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: 22

Answer

$\dfrac{38}{15}$

Work Step by Step

In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $ u=5x-1$ $ du = 5dx$ $ \dfrac{1}{5}du=dx$ Change limits: $x_1 =1$ $u_1=5(1)-1=4$ $x_2 =2$ $u_2=5(2)-1=9$ Then substitute: \[ \dfrac{1}{5} \int_{4}^{9} \sqrt{u} du = \dfrac{1}{5} \int_{4}^{9} u^{1/2}du \] \[ \] Then integrate \[ \dfrac{1}{5} \bigg [\dfrac{2}{3}u^{2/3} \bigg ]_{4}^{9} = \dfrac{2}{15} \bigg [u^{2/3} \bigg ]_{4}^{9} \] Apply calculus 2nd theorem and reduce: \[ \dfrac{2}{15} [ 9^{3/2} -4^{3/2}] = \dfrac{2}{15}[27-8]=\dfrac{38}{15}\]
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