Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 28

Answer

$$\pi$$

Work Step by Step

\begin{align*} \int_{0}^{1} \frac{4 r d r}{\sqrt{1-r^{4}}}&=\lim _{b \rightarrow 1}\int_{0}^{b} \frac{4 r d r}{\sqrt{1-r^{4}}}\\ &=\lim _{b \rightarrow 1}\left[2 \sin ^{-1}\left(r^{2}\right)\right]_{0}^{b}\\ &=\lim _{b \rightarrow 1^{-}}\left[2 \sin ^{-1}\left(b^{2}\right)-2 \sin ^{-1} 0\right]\\ &=2 \cdot \frac{\pi}{2}-0=\pi \end{align*}
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