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

Apr 15, 2018

$1 - \frac{1}{4} \ln 5$

#### Explanation:

The average value of a function $f$ on the interval $a < = x < = b$ is given by the integral $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$.

So here, we wish to find:

$\frac{1}{4 - 0} {\int}_{0}^{4} \frac{x}{x + 1} \mathrm{dx}$

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

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

$= \frac{1}{4} {\int}_{1}^{5} \frac{u - 1}{u} \mathrm{du}$

Split up this integral:

$= \frac{1}{4} {\int}_{1}^{5} \left(1 - \frac{1}{u}\right) \mathrm{du}$

Both of these are common integrals:

$= \frac{1}{4} {\left[\left(u - \ln \left\mid u \right\mid\right)\right]}_{u = 1}^{u = 5}$

Evaluating:

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

Note that $\ln 1 = 0$:

$= \frac{1}{4} \left(4 - \ln 5\right)$

$= 1 - \frac{1}{4} \ln 5$

$\approx 2.39056$