Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - Review Exercises: 21

Answer

$\frac{7}{6}$

Work Step by Step

$\int_{0}^{1}\sqrt x(\sqrt x + 1) dx$ = $\int_{0}^{1} x+ \sqrt xdx$ = $\frac{1}{2} x^2 + \frac{2}{3}x^\frac{3}{2}|_{0}^{1}$ =$(\frac{1}{2} (1)^2 + \frac{2}{3}(1)^\frac{3}{2})-(\frac{1}{2} (0)^2 + \frac{2}{3}(0)^\frac{3}{2})$ =$(\frac{1}{2}+\frac{2}{3}) + 0$ =$\frac{7}{6}$
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