Answer: Yes, Mean Value Theorem for Integrals applies for the function [tex]f(x)=3-x^{2}[/tex] and values of x are 1 and -1
Step-by-step explanation: Mean Value Theorem for Integrals states that if a function is continuous on a closed interval, there is a value c on the interval such that
[tex]f(c)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx[/tex]
where a and b are the closed interval given
As the graph of [tex]f(x)=3-x^{2}[/tex] shown below, the function is continuous on the interval [0,[tex]\sqrt{3}[/tex]], so the theorem applies.
To find the x-coordinates, first determine value of f(c) at [0,[tex]\sqrt{3}[/tex]]:
[tex]f(c)=\frac{1}{\sqrt{3} -0} \int\limits {3-\sqrt{x^{2}} } \, dx[/tex]
[tex]f(c)=\frac{1}{\sqrt{3}}[3x-\frac{x^{3}}{3} ][/tex]
[tex]f(c)=\frac{1}{\sqrt{3}}[ 3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3} ][/tex]
[tex]f(c)=\frac{1}{\sqrt{3} } [2\sqrt{3} ][/tex]
f(c) = 2
The x-coordinates will be:
[tex]f(x)=3-x^{2}[/tex]
[tex]2=3-x^{2}[/tex]
[tex]x^{2}=1[/tex]
[tex]x=\sqrt{1}[/tex]
x = ±1
The values for x guaranteed by the Mean Value Theorem are +1 and -1.
