# Mean value theorem for definite integrals/Riemann/Section

For a Riemann-integrable function ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} }$, one may consider

${\displaystyle {\frac {\int _{a}^{b}f(t)\,dt}{b-a}}}$

as the mean height of the function, since this value, multiplied with the length ${\displaystyle {}b-a}$ of the interval, yields the area below the graph of ${\displaystyle {}f}$. The Mean value theorem for definite integrals claims that, for a continuous function, this mean value is in fact obtained by the function somewhere.

## Theorem

Suppose that ${\displaystyle {}[a,b]}$ is a compact interval, and let

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

be a continuous function. Then there exists some ${\displaystyle {}c\in [a,b]}$ such that

${\displaystyle {}\int _{a}^{b}f(t)\,dt=f(c)(b-a)\,.}$

### Proof

On the compact interval, the function ${\displaystyle {}f}$ is bounded from above and from below, let ${\displaystyle {}m}$ and ${\displaystyle {}M}$ denote the minimum and the maximum of the function. Due to fact, they are both obtained. Then, in particular, ${\displaystyle {}m\leq f(x)\leq M}$ for all ${\displaystyle {}x\in [a,b]}$, and so

${\displaystyle {}m(b-a)\leq \int _{a}^{b}f(t)\,dt\leq M(b-a)\,.}$

Therefore, ${\displaystyle {}\int _{a}^{b}f(t)\,dt=d(b-a)}$ with some ${\displaystyle {}d\in [m,M]}$. Due to the Intermediate value theorem, there exists a ${\displaystyle {}c\in [a,b]}$ such that ${\displaystyle {}f(c)=d}$.

${\displaystyle \Box }$