# What is the average value of a function  y = x^2 (x^3 + 1)^(1/2)  on the interval [0, 2]?

Jun 2, 2016

$\frac{26}{9}$

#### Explanation:

The average value of the differentiable function $f \left(x\right)$ on the interval $\left[a , b\right]$ can be found through

$\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

For $f \left(x\right) = {x}^{2} {\left({x}^{3} + 1\right)}^{\frac{1}{2}}$ on $\left[0 , 2\right]$, this yields an average value of

$\frac{1}{2 - 0} {\int}_{0}^{2} {x}^{2} {\left({x}^{3} + 1\right)}^{\frac{1}{2}} \mathrm{dx}$

Use substitution: $u = {x}^{3} + 1$ and $\mathrm{du} = 3 {x}^{2} \mathrm{dx}$.

$= \frac{1}{6} {\int}_{0}^{2} 3 {x}^{2} {\left({x}^{3} + 1\right)}^{\frac{1}{2}} \mathrm{dx}$

Note that the bounds will change--plug the current bounds into $u = {x}^{3} + 1$.

$= \frac{1}{6} {\int}_{1}^{9} {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{1}{6} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{1}^{9} = \frac{1}{6} \left[\frac{2}{3} {\left(9\right)}^{\frac{3}{2}} - \frac{2}{3} {\left(1\right)}^{\frac{3}{2}}\right] = \frac{1}{6} \left[\frac{2}{3} \left(27\right) - \frac{2}{3} \left(1\right)\right]$

$= \frac{1}{6} \left[\frac{54}{3} - \frac{2}{3}\right] = \frac{26}{9}$