Q. 82 - Calculus

Question

Prove the Mean Value Theorem for Integrals by following these steps:

(a) Use the Extreme Value Theorem to argue that f has a maximum value M and a minimum value m on the interval [a, b].

(b) Use an upper sum and a lower sum with one rectangle to argue that $m(b-a) ⩽ {\int }_a^{b}f\left(x\right)dx ⩽ M(b-a)$ and thus that the average value of f on [a, b] is between m and M.

Step-by-Step Solution

Answer

Hence, proved.

Step 1. Given Information.

A function f(x) in the interval [a, b] has maximum value M and minimum value m.

Step 2. part (a) Extreme value theorem.

$$\begin{gathered} The extreme value theorem states that if a real valued function f is continuous in \\ the closed and bounded interval [a,b], then f must attain its maximum and minimum, \\ each at least once. \\ Therefore from this theorem f\left(x\right) = M which is its maximum value \\ and f\left(x\right) =m which is its minimum value in the interval [a,b] at least once. \end{gathered}$$

Step 3. part (b) Proof.

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx = \underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)∆x, \\ So, \\ \sum _{k=1}^{n}f\left(x_k\right)∆x ⩽ \sum _{k=1}^{n}M∆x, \\ \sum _{k=1}^{n}f\left(x_k\right)∆x ⩽ M\left(\sum _{k=1}^n∆x\right), \\ Since, \sum _{k=1}^n∆x = b-a, we get, \\ \sum _{k=1}^{n}f\left(x_k\right)∆x ⩽ M(b-a), \\ Taking limit we get, \\ {\int }_a^{b}f\left(x\right)dx ⩽ M(b-a). \\ Similarly it can be shown that, \\ {\int }_a^{b}f\left(x\right)dx ⩾ m(b-a). \\ Therefore, it is proved that, \\ m(b-a)⩽{\int }_a^{b}f\left(x\right)dx ⩽ M(b-a). \\ Hence, proved. \end{gathered}$$