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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

 

Approved by eNotes Editorial Team
Soaring plane image

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial