Calculus 8th Edition

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

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 43


$$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}=3$$

Work Step by Step

To evaluate the integral $$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}$$ we will use substitution $1+2x=t$ which gives us $2dx=dt\Rightarrow dx=\frac{dt}{2}$ and the integration bounds would be: for $x=0$ we have $t=1$ and for $x=13$ we have $t=27,$ so we get: $$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}=\int_1^{27}t^{-2/3}\frac{dt}{2}=\frac{1}{2}\left.\frac{t^{1/3}}{\frac{1}{3}}\right|_1^{27}=\frac{1}{2}\cdot3(27^{1/3}-1^{1/3})=\frac{3}{2}(3-1)=3$$
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