Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 73


$$\int^1_0\frac{dx}{(1+\sqrt x)^4}=\frac{1}{6}$$

Work Step by Step

$$A=\int^1_0\frac{dx}{(1+\sqrt x)^4}$$ Let $u=1+\sqrt x$ Then we would have $du=(1+\sqrt x)'dx=(x^{1/2})'dx=\frac{1}{2\sqrt x}dx$. so $dx=(2\sqrt x)du$ However, since $u=1+\sqrt x$, we also have $\sqrt x=u-1$. Therefore $dx=2(u-1)du$ When $x=0$, $u=1$ and when $x=1$, $u=2$. Now we would substitute into $A$: $$A=\int^2_1\frac{2(u-1)}{u^4}du$$ $$A=2\int^2_1\frac{u-1}{u^4}du$$ $$A=2\int^2_1(u^{-3}-u^{-4})du$$ $$A=2(\frac{u^{-2}}{-2}-\frac{u^{-3}}{-3})\Bigg]^2_1$$ $$A=2(\frac{1}{-2u^2}+\frac{1}{3u^3})\Bigg]^2_1$$ $$A=(-\frac{1}{u^2}+\frac{2}{3u^3})\Bigg]^2_1$$ $$A=(-\frac{1}{4}+\frac{2}{3\times8})-(-1+\frac{2}{3\times1^3})$$ $$A=-\frac{1}{6}-(-\frac{1}{3})$$ $$A=\frac{1}{6}$$
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