# Find the average value of the cost function C(x)=3x√x+10 on the interval [0,81].?

May 4, 2018

$V a l u {e}_{a v g} \approx 959.1$

#### Explanation:

The average value of a function $f \left(x\right)$ is given by:

$V a l u {e}_{a v g} = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

In this case:

$V a l u {e}_{a v g} = \frac{1}{81 - 0} {\int}_{0}^{81} 3 x \sqrt{x + 10} \mathrm{dx}$

$= \frac{3}{81} {\int}_{0}^{81} x \sqrt{x + 10} \mathrm{dx}$

Let: $u = x + 10 , \mathrm{du} = \mathrm{dx} , x = u - 10$:

$= \frac{1}{27} {\int}_{0}^{81} \left(u - 10\right) \cdot {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{1}{27} \cdot \frac{2}{15} {u}^{\frac{3}{2}} \left(3 u - 50\right) \left[0 , 81\right]$

$= \frac{1}{27} \cdot \frac{2}{15} {\left(x + 10\right)}^{\frac{3}{2}} \left(3 x - 20\right) \left[0 , 81\right]$

$= \frac{2}{405} \left[\left(223\right) {\left(91\right)}^{\frac{3}{2}} - \left(- 20\right) {\left(10\right)}^{\frac{3}{2}}\right]$

$= \frac{2}{405} \left[20293 \sqrt{91} + 200 \sqrt{10}\right]$

$\approx 959.1$