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: 16



Work Step by Step

Re-arrange the given integral as follows: $\int_{0}^{2} \dfrac{x}{\sqrt {4-x^2}}+\int_{0}^{2} \dfrac{1}{\sqrt {4-x^2}}=$ ....(1) Plug $u(\theta)=4-x^2 \implies du=-2xdx$ Then, we have $\sin^{-1} (x/2) dx|_0^2+(\dfrac{-1}{2} \int_{0}^{4}\dfrac{du}{\sqrt u}=\sin^{-1}(1)-\sin^{-1}(0)- (\dfrac{1}{2}) (2) u^{1/2}$ Thus, equation (1) becomes $\dfrac{\pi}{2}-0-(-2)=2+\dfrac{\pi}{2}$
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