Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9 - Page 340: 21



Work Step by Step

In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $ u=2x-1$ $ du=2dx$ $ \dfrac{1}{2}du=dx$ Change limits: $x_1 =1$ $u_1=2(1)-1=1$ $x_2 =5$ $u_2=2(5)-1=9$ Then substitute: \[ = \dfrac{1}{2}\int_{1}^{9} \dfrac{du}{\sqrt{u}} =\dfrac{1}{2}\int_{1}^{9} u^{-1/2} du\] Then integrate \[ \bigg [ u^{1/2} \bigg ]_{1}^{9} \] Apply calculus 2nd theorem and simplify: \[ 9^{1/2}-1^{1/2}=3-1=2\]
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