Answer
$$\bar f = \frac{1}{{2\ln 2}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = {e^{ - y}};{\text{ }}R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 6,{\text{ 0}} \leqslant y \leqslant \ln {\text{2}}} \right\} \cr
& {\text{The average value of an integrable function }}f{\text{ over a region }}R \cr
& {\text{is:}} \cr
& \bar f = \frac{1}{{{\text{area of }}R}}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& {\text{The area of the region }}R{\text{ is:}} \cr
& {\text{Area of }}R = \int_0^{\ln 2} {\int_0^6 {dx} dy} \cr
& {\text{Area of }}R = 6\ln 2 \cr
& \cr
& {\text{Therefore}}{\text{, }} \cr
& \bar f = \frac{1}{{6\ln 2}}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& \bar f = \frac{1}{{6\ln 2}}\int_0^{\ln 2} {\int_0^6 {{e^{ - y}}dx} dy} \cr
& {\text{Integrate with respect to }}x \cr
& \bar f = \frac{1}{{6\ln 2}}\int_0^{\ln 2} {\left[ {x{e^{ - y}}} \right]_0^6dy} \cr
& \bar f = \frac{6}{{6\ln 2}}\int_0^{\ln 2} {{e^{ - y}}dy} \cr
& \bar f = \frac{1}{{\ln 2}}\int_0^{\ln 2} {{e^{ - y}}dy} \cr
& {\text{Integrate and evaluate}} \cr
& \bar f = - \frac{1}{{\ln 2}}\left[ {{e^{ - y}}} \right]_0^{\ln 2} \cr
& \bar f = - \frac{1}{{\ln 2}}\left[ {{e^{ - \ln 2}} - {e^0}} \right] \cr
& \bar f = - \frac{1}{{\ln 2}}\left( { - \frac{1}{2}} \right) \cr
& \bar f = \frac{1}{{2\ln 2}} \cr} $$
