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: 36



Work Step by Step

\begin{equation} \sum_{k=1}^{30}(4k-3)=4\sum_{k=1}^{30}k-\sum_{k=1}^{30}3\end{equation}\begin{equation}=4\times\frac{30(30+1)}{2}-(30\times3)=1770\end{equation}
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