How do you find the average value of the function for f ( x ) = <msqrt> x </msqrt

How do you find the average value of the function for $f\left(x\right)=\sqrt{x}+\frac{1}{\sqrt{x}},1\le x\le 9$?
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Zackery Harvey
Use the formula $\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx$ and the Fundamental Theorem of Calculus (FTC) to get an average value equal to $\frac{8}{3}$.
Explanation:
The average value is $\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx=\frac{1}{9-1}{\int }_{1}^{9}\left({x}^{\frac{1}{2}+{x}^{-\frac{1}{2}}}\right)dx$.
The FTC then leads us to say this equals
$\frac{1}{8}\left(\frac{2}{3}{x}^{\frac{3}{2}}+2{x}^{\frac{1}{2}}\right){|}_{1}^{9}=\frac{1}{8}\left(\left(\frac{2}{3}\ast 27+2\ast 3\right)-\left(\frac{2}{3}\ast 1+2\ast 1\right)\right)$
This simplifies to
$\frac{1}{8}\left(24-\frac{8}{3}\right)=\frac{1}{8}\left(\frac{64}{3}\right)=\frac{8}{3}$