Chapter 14 - Section 14.2 - Area between Two Curves and Applications - Exercises - Page 1031: 36

$3 \ln (3)-2$

Here, we have: $y=e^{-x}$ and $y=3$ Suppose that $y=y \implies e^{-x}=3$ or, $\ln e^{-x} =\ln 3 \implies x=-\ln (3)$ Here, the area can be expressed as: $A=\int_{-\ln 3}^{0} (3-e^{-x}) \ dx$ or, $=[3x+e^{-x}]_{-\ln (3)}^{0}$ or, $=[3(0)+e^{-0}]-[3(-\ln (3))+e^{-(-\ln (3))}]$ Therefore, the required area is: $Area=3 \ln (3)-2$