Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 513: 11

Answer

$$945$$

Work Step by Step

$$\eqalign{ & \int_1^2 {\int_0^5 {\left( {{x^4}y + y} \right)} } dxdy \cr & = \int_1^2 {\left[ {\int_0^5 {\left( {{x^4}y + y} \right)dx} } \right]} dy \cr & {\text{solve the inner integral}} \cr & \int_0^5 {\left( {{x^4}y + y} \right)dx} = \left[ {y\left( {\frac{{{x^5}}}{5}} \right) + y\left( x \right)} \right]_0^5 \cr & {\text{then}} \cr & = \left[ {\frac{{{x^5}y}}{5} + xy} \right]_0^5 \cr & {\text{evaluating the limits in the variable }}x \cr & = \left[ {\frac{{{{\left( 5 \right)}^5}y}}{5} + \left( 5 \right)y} \right] - \left[ {\frac{{{{\left( 0 \right)}^5}y}}{5} + \left( 0 \right)y} \right] \cr & {\text{simplifying}} \cr & = {5^4}y + 5y \cr & = 625y + 5y \cr & = 630y \cr & \cr & = \int_1^2 {630y} dy \cr & {\text{integrating}} \cr & = \left( {\frac{{630{y^2}}}{2}} \right)_1^2 = \left( {315{y^2}} \right)_1^2 \cr & {\text{evaluate}} \cr & = 315{\left( 2 \right)^2} - 315{\left( 1 \right)^2} \cr & = 945 \cr} $$
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