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: 38

Answer

378,507,500

Work Step by Step

\begin{equation} \sum_{k=101}^{200}k^{3}=\sum_{k=1}^{200}k^{3}-\sum_{k=1}^{100}k^{3}\end{equation}\begin{equation}=\frac{200^{2}(200+1)^{2}}{4}-\frac{100^{2}(100+1)^{2}}{4}\end{equation}\begin{equation}=378,507,500 \end{equation}
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