Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 5 − x2 on the interval the closed interval from 0 to the square root of 5. If so, find the x-coordinates of the point(s) guaranteed by the theorem. No, the Mean Value Theorem for Integrals does not apply Yes, x equals 10 thirds Yes, x equals the quotient of square root of 5 and 3 Yes, x equals the quotient of the square root of 15 and 3

The mean value theorem for integrals applies as long as the function is continuous and the interval is closed.

The theorem states that there is a y value that is the average of the integral over the interval.

The average of the integral is: [tex] \frac{1}{ \sqrt{5}} \int\limits^ {\sqrt{5}}_0 {5-x^{2}} \, dx = \frac{10}{3}[/tex]

So, there is a point f(x) = 10/3.

To find x, we substitute that into the original function:

[tex]\frac{10}{3} = 5 - x^{2} [/tex]

[tex]x^{2} = \frac{5}{3}[/tex]

[tex]x = \sqrt{\frac{5}{3}} [/tex]

The answer is C. "Yes, x equals the quotient of square root of 5 and 3"

Edit:

The average of an integral of f(x) from a to b is: 

[tex]\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx [/tex]