Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 415: 38

Answer

$\frac{{\pi}^{3/2}}{3\sqrt 2} - 3$

Work Step by Step

$$\eqalign{ &\text{Let }I= \int_0^{\pi /2} {\left( {\sqrt t - 3\cos t} \right)} dt \cr & = \int_0^{\pi /2} {\left( {{t^{1/2}} - 3\cos t} \right)} dt \cr & {\text{Integrate by the power rule and }}\int {\cos x} dx = \sin x + C \cr &I = \left[ {\frac{{{t^{3/2}}}}{{3/2}} - 3\sin t} \right]_0^{\pi /2} \cr & = \left[ {\frac{{2{t^{3/2}}}}{3} - 3\sin t} \right]_0^{\pi /2} \cr & {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr & I = \left[ {\frac{{2{{\left( {\pi /2} \right)}^{3/2}}}}{3} - 3\sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ {\frac{{2{{\left( 0 \right)}^{3/2}}}}{3} - 3\sin \left( 0 \right)} \right] \cr & {\text{Simplify}} \cr & I= \left[ {\frac{{2{{\left( {\pi /2} \right)}^{3/2}}}}{3} - 3\left( 1 \right)} \right] - \left[ 0 \right] \cr & = \frac{{2{{\left( {\pi /2} \right)}^{3/2}}}}{3} - 3 \cr & = \frac{{\pi}^{3/2}}{3\sqrt 2} - 3 \cr} $$
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