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: 47

Answer

$$\int_1^2x\sqrt{x-1}dx=\frac{16}{15}$$

Work Step by Step

To evaluate the integral $$\int_1^2x\sqrt{x-1}dx$$ we will use substitution $x-1=t$ which gives us $dx=dt$ and $x=t+1$. New integration bounds are: for $x=1$ we have $t=0$ and for $x=2$ we have $t=1$. Putting this into the integral we get: $$\int_1^2x\sqrt{x-1}dx=\int_0^1(t+1)\sqrt t dt=\int_0^1t^{3/2}dt+\int_0^1t^{1/2}dt= \left.\frac{t^{5/2}}{\frac{5}{2}}\right|_0^1+\left.\frac{t^{3/2}}{\frac{3}{2}}\right|_0^1= \frac{2}{5}(1^{5/2}-0^{5/2})+\frac{2}{3}(1^{3/2}-0^{3/2})=\frac{2}{5}\cdot1+\frac{2}{3}\cdot1= \frac{16}{15}$$
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