Calculus (3rd Edition)

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

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 846: 27

Answer

$$4$$

Work Step by Step

\begin{align*} \int_{0}^{1} \int_{0}^{2}\left(x+4 y^{3}\right) d x d y&=\int_{0}^{1} \int_{0}^{2} x d x d y+\int_{0}^{1} \int_{0}^{2} 4 y^{3} d x d y\\ &= \left(\int_{0}^{2} x d x\right)\left(\int_{0}^{1} 1 d y\right)+ \left(\int_{0}^{1} 4 y^{3} d y\right)\left(\int_{0}^{2} 1 d x\right) \\ &=\left(\left.\frac{1}{2} x^{2}\right|_{0} ^{2}\right)\left(\left.y\right|_{0} ^ { 1 }\right) + \left(\left.y^{4}\right|_{0} ^{1 }\right)\left(\left.x\right|_{0} ^{2}\right)\\ &=4 \end{align*}
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