Question

Find the average value of the function on the given interval.
f(x)=x^2e^2x; [0,2]

Answer 1

`f_(avg)=(1)/(8)(5e^4-1)`

Explanation:

We are given that a function

`f(x)=x^2e^(2x)`

We have to find the average value of function on the given interval [0,2]

Average value of function on interval [a,b] is given by

`(1)/(b-a)\int_(a)^(b)f(x)dx`

Using the formula

`f_(avg)=(1)/(2-0)\int_(0)^(2)x^2e^(2x) dx`

By Parts integration formula

`\int(uv)dx=u\int vdx-\int((du)/(dx)\int vdx)dx`

u=
`x^2` and v=
`e^(2x)`

Apply by parts integration

`f_(avg)=(1)/(2-0)([(x^2e^(2x))/(2)]^(2)_(0)-\int_(0)^(2)(2x* (e^(2x))/(2)dx)`

`f_(avg)=(1)/(2)(2e^4-(\int_(0)^(2)xe^(2x))dx`

`f_(avg)=(1)/(2)(2e^4-0-([(xe^(2x))/(2)]^(2)-(1)/(4)[e^(2x)]^(2)_(0)))`

`f_(avg)=(1)/(2)(2e^4-e^4+(1)/(4)(e^4-1))=(1)/(2)(e^4+(1)/(4)e^4-(1)/(4))`

`f_(avg)=(1)/(8)(5e^4-1)`