Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.1 Approximating and Computing Area - Exercises - Page 236: 35



Work Step by Step

\begin{equation}\sum_{j=1}^{20}8j^{3}=8\sum_{j=1}^{20}j^{3}=8\times\frac{20^{2}(20+1)^{2}}{4}=352,800\end{equation} Formula (5) is used.
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