Calculus, 10th Edition (Anton)

Published by Wiley

Chapter 14 - Multiple Integrals - 14.2 Double Integrals Over Nonrectangular Regions - Exercises Set 14.2 - Page 1015: 8

Answer

$$\frac{7}{3}\left( {e - 1} \right)$$

Work Step by Step

\eqalign{ & \int_1^2 {\int_0^{{y^2}} {{e^{x/{y^2}}}} dxdy} \cr & = \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}} dx} \right]dy} \cr & {\text{solve the inner integral and treat }}y{\text{ as a constant}} \cr & = \int_0^{{y^2}} {{e^{x/{y^2}}}} dx \cr & {\text{using }}\int {{e^{kx}}} dx = \frac{1}{k}{e^{kx}} + C \cr & = \frac{1}{{1/{y^2}}}\left( {{e^{x/{y^2}}}} \right)_0^{{y^2}} \cr & = {y^2}\left( {{e^{x/{y^2}}}} \right)_0^{{y^2}} \cr & {\text{evaluating the limits in the variable }}x \cr & = {y^2}\left( {{e^{{y^2}/{y^2}}} - {e^{0/{y^2}}}} \right) \cr & {\text{simplifying}} \cr & = {y^2}\left( {e - 1} \right) \cr & {\text{then}} \cr & \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}} dx} \right]dy} = \int_1^2 {{y^2}\left( {e - 1} \right)dy} \cr & = \left( {e - 1} \right)\int_1^2 {{y^2}dy} \cr & {\text{integrating by the power rule}} \cr & = \left( {e - 1} \right)\left[ {\frac{{{y^3}}}{3}} \right]_1^2 \cr & = \frac{{e - 1}}{3}\left[ {{y^3}} \right]_1^2 \cr & = \frac{{e - 1}}{3}\left( {{{\left( 2 \right)}^3} - {{\left( 1 \right)}^3}} \right) \cr & {\text{simplifying}} \cr & = \frac{{e - 1}}{3}\left( {8 - 1} \right) \cr & = \frac{{e - 1}}{3}\left( 7 \right) \cr & = \frac{7}{3}\left( {e - 1} \right) \cr}

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