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Before considering the Mean Value Theorem for Integrals, let us observe that if $f(x) \ge g(x)$ on $[a,b]$, then
$$\displaystyle{\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx}$$This is known as the Comparison Property of Integrals and should be intuitively reasonable for non-negative functions $f$ and $g$, at least. Note that the definite integrals above give the area under these functions, respectively -- and if $f(x) \ge g(x)$ on $[a,b]$, then the area under $f$ on this interval should be greater than or equal to the area under $g$ on the same interval.
Now, suppose $f$ is continuous on $[a,b]$. By the Extreme Value Theorem, we know that $f$ attains its absolute maximum and minimum values on this interval. Suppose $M$ denotes the absolute maximum value and $m$ denotes the absolute minimum value. Further, let $x_{max}$ and $x_{min}$ be those values in $[a,b]$ where $f(x_{max}) = M$ and $f(x_{min})$, respectively.
The Comparison Property of Integrals then tells us that
$$\int_a^b m\,dx \le \int_a^b f(x)\,dx \le \int_a^b M\,dx$$Of course, the definite integrals on the left and right can be easily evaluated, yielding
$$m(b-a) \le \int_a^b f(x)\,dx \le M(b-a)$$which is equivalent to
$$m \le \frac{1}{b-a} \int_a^b f(x)\,dx \le M$$Thus $\frac{1}{b-a} \int_a^b f(x)\,dx$ is a value between $f(x_{min})$ and $f(x_{max})$. Since $x_{min}$ and $x_{max}$ are contained in $[a,b]$ and $f$ is continuous on $[a,b]$, it follows that $f$ is continuous on $[x_{min},x_{max}]$.
But then the Intermediate Value Theorem applies! There must consequently be some $c$ in $(x_{min},x_{max})$ where
$$f(c) = \frac{1}{b-a} \int_a^b f(x)\,dx$$Putting this all together, we have the following important result:
The Mean Value Theorem for Integrals
If $f$ is continuous on $[a,b]$, then there exists some $c$ in $[a,b]$ where $\displaystyle{f(c) = f_{avg} = \frac{1}{b-a} \int_a^b f(x)\,dx}$