Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 760: 34

$e^3-4$

Re-arrange the given integral as follows: $\int_{0}^{1}\int_{0}^{3} xe^{xy}dx dy=\int_{0}^{3}\int_{0}^{1} xe^{xy}dx dy$ Then, we have $\int_{0}^{3} [x \dfrac{e^{xy}}{x}]_0^1dx=\int_{0}^{3} [e^{xy}]_0^1dx$ Hence, $\int_{0}^{3} [e^x -1]dx=|e^x-x]_0^3=e^3-4$