How do you find the average value of #f(x)=1/(x^2+1)# as x varies between #[-1,1]#?

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Answer 1 · 2016-11-07T00:27:29

Evaluate #1/(1-(-1)) int_-1^1 1/(x^2+1) dx = pi/4#

The average value of #f# on the interval #[a,b]# is

#1/(b-a) int_a^b f(x) dx#

We also need:

#int 1/(x^2+1) dx = tan^-1 x +C#

or, you may prefer the notation

#int 1/(x^2+1) dx = arctan x +C#.

The average value is

#1/(1-(-1)) int_-1^1 1/(x^2+1) dx =1/2 [tan^-1 x]_-1^1#

# = 1/2[pi/4 - (-pi/4)]#

# = pi/4#