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: 44



Work Step by Step

To solve this integral we will use a substitution $a^2-x^2=t$ giving $dt=t'dx=(a^2-x^2)'dx=-2xdx.$ Now we have $xdx=-\frac{1}{2}dt$. For the bounds we have $x_1=0\to t_1=a^2-0^2=a^2$ $x_2=a\to t_2=a^2-a^2=0$ so the integral becomes $$\int_{0}^ax\sqrt{a^2-x^2}dx=-\frac{1}{2}\int_{a^2}^0\sqrt{t}dt=\frac{1}{2}\int_0^{a^2}t^{\frac{1}{2}}dt=\frac{1}{2}\left.\frac{t^\frac{3}{2}}{\frac{3}{2}}\right|_{t=0}^{t=a^2}=\frac{1}{3}\left((a^2)^{\frac{3}{2}}-0^\frac{3}{2}\right)=\frac{1}{3}\bigg(\sqrt{a^2}\bigg)^3=\frac{1}{3}|a|^3.$$
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