The average value of the function f(x) on the interval [a,b] is

`1/(b-a) int_a^b f(x) dx`

Thus, the average value for your question is:

`1/(2+1) int_(-1)^2 x^2 dx`

`=1/3 [1/3 x^3 ]|_(-1)^2`

`=1/9 (8- (-1))`

=1

now, if f(x) is continuous on [a,b], then there is a point c in [a.b] where the average of f(x) equals f(c)

here, f(x) is a polynomial, which is continuous. So, there is some c in [-1,2] at which f(c) =1

`f(c)=c^2 = 1`

`c= +- 1`

The gist is:

The area under the curve f(x) on [-1,2] is the same as the area of the rectangle that has width 2+1 and height f(c):

The area under the parabola is the same as the area under the straight line.

This is different from the mean value theorem.

The mean value theorem states:

if f(x) is continuous on [a,b], and differentiable on (a,b) , then there is some c in (a,b) where

`f'(c) = (f(b)-f(a))/(b-a)`

The gist is this:

If you drew a line through the points (a,f(a)) and (b,f(b)), then somewhere between (a,b) there is a place where the tangent line has the same slope as the line you drew

So, for us, f(a) = f(-1) = 1, f(b) = f(2) = 4

`(f(b)-f(a))/(b-a) = (3-1)/(2+1) = 3/3=1`

`f'(c) = 2c`

2c = 1

c = 1/2

That is, at c=1/2, f(c) has the same tangent slope as the slope of the line connecting (-1,1) to (2,4):