# Notes - Analysis III TT23, Integral mean value theorems

> Source: https://ollybritton.com/notes/uni/prelims/tt23/analysis-iii/notes/integral-mean-value-theorems/ · Updated: 2023-05-03 · Tags: uni, notes

- [Course - Analysis III TT23](https://ollybritton.com/notes/uni/prelims/tt23/analysis-iii/)

### Flashcards
@State the most basic mean value theorem for integration.::
Suppose $f : [a, b] \to \mathbb R$ is a continuous function with $a \ne b$. Then $\exists c \in [a, b]$ such that
$$
f(c) = \frac{1}{(b-a)}\int_a^b f
$$

@State the “weighted” version of the mean value theorem for integration.::
Suppose $f : [a, b] \to \mathbb R$ is a continuous function and $w : [a, b] \to \mathbb R$ is a nonnegative continuous function. Then $\exists c \in [a, b]$ such that
$$
f(c) \int^b_a w = \int^b_a fw
$$

If $m$ and $M$ are the minimum and maximum value of some continuous function $f: [a, b] \to \mathbb R$, @Prove that $\exists c \in [a, b]$ such that
$$
f(c) = \frac{1}{(b-a)}\int^b_a f
$$
::
We have
$$
\begin{aligned}
m(b - a) \le \int^b_a f \le M(b -a) &\implies m \le \frac{1}{b-a} \int^b_a f \le M
\end{aligned}
$$
and so the result follows from the intermediate value theorem on $f$ since it attains every value in $[m, M]$.

### Proofs

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