University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 7



Work Step by Step

$\int_{0}^{1} (x^2+\sqrt x)dx=[\frac{x^3}{3}+\frac{2}{3}\cdot x^{\frac{3}{2}}]_{0}^{1}=[\frac{1^3}{3}+\frac{2}{3}\cdot 1^{\frac{3}{2}}]-[\frac{0^3}{3}+\frac{2}{3}\cdot 0^{\frac{3}{2}}]=[\frac{1}{3}+\frac{2}{3}\cdot 1]-0=[\frac{1}{3}+\frac{2}{3}]=1$
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