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

Question

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

Christopher Allers · Accepted Answer

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.#

Axel Alvarez · Answer

undefined To find the average value of the function $ f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} $ over the interval $ 1 \leq x \leq 9 $, you need to compute the definite integral of the function over that interval and then divide by the length of the interval.

The definite integral of $ f(x) $ over the interval $ [1, 9] $ is calculated as follows:

$$ \int_{1}^{9} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \, dx $$

After evaluating this integral, divide the result by the length of the interval, which is $ 9 - 1 = 8 $.

$$ \text{Average value} = \frac{1}{8} \int_{1}^{9} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \, dx $$

Once you evaluate the integral, you will have the average value of the function over the interval $ [1, 9] $.

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

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