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: 3

Answer

$$\frac{2}{9}x\left( {{{\left( {{x^2} + 15} \right)}^{3/2}} - {{\left( {{x^2} + 12} \right)}^{3/2}}} \right)$$

Work Step by Step

$$\eqalign{ & \int_4^5 {x\sqrt {{x^2} + 3y} dy} \cr & {\text{the notation }}dy{\text{ indicates integration with respect to }}y,{\text{ so we treat }}y{\text{ as variable and}} \cr & x{\text{ as a constant}}{\text{. write the integrand as}} \cr & = \frac{1}{3}x\int_4^5 {{{\left( {{x^2} + 3y} \right)}^{1/2}}\left( 3 \right)dy} \cr & {\text{using the power rule }} \cr & = \frac{1}{3}x\left( {\frac{{{{\left( {{x^2} + 3y} \right)}^{3/2}}}}{{3/2}}} \right)_4^5 \cr & {\text{then}} \cr & = \frac{2}{9}x\left( {{{\left( {{x^2} + 3y} \right)}^{3/2}}} \right)_4^5 \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{2}{9}x\left( {{{\left( {{x^2} + 3\left( 5 \right)} \right)}^{3/2}} - {{\left( {{x^2} + 3\left( 4 \right)} \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{9}x\left( {{{\left( {{x^2} + 15} \right)}^{3/2}} - {{\left( {{x^2} + 12} \right)}^{3/2}}} \right) \cr} $$
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