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

1 Answer
Apr 15, 2018

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