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: 2

Answer

$$\frac{{11x}}{4}$$

Work Step by Step

$$\eqalign{ & \int_1^2 {\left( {x{y^3} - x} \right)dy} \cr & {\text{the notation }}dy{\text{ indicates integration with respect to }}y,{\text{ so we treat }}y{\text{ as variable and}} \cr & x{\text{ as a constant}}{\text{. using the power rule for antiderivatives we obtain}} \cr & \int_1^2 {\left( {x{y^3} - x} \right)dy} = \left[ {x\left( {\frac{{{y^4}}}{4}} \right) - x\left( y \right)} \right]_1^2 \cr & {\text{then}} \cr & = \left[ {\frac{{x{y^4}}}{4} - xy} \right]_1^2 \cr & {\text{evaluating the limits in the variable }}y \cr & = \left[ {\frac{{x{{\left( 2 \right)}^4}}}{4} - x\left( 2 \right)} \right] - \left[ {\frac{{x{{\left( 1 \right)}^4}}}{4} - x\left( 1 \right)} \right] \cr & {\text{simplifying}} \cr & = \left( {4x - 2x} \right) - \left( {\frac{1}{4}x - x} \right) \cr & = 2x + \frac{3}{4}x \cr & = \frac{{11x}}{4} \cr} $$
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