Mean Value Theorem for Integrals

The mean value theorem for integrals states that if a function f(x) is continuous on a closed interval [a, b], there exists a point ‘c’ on [a, b] such that f(x) at c equals the average value of f(x) on the given interval.

Mathematically, it is generalized as,

$f\left(c\right)={\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(x\right)dx$

or, 

${\int }_a^{b}f\left(x\right)dx=f\left(c\right)(b-a)$

In the above graph, the shaded rectangular part of the curve is the average value of the function f(x).

Proof

Since f(x) is continuous on [a, b]

By the extreme value theorem, f(x) attains its minimum and maximum values. 

Let ‘m’ and ‘M’ be the minimum and the maximum values on [a, b]

Thus, ∀ x Є [a, b], m ≤ f(x) ≤ M

Now, by the comparison theorem, we get

$m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le M(b-a)$

On dividing by (b – a), we get

$m\le {\displaystyle \frac{1}{(b-a)}}{\int }_a^{b}f\left(x\right)dx\le M$

Since f(x) is continuous on [a, b] and $m\le {\displaystyle \frac{1}{(b-a)}}{\int }_a^{b}f\left(x\right)dx\le M$, for m, M Є [a, b], then by the intermediate value theorem, a number ‘c’ exists on [a, b] such that

$f\left(c\right)={\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(x\right)dx$

⇒ ${\int }_a^{b}f\left(x\right)dx=f\left(c\right)(b-a)$

Hence, the mean value theorem for integrals is proved.

This relationship can also be represented as:

If f(x) is integrable on [a, b], the average value of f(x) on [a, b] is $f_{avg}={\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(x\right)dx$

Solved Examples