# How do you find the average value of the function for f(x)=sqrtx+1/sqrtx, 1<=x<=9?

##### 1 Answer
Sep 14, 2017

Use the formula $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx}$ and the Fundamental Theorem of Calculus (FTC) to get an average value equal to $\frac{8}{3}$.

#### Explanation:

The average value is $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx} = \frac{1}{9 - 1} {\int}_{1}^{9} \left({x}^{\frac{1}{2}} + {x}^{- \frac{1}{2}}\right) \setminus \mathrm{dx}$.

The FTC then leads us to say this equals

$\frac{1}{8} \left(\frac{2}{3} {x}^{\frac{3}{2}} + 2 {x}^{\frac{1}{2}}\right) {|}_{1}^{9} = \frac{1}{8} \left(\left(\frac{2}{3} \cdot 27 + 2 \cdot 3\right) - \left(\frac{2}{3} \cdot 1 + 2 \cdot 1\right)\right)$.

This simplifies to

$\frac{1}{8} \left(24 - \frac{8}{3}\right) = \frac{1}{8} \left(\frac{64}{3}\right) = \frac{8}{3.}$