Question

What is the average value of the function `f(x)=e^(-x)*sin(x)` on the interval `[1, pi]`?

Answer 1

I get `1/(pi-1) int_1^pi e^-x * sinx dx = (e+e^pi(sin(1)+cos(1)))/(2(pi-1)e^(pi+1))`

Explanation:

From the definition of average value of a function, we need to find

`1/(pi-1) int_1^pi e^-x * sinx dx`.

Evaluate `int e^-x * sinx dx` by parts.

Use `u = sinx` and `dv = e^-x dx` (Or the other way around. In this case, either will work.)

We get `du = cosx dx` and `v = -e^-x`.

`uv-int v du = -e^-xsinx + int e^-xcosxdx`

Repeat parts with `u = cosx` and `dv = e^-x dx` for this integral to get

`int e^-x * sinx dx = -e^-xsinx - e^-x-inte^-xsinx dx`.

Add the integral to both sides and divide by `2` then factor, to finish with

`int e^-x sinx dx = -1/(2e^x)(sinx+cosx)`.

Evaluating from `1` to `pi`, we get:

`int_1^pi e^-x sinx dx = -1/(2e^x)(sinx+cosx)]_1^pi`

`= 1/(2e^pi) + (sin(1)+cos(1))/(2e)`

`= (e+e^pi (sin(1)+cos(1)))/(2e^(pi+1))`.

Finally, divide by the length of the interval, to answer:

`1/(pi-1) int_1^pi e^-x sinx dx = (e+e^pi(sin(1)+cos(1)))/(2(pi-1)e^(pi+1))`.