Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 513: 61



Work Step by Step

Use u-substitution to integrate the function $\int_0^8 (\frac{2x}{\sqrt {x^2 +36}}) dx$, let $u=x^2+36$ $\int_0^8 u^{-\frac{1}{2}} du$ $[2u^{\frac{1}{2}}]_0^8$ $[2\sqrt {x^2 +36} ]_0^8$ $(2\sqrt {100} - 2\sqrt {36})$ $20-12= 8$
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