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: 30



Work Step by Step

Here, we have $\int_{2}^{4} \dfrac{dt}{t \sqrt {t^2-4}}=\lim\limits_{a \to 2^{+}}\int_{a}^{4} \dfrac{dt}{t \sqrt {t^2-4}}$ This implies that $\lim\limits_{a \to 2^{+}}\int_{a}^{4} \dfrac{dt}{t \sqrt {t^2-4}}=\lim\limits_{a \to 2^{+}}[(\dfrac{1}{2})\sec^{-1} \dfrac{t}{2}]_{b}^{4}$ Thus, $\lim\limits_{a \to 2^{+}}(\dfrac{1}{2})\sec^{-1} \dfrac{4}{2}-\lim\limits_{a \to 2^{+}}(\dfrac{1}{2})\sec^{-1} \dfrac{a}{2}=\dfrac{\pi}{6}$
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