Answer
3
Work Step by Step
$$\eqalign{
& \int_0^{\ln 4} {\int_0^{\ln 3} {\int_0^{\ln 2} {{e^{ - x + y + z}}} dx} dy} dz \cr
& = \int_0^{\ln 4} {\int_0^{\ln 3} {\int_0^{\ln 2} {{e^{ - x + y + z}}} dx} dy} dz \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^{\ln 4} {\int_0^{\ln 3} {\left[ { - {e^{ - x + y + z}}} \right]} _0^{\ln 2}dy} dz \cr
& = \int_0^{\ln 4} {\int_0^{\ln 3} {\left[ { - {e^{ - \ln 2 + y + z}} + {e^{y + z}}} \right]} dy} dz \cr
& = \int_0^{\ln 4} {\int_0^{\ln 3} {\left( { - \frac{1}{2}{e^{y + z}} + {e^{y + z}}} \right)} dy} dz \cr
& = \frac{1}{2}\int_0^{\ln 4} {\int_0^{\ln 3} {{e^{y + z}}} dy} dz \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{1}{2}\int_0^{\ln 4} {\left[ {{e^{y + z}}} \right]_0^{\ln 3}} dz \cr
& = \frac{1}{2}\int_0^{\ln 4} {\left[ {{e^{\ln 3 + z}} - {e^{0 + z}}} \right]} dz \cr
& = \frac{1}{2}\int_0^{\ln 4} {\left( {3{e^z} - {e^z}} \right)} dz \cr
& = \int_0^{\ln 4} {{e^z}} dz \cr
& {\text{Integrate with respect to }}z \cr
& = \left[ {{e^z}} \right]_0^{\ln 4} \cr
& = {e^{\ln 4}} - {e^0} \cr
& = 3 \cr} $$
