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

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Answer 1 · 2017-09-14T00:38:14

Use the formula $1/(b-a)int_{a}^{b}f(x)\ dx$ and the Fundamental Theorem of Calculus (FTC) to get an average value equal to $8/3$ .

The average value is $1/(b-a)int_{a}^{b}f(x)\ dx=1/(9-1)int_{1}^{9}(x^{1/2}+x^{-1/2})\ dx$ .

The FTC then leads us to say this equals

$1/8(2/3x^{3/2}+2x^{1/2})|_{1}^{9}=1/8((2/3*27+2*3)-(2/3*1+2*1))$ .

This simplifies to

$1/8(24-8/3)=1/8(64/3)=8/3.$

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