Average value of a function Maybe I am just bugging out, but I need a sanity check

The question:


  If $\int_0^4 f(x)dx = C$, what is the average value of $f(x)$?


The question does not include a body for $f(x)$ or a table of values/graph for the output of f(x), so can this question be answered?

I know that:
$$\int_a^b f(x)\,dx = F(b)-F(a),$$
will evaluate the integral across a given interval, but the question is asking what the average value of f(x) is, not its integral.

If I take the derivative of the above integral, I can get the function itself, but what do I need to do about the interval $[0, 4]$? 

If I take the derivative of the integral, will I be left with $f(x) = 0$ (since the constant will disappear).

Am I thinking too hard about this?

Thanks for the help.
Remember that if $f(x)$ is continuous on $[a,b]$, then the average value of $f(x)$ on $[a,b]$ is defined to be
$$\text{average value of }f\text{ on }[a,b]  = \frac{1}{b-a}\int_a^b f(x)\,dx.$$

So if you need to know the average value of $f(x)$ on $[0,4]$ (as opposed to "the value of $f(x)$"), then just use that formula.
The $\int_0^4f(x).dx$ gives you the area, which is the same thing as the sum of all the values of f from 0 to 4,  multiplied by the dx (the horizontal distance between the values):

$\int_0^4f(x).dx$ = F(4)-F(0) = $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx$

divide this number by dx multiplied by the number of f's and you get the mean value.

$\frac{[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx}{(ndx)}$ where n is the number of f's (or points)

Remember, dx is just the (horizontal) length of the area you're integrating, divided by the number of points (as the number of points goes to infinity). So, ndx=n(horizontal length)/n is just the horizontal length of the area you're integrating over.

Simplifying the previous thing, you get $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx/(ndx) = [F(4)-F(0)]/\text{(horizontal length)}= [F(4)-F(0)]/(4-0) = C/4$