Answer to Question #189077 in Calculus for Joshua

Question

The region bounded by y=3-e^-x, the x-axis, x=2 and the y-axis.

Answer 1

Solution.

$$y=3-e^{-x}, y=0, x=0, x=2.$$

$$S=\int_0^2 (3-e^{-x})dx=\newline =(3x+e^{-x})|_0^2=\newline =6+e^{-2}-e^0=5+\frac{1}{e^2}=\frac{5e^2+1}{e^2}.$$