Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 2



Work Step by Step

$\int_0^1 (3x+1)^\sqrt{2}dx$ Let $u=3x+1$. Then $du=3dx$, and $dx=\frac{du}{3}$. The bounds are $3*0+1=1$ and $3*1+1=4$. $=\int_{3*0+1}^{3*1+1}u^\sqrt{2}*\frac{du}{3}$ $=\frac{1}{3}\int_1^4 u^\sqrt{2}du$ $=\frac{1}{3}(\frac{u^{1+\sqrt{2}}}{1+\sqrt{2}})|_1^4$ $=\frac{1}{3}(\frac{4^{1+\sqrt{2}}}{1+\sqrt{2}}-\frac{1^{1+\sqrt{2}}}{1+\sqrt{2}})$ $=\frac{1}{3}(\frac{4^{1+\sqrt{2}}-1}{1+\sqrt{2}})$ $=\frac{4^{1+\sqrt{2}}-1}{3(1+\sqrt{2})}$
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