Notes - Analysis III TT23, Integral mean value theorems
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 maximum and minimum 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 $[m, M]$.