Proving an Integral Theorem Theorem: Let $f$ be a continuous function on $[a,b] \in \Bbb R$. Prove that there exists a $c \in [a,b]$ so that $$ f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx  $$
I cannot use derivatives or any other integral theorems, I am doing these proofs using Riemann Sums and continuity. 
My Thoughts:
I am planning on approaching this problem using the following theorem:

Theorem: Let $f$ be a continuous function on the interval $[a,b]$. Then $$\left| \int_a^bf(x)\,dx \right| \  \le \  (b-a)\sup\limits_{[a,b]}|f(x)|  $$

Or I could show that this $f(c)$ exists, but then I am unsure as to how I would prove that $c \in [a,b]$. Hints would be appreciated. 
Edit: So I have realized that I have the following inequality for $m := \inf\limits_{[a,b]}f(x)(b-a)$, $M := (b-a)\sup\limits_{[a,b]}|f(x)|$:
$$ m \le \frac{1}{b-a}\left| \int_a^bf(x)\,dx\right| \le M $$
So now how do I relate this to $f(c)$?

Edit 2: I changed the title as upon considering the existing MVT, I don't think this is the same thing.
 A: Let $I=\frac1{b-a}\int\limits_a^bf$. Assume that the conclusion does not hold, then either $f(x)\lt I$ for every $x$ in $[a,b]$, or $f(x)\gt I$ for every $x$ in $[a,b]$ (otherwise, $f(x)\lt I\lt f(y)$ for some $x$ and $y$ in $[a,b]$ and, by the intermediate value theorem, $f(z)=I$ for some $z$ between $x$ and $y$). 
Without loss of generality, the first case happens. Then $\int\limits_a^bf(x)\,\mathrm dx\lt\int\limits_a^bI\,\mathrm dx=(b-a)I=\int\limits_a^bf$, this is absurd.
A: Define $F(y)=\int_{a}^{y}f(x)dx$ use the theorem fundamental calculus and mean value theorem
A: If you have the extreme value theorem that says that a continuous function on a closed bounded interval actually reaches its sup and its inf, then you have $f(x_0)=M$ and $f(x_1)=m$ for some $x_0,x_1\in[a,b]$.  Then the intermediate value theorem implies $f(x_2)=\text{the desired value}$, since you say you've already shown that the desired value is intermediate.
A: Let's consider $M=\max{f(x)}$, $m=\min{f(x)}$ and then we have that
$$(b-a) m\leq f(c) ({b-a})= \int_a^b f(x)\,dx \leq (b-a) M$$
If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c. Hence, we are done.
Q.E.D. 
