Mean value theorem for definite integrals/Riemann/Section

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

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

as the mean height of the function, since this value, multiplied with the length ${}b-a$ of the interval, yields the area below the graph of ${}f$. 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 ${}[a,b]$ is a compact interval, and let

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

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

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

Proof

On the compact interval, the function ${}f$ is bounded from above and from below, let ${}m$ and ${}M$ denote the minimum and the maximum of the function. Due to

they are both obtained. Then, in particular, ${}m\leq f(x)\leq M$ for all ${}x\in [a,b]$, and so

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

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

$\Box$