Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 45

Answer

$= \frac{16}{3}$
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Work Step by Step

$\int^{4}_0 (\sqrt x - 0) dx$ $= \int^{4}_0 \sqrt x dx$ $= \int^{4}_0 x^{\frac{1}{2}} dx$ $= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} |^{4}_0$ $= [\frac{4^{\frac{3}{2}}}{\frac{3}{2}}]-[\frac{0^{\frac{3}{2}}}{\frac{3}{2}}]$ $= [\frac{4^{\frac{3}{2}}}{\frac{3}{2}}]-0$ $= [\frac{8}{\frac{3}{2}}]$ $= \frac{2\times8}{3}$ $= \frac{16}{3}$
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