Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 760: 19

$\dfrac{1}{2}$

Re-arrange the given integral as follows: $\int_{0}^{\ln 2} [e^x(-e^{-y}]_0^{\ln 2} dx=\int_{0}^{\ln 2} (1-\dfrac{1}{2})e^x dx$ This implies that $[\dfrac{1}{2} e^x]_{0}^{\ln 2}=(\dfrac{1}{2}) (e^{\ln 2} -1)$ Hence, $(\dfrac{1}{2}) (2-1)=\dfrac{1}{2}$