University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 27



Work Step by Step

Here, we have $\lim\limits_{a \to 2^{-}}\int_{0}^{a} \dfrac{ds}{\sqrt {4-s^2}}=\lim\limits_{a \to 2^{-}}[\sin^{-1} [\dfrac{s}{2}]_0^a$ This implies that $\lim\limits_{a \to 2^{-}}\sin^{-1} (\dfrac{a}{2})-\lim\limits_{a \to 2^{-}}\sin^{-1} (0)=\lim\limits_{a \to 2^{-}} \sin^{-1} \dfrac{a}{2}$ or, $\lim\limits_{a \to 2^{-}} \sin^{-1} \dfrac{a}{2}=\sin^{-1}(1)$ or, $\sin^{-1}(\sin \dfrac{\pi}{2})=\dfrac{\pi}{2}$
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