# How do you find the average value of the function for f(x)=sqrt(2x-1), 1<=x<=5?

##### 1 Answer
Oct 31, 2017

The average value is $= \frac{1}{2}$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$

The average value is

$\overline{y} = \frac{1}{5 - 1} {\int}_{1}^{5} \left(\frac{1}{\sqrt{2 x - 1}}\right) \mathrm{dx}$

Perform first the integral by substitution

Let $u = 2 x - 1$, $\implies$, $\mathrm{du} = 2 \mathrm{dx}$

Therefore,

$\int \left(\frac{1}{\sqrt{2 x - 1}}\right) \mathrm{dx} = \frac{1}{2} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{2} \cdot 2 \sqrt{u} = \sqrt{u} = \sqrt{2 x - 1}$

So,

$\overline{y} = \frac{1}{4} {\left[\sqrt{2 x - 1}\right]}_{1}^{5} = \frac{1}{4} \left(\left(\sqrt{9}\right) - \left(\sqrt{1}\right)\right) = \frac{1}{4} \cdot 2 = \frac{1}{2}$