Question

How do you find the average value of the function for `f(x)=x/(x+1), 0<=x<=4`?

Answer 1

`1-1/4ln5`

Explanation:

The average value of a function `f` on the interval `alt=xlt=b` is given by the integral `1/(b-a)int_a^bf(x)dx`.

So here, we wish to find:

`1/(4-0)int_0^4x/(x+1)dx`

There are a lot of ways to solve this integral. I would try the substitution `u=x+1`. This substitution implies that `du=dx` and when we examine the bounds of integration, notice that `x=0=>u=1` and `x=4=>u=5`.

We also should realize that `x=u-1`. We can then transform the integral as follows:

`=1/4int_1^5(u-1)/udu`

Split up this integral:

`=1/4int_1^5(1-1/u)du`

Both of these are common integrals:

`=1/4[(u-lnabsu)]_(u=1)^(u=5)`

Evaluating:

`=1/4[(5-lnabs5)-(1-lnabs(1))`

Note that `ln1=0`:

`=1/4(4-ln5)`

`=1-1/4ln5`

`approx2.39056`