`f(x) = (2x)/(1 + x^2)^2, [0, 2]` Find the average value of the function on the given interval.

Given `f(x)=(2x)/(1+x^2)^2, [0, 2]`

Average Value Formula=`1/(b-a)int_a^bf(x)dx`

`f_(ave)=1/(2-0)int_0^2(2x)/(1+x^2)^2dx`

`=(1/2)*2int_0^2x/(1+x^2)^2dx`

`=int_0^2x/(1+x^2)^2dx`

Integrate using the u-subsitution method.

Let `u=1+x^2`

`(du)/dx=2x`

`dx=(du)/(2x)`

`=int_0^2x/u^2*(du)/(2x)`

`=(1/2)int_0^2(1/u^2)du`

`=(1/2)int_0^2(u^-2)du`

`=(1/2)(u^-1)/(-1)|_0^2`

`=(-1/2)[1/u]_0^2`

`=(-1/2)[1/(1+x^2)]_0^2`

`=(-1/2)[1/(1+(2)^2)-1/(1-0^2)]`

`=(-1/2)[1/5-1]`

`=(-1/2)(-4/5)`

`=2/5`

The average value is 2/5.

Given `f(x)=(2x)/(1+x^2)^2, [0, 2]`

Average Value Formula=`1/(b-a)int_a^bf(x)dx`

`f_(ave)=1/(2-0)int_0^2(2x)/(1+x^2)^2dx`

`=(1/2)*2int_0^2x/(1+x^2)^2dx`

`=int_0^2x/(1+x^2)^2dx`

Integrate using the u-subsitution method.

Let `u=1+x^2`

`(du)/dx=2x`

`dx=(du)/(2x)`

`=int_0^2x/u^2*(du)/(2x)`

`=(1/2)int_0^2(1/u^2)du`

`=(1/2)int_0^2(u^-2)du`

`=(1/2)(u^-1)/(-1)|_0^2`

`=(-1/2)[1/u]_0^2`

`=(-1/2)[1/(1+x^2)]_0^2`

`=(-1/2)[1/(1+(2)^2)-1/(1-0^2)]`

`=(-1/2)[1/5-1]`

`=(-1/2)(-4/5)`

`=2/5`

The average value is 2/5.