Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 32

Answer

$\displaystyle \frac{\pi}{2}$ (converges)

Work Step by Step

Discontinuity at x=1. Type 1. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\lim_{t\rightarrow 1^{-}}\int_{0}^{t}\frac{dx}{\sqrt{1-x^{2}}}$ $=\displaystyle \lim_{t\rightarrow 1-}\left[\arcsin x\right]_{0}^{t}$ $=\displaystyle \lim_{t\rightarrow 1^{-}}\arcsin t-\lim_{t\rightarrow 1^{-}}\arcsin 0$ $=\displaystyle \frac{\pi}{2}-0$ $=\displaystyle \frac{\pi}{2}$ (converges)
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