Prove that $\int_{a}^{b}f(x)dx = (b-a)f(c)$ Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Prove that for all $a,b \in \mathbb{R}$ there is a $c$ between $a \mbox{ and } b$ such that $\int_{a}^{b}f(x)dx = (b-a)f(c)$
Is this the mean value theorem for definite integrals? I am not sure as to how to prove it.  
 A: Yes it is the MVT. You just write

$${1\over b-a}\int_a^bf(x)\,dx = {F(b)-F(a)\over b-a}$$

using the FTC, where $F$ is an anti-derivative for $f$. but then by the regular MVT, this gives
$${F(b)-F(a)\over b-a}=F'(c) = f(c)$$
clearing denominators gives your result.
A: It is simpler to treat this as a theorem of integral calculus and here is a proof which does not use Mean Value Theorem. Since $f$ is continuous on $[a, b]$ there is a minimum value $m$ of $f$ and a maximum value $M$ of $f$ on $[a, b]$. And by intermediate value theorem $f$ takes every values between $m$ and $M$ at least once in interval $[a, b]$.
We have $$m \leq f(x) \leq M$$ for all $x \in [a, b]$ and hence on integrating the above equation we get $$m(b - a) \leq \int_{a}^{b}f(x)\,dx \leq M(b - a)$$ so that $$m \leq \frac{1}{b - a}\int_{a}^{b}f(x)\,dx \leq M$$ and hence $f$ takes this value $$k = \frac{1}{b - a}\int_{a}^{b}f(x)\,dx$$ at least once in interval $[a, b]$.
Therefore we have a number $c \in [a, b]$ such that $$f(c) = \frac{1}{b - a}\int_{a}^{b}f(x)\,dx$$ or $$\int_{a}^{b}f(x)\,dx = (b - a)f(c)$$
BTW, the result in question is also called "First Mean Value Theorem for Integrals".
