Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Problems Plus - Problems - Page 433: 2

Answer

$\int_0^1f^{-1}(y)dy=\frac{2}{3}$

Work Step by Step

Let $x=f^{-1}(y)$ or $y=f(x)$. Then, i) $\frac{dy}{dx}=f'(x)$ or $dy=f'(x)dx$; ii) for $y_1=0\Rightarrow x_1=f^{-1}(y_1)=f^{-1}(0)=0$ (since $f(0)=0\Leftrightarrow f^{-1}(1)=1$); iii) for $y_2=1\Rightarrow x_2=f^{-1}(y_2)=f^{-1}(1)=1$ (since $f(1)=1\Leftrightarrow f^{-1}(1)=1$). Using the substitution rule, $\int_{y_1=0}^{y_2=1}f^{-1}(y)dy=\int_{x_1=0}^{x_2=1}xf'(x)dx$ $=\int_{0}^{1}(xf'(x)+f(x))-f(x)dx$ (Apply the Difference property for integrals) $=\int_0^1(xf'(x)+f(x))dx-\int_0^1f(x)dx$ $=[xf(x)]_0^1-\frac{1}{3}$ (Since $\frac{d}{dx}(xf(x))=xf'(x)+f(x)$) $=1\cdot f(1)-0\cdot f(0)-\frac{1}{3}$ $=1\cdot 1-0\cdot 0-\frac{1}{3}$ $=\frac{2}{3}$ Thus, $\int_0^1f^{-1}(y)dy=\frac{2}{3}$
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