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 347: 48



Work Step by Step

To evaluate the integral $$\int_0^4\frac{x}{\sqrt{1+2x}}dx$$ we will use substitution $1+2x=t$ which gives us $2dx=dt\Rightarrow dx=\frac{dt}{2}$ and $x=\frac{t-1}{2}$. New integration bounds would be: for $x=0$ we have $t=1$ and for $x=4$ we have $t=9.$ Putting this into the integral we get: $$\int_0^4\frac{x}{\sqrt{1+2x}}dx=\int_1^9\frac{1}{\sqrt t}\frac{t-1}{2}\frac{dt}{2}=\frac{1}{4}\int_1^9\frac{t}{\sqrt t}dt-\frac{1}{4}\int_1^9\frac{1}{\sqrt t}dt= \frac{1}{4}\int_1^9 t^{1/2}dt-\frac{1}{4}\int_1^9 t^{-1/2}dt= \left.\frac{1}{4}\frac{t^{3/2}}{\frac{3}{2}}\right|_1^9-\left.\frac{1}{4}\frac{t^{1/2}}{\frac{1}{2}}\right|_1^9= \frac{1}{4}\frac{2}{3}(9^{3/2}-1^{3/2})-\frac{1}{4}\cdot2(9^{1/2}-1^{1/2})= \frac{1}{6}(27-1)-\frac{1}{2}(3-1)= \frac{13}{3}-1=\frac{10}{3}$$
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