calculus II J Robert Buchanan Department of Mathematics Summer 2008 Introduction 0 If necessary refresh your skills with basic integration including integration by substitution by reviewing Section 61 and working the exercises at the end of the section 0 Today s discussion will focus on the second major technique of integration integration by parts 0 We will see that integration by parts is related to the product rule for derivatives Product Rule for Derivatives 7 Suppose that u and v are functions of X then vdu udv vduudv uvivdu Integration by Parts Formula If u fx and v gx and f and g are continuous then udvuv7vdu or alternatively fxg xdx fxgx e f xgxdx Integration by Parts Formula If u fx and v gx and f and g are continuous then udvuv7vdu or alternatively fxg xdx noun 7 f xgxdx Remark the derivative of gx in the integral on the lefthand side has moved over to fx in the integral on the righthand side Further Remarks 0 When trying to apply integration by parts we must designate part of the integrand to be dv and the rest u 0 Usually dv is the most complicated part ofthe integrand that we can integrate using an elementary technique integrals OXsinxdx 9 Inde 9 cos 1xdx Use integration by parts to evaluate the following indefinite Repeated Integration by Parts i Remark Sometimes integration by parts must be used more than once Use integration by parts to evaluate the following indefinite integrals Ochosxdx Qexsinxdx Tabular Integration Consider X2 cosx dx and suppose we arrange our work in the following table u dv sign cosx X2 sin X 2X 7 cosx 7 2 7 sin X O cosx 7 If we multiply across the completed rows and add these products we get X2COSXdX X2sinx2xcosx72sinx C Note differentiate down the first column antidifferentiate down the second column alternate signs starting with in the third column 7 r Use tabular integration or repeated integration by parts to evaluate X3equot dx Reduction Formulas 7 Often common integration formulas are expressed recursively For example cos xdx cos 1xcosxdx using integration by parts with ucos 1x vsinx du7n71sinxcos 2xdx dvcosxdx we obtain cos x dx sinxcos 7 1 X n 71sinzxcos 2xdx Simplification cos X dX ncos XdX cos X dX sin X cos 7 1 X n 71sin2 X cos 7 2 X dX sinxcos 7 1 X n7117 c052Xcos 2XdX sinxcos 7 1 X n 71cos 2XdX 7n71cos XdX sinxcos 7 1 X n 71cos 2XdX cos 7 2 X dX 71 1Xn 1 7 7 sm X cos n Definite Integrals Integration by parts is easily adapted to definite integrals xb b xb udv uvi 7 vdu xa xa Use integration by parts to evaluate the following definite integrals 7r2 0 xsinxdx o o 9 3xe2XdX 71 e 9 lnxdx 1 Homework Read Section 62 and work exercises 1 57 odd

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