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: 49


$\int_{0}^{27}\sqrt[3] x~dx = \frac{243}{4}$

Work Step by Step

We can find the area under the graph: $\int_{0}^{27}\sqrt[3] x~dx$ $=\frac{3}{4}x^{4/3}~\vert_{0}^{27}$ $=\frac{3}{4}(27^{4/3} -0^{4/3})$ $=\frac{3}{4}(81 -0)$ $=\frac{243}{4}$
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