Area — Question 8

Question 8

Find the exact area of the region enclosed by the curves $$y=e^{-x} \quad\text{and}\quad y=1+\left(\frac{1}{e}-1\right)x.$$

Question 8 - Solution

First find the intersection points: $$e^{-x}=1+\left(\frac{1}{e}-1\right)x.$$ At $x=0$, $$e^{0}=1=1+\left(\frac{1}{e}-1\right)\cdot 0.$$ At $x=1$, $$e^{-1}=\frac{1}{e}=1+\left(\frac{1}{e}-1\right)\cdot 1.$$ So the curves intersect at $x=0$ and $x=1$.

Since $e^{-x}$ is convex, the chord (the line) lies above $e^{-x}$ on $(0,1)$, so the area is $$A=\int_{0}^{1}\left(1+\left(\frac{1}{e}-1\right)x-e^{-x}\right)\,dx.$$

Compute each piece: $$\int_{0}^{1}\left(1+\left(\frac{1}{e}-1\right)x\right)\,dx =\left[x+\left(\frac{1}{e}-1\right)\frac{x^{2}}{2}\right]_{0}^{1} =1+\frac{1}{2}\left(\frac{1}{e}-1\right) =\frac{1}{2}+\frac{1}{2e},$$ $$\int_{0}^{1}e^{-x}\,dx=\left[-e^{-x}\right]_{0}^{1}=1-\frac{1}{e}.$$

Thus $$A=\left(\frac{1}{2}+\frac{1}{2e}\right)-\left(1-\frac{1}{e}\right) =-\frac{1}{2}+\frac{3}{2e} =\frac{1}{2}\left(\frac{3}{e}-1\right).$$

$$\boxed{\frac{1}{2}\left(\frac{3}{e}-1\right)}$$