Mean Value Theorem for Integrals
Theorem
If a function $f$ is continuous on the closed interval $[a,b]$, then there exists at least one $c$ in $(a,b)$ satisfying $\displaystyle f(c) = {{1}\over {b-a} } \int_{a}^{b} f(x) dx$.
Explanation
It is similar to the mean value theorem, but as the name suggests, it is used for integrals, hence the name. Its usage is also very similar, and its usefulness is by no means inferior to that of the mean value theorem. Meanwhile, considering that the mean value of a function is defined as the right-hand side, it might actually be more appropriate to call this one the mean value theorem, and to call the originally well-known mean value theorem the ‘mean value theorem for derivatives’.
Proof
Strategy: Since the continuity of $f$ is assumed, we use the extreme value theorem and the intermediate value theorem.
Since $f$ is continuous on $[a,b]$, by the extreme value theorem there exist a minimum value $m$ and a maximum value $M$, so
$$ \int_{a}^{b} m dx \le \int_{a}^{b} f(x) dx \le \int_{a}^{b} M dx $$
$$ \implies m \le {{1}\over {b-a} } \int_{a}^{b} f(x) dx \le M $$
Once again, since $f$ is continuous on $[a,b]$, by the intermediate value theorem, for $\displaystyle {{1}\over {b-a} } \int_{a}^{b} f(x) dx$ between $m$ and $M$, there exists at least one $c$ between $a$ and $b$ satisfying $f(c) = \displaystyle {{1}\over {b-a} } \int_{a}^{b} f(x) dx$.
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Meanwhile, in exactly the same way, it can be generalized with respect to a weight $w$ as follows. The form introduced above is the case $w(x) = 1$, where $\displaystyle \int_{a}^{b} dx = b - a$, so it is well covered by the theorem below.
Corollary
If a function $f$ is continuous on the closed interval $[a,b]$ and $w(x) \ge 0$ is integrable, then there exists at least one $\xi$ in $(a,b)$ satisfying $\displaystyle \int_{a}^{b} f(x) w(x) dx = f( \xi ) \int_{a}^{b} w(x) dx$.