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: 24

Answer

$1$

Work Step by Step

In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $ u = \cos x +1$ $ du = -\sin x dx$ $ -du = \sin x dx$ Change limits: $x_1 =\pi/2$ $u_1= \cos (\pi/2) +1=1$ $x_2 =\pi$ $u_2= \cos (\pi) +1=0$ Then substitute: \[ = -\int_{1}^{0} 6u^5 du = \int_{0}^{1} 6u^5 du\] Then integrate \[ =u^6 \bigg ]_{0}^{1} \] Apply calculus 2nd theorem and reduce: \[ =(1)^6-(0)^6=1 \]
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