Why is $f(c)$ in the Mean-value theorem for integrals the average of $f$ on $[a,b]$? The mean-value theorem for integrals is:
$$\int_a^bf(x)dx=(b-a)f(c)$$
My calculus-book merely "observes" that $f(c)$ is also the average value of $f$ on $[a,b]$:
$$\overline{f}=f(c)=\frac{1}{b-a}\int_a^bf(x)dx$$
But why is this so? How can one prove this?
 A: From the definition of the integral as limit of Riemann sums it becomes natural to call
$${1\over b-a}\int_a^b f(x)\>dx$$
the average of $f$ on $[a,b]$. The MVT then says that there is a point $c\in[a,b]$ such that $f(c)$ is exactly equal to this average value.
A: Hint:
Suppose $f(x)$ is continues on $[a,b]$ so it has maximum and minimum 
assume $max f(x)=M ,min f(x)=n$ so 
$$\forall x \in [a,b] :n \leq f(x)\leq M$$ so avegrage of $f(x)$ is between max,min 
$$average=\overline {f} \to \\n \leq \overline {f}\leq M\\$$
$f(x) $ is continues ,so $F(x)=\int_a^x f(t)dt$ is continues too . So $ n(b-a)\leq F(x) \leq M(b-a)$ now $F(x)$ now use intermediate point theorem for  $F(x)$ so ,must be $\exists c \in (a,b) \to F'(c)=\frac{F(b)-F(a)}{b-a}\to f(c)=\frac{F(b)-F(a)}{b-a}=\frac{\int_a^b f(x)dx-0}{b-a}=\overline {f}$   
A: If $f$ is Riemann-integrable, you know that the integral is a limit of Riemann sums, so that for example when $n$ is large
$$
\int_a^b f(t) d t \approx \Delta x \big(f(a) + f(a + \Delta x) + f(a + 2 \Delta x)+ \cdots+ f(b-\Delta x)\big)
$$
with $\Delta x = \frac{b-a}{n}$. You see that
$$
\frac{1}{b-a}\int_a^b f(t) d t \approx \frac{1}{n} \big(f(a) + f(a + \Delta x) + f(a + 2 \Delta x)+ \cdots+ f(b-\Delta x)\big)
$$
is actually close to the average of $n$ values of $f$ at regularly separated points.
