Average Function Value — Question 10

Question 10 - Solution

The average value of a function on $[a,b]$ is $$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$

Here, $$f_{\text{avg}}=\frac{1}{2}\int_{0}^{2} x e^{-x}\,dx.$$

Evaluate the integral using integration by parts.

Let $$u=x, \qquad dv=e^{-x}\,dx.$$ Then $$du=dx, \qquad v=-e^{-x}.$$

Apply integration by parts: $$\int x e^{-x}\,dx = -xe^{-x}+\int e^{-x}\,dx = -xe^{-x}-e^{-x}.$$

Thus, $$\int x e^{-x}\,dx = -(x+1)e^{-x}.$$

Evaluate from $0$ to $2$: $$\left[-(x+1)e^{-x}\right]_{0}^{2} = -3e^{-2}-(-1) = 1-3e^{-2}.$$

Now compute the average value: $$f_{\text{avg}}=\frac{1}{2}\left(1-3e^{-2}\right).$$

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