# 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 â€˜câ€™ 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,

${f\left( c\right) =\dfrac{1}{b-a}\int ^{b}_{a}f\left( x\right) dx}$

or,

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

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 â€˜mâ€™ and â€˜Mâ€™ be the minimum and the maximum values on [a, b]

Thus, âˆ€ x Ð„ [a, b], m â‰¤ f(x) â‰¤ M

Now, by the comparison theorem, we get

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

On dividing by (b – a), we get

${m\leq \dfrac{1}{\left( b-a\right) }\int ^{b}_{a}f\left( x\right) dx\leq M}$

Since f(x) is continuous on [a, b] and ${m\leq \dfrac{1}{\left( b-a\right) }\int ^{b}_{a}f\left( x\right) dx\leq M}$, for m, M Ð„ [a, b], then by the intermediate value theorem, a number â€˜câ€™ exists on [a, b] such that

${f\left( c\right) =\dfrac{1}{b-a}\int ^{b}_{a}f\left( x\right) dx}$

â‡’ ${\int ^{b}_{a}f\left( x\right) dx=f\left( c\right) \left( b-a\right)}$

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 ${f_{avg}=\dfrac{1}{b-a}\int ^{b}_{a}f\left( x\right) dx}$

## Solved Examples

Find the point â€˜câ€™ to satisfy the mean value theorem for the integrals of a function f(x) = 12x – 8 on [2, 6].

Solution:

Since f(x) = 12x – 8 is a polynomial, it is continuous on [2, 6]
By the mean value theorem for integrals, we get
${\int ^{b}_{a}f\left( x\right) dx=f\left( c\right) \left( b-a\right)}$
â‡’ ${\int ^{6}_{2}\left( 12x-8\right) dx=\left( 12c-8\right) \left( 6-2\right)}$
â‡’ ${12\int ^{6}_{2}xdx-8\int ^{6}_{2}dx=48c-32}$
â‡’ ${12\left[ \dfrac{6^{2}}{2}-\dfrac{2^{2}}{2}\right] -8\left[ 6-2\right] =48c-32}$
â‡’ ${6\times 32-8\times 4=48c-32}$
â‡’ 48c = 192
â‡’ c = 4 Ð„ [2, 6]
Thus, the only solution of â€˜câ€™ is 4.

Find the average value of f(x) = 3x2 – 5x on [0, 4] and find â€˜câ€™ such that f(c) = favg on [0, 4].

Solution:

Since f(x) = 3x2 – 5x is a polynomial, it is continuous on [0, 4]
By the mean value theorem for integrals, we get
${f\left( c\right) =\dfrac{1}{b-a}\int ^{b}_{a}f\left( x\right) dx}$
â‡’ ${f\left( c\right) =\dfrac{1}{4-0}\int ^{4}_{0}\left( 3x^{2}-5x\right) dx}$
â‡’ ${f\left( c\right) =\dfrac{1}{4}\int ^{4}_{0}3x^{2}dx-\dfrac{5}{4}\int ^{4}_{0}xdx}$
â‡’ ${f\left( c\right) =\dfrac{1}{4}\left[ \left( 4\right) ^{3}-\left( 0\right) ^{3}\right] -\dfrac{5}{4}\left[ \dfrac{\left( 4\right) ^{2}}{2}-\dfrac{\left( 0\right) ^{2}}{2}\right]}$
â‡’ ${f\left( c\right) =\dfrac{1}{4}\times 64-\dfrac{5}{4}\times \dfrac{16}{2}}$
â‡’ ${f\left( c\right) =16-10}$
â‡’ ${f\left( c\right) =6}$, which is the average value of f(x) on [0, 4].
Now, f(c) = 3c2 – 5c = 6
â‡’ 3c2 – 5c – 6 = 0
On applying the quadratic formula, we get
either c = ${\dfrac{5+\sqrt{97}}{6}}$ or c = ${\dfrac{5-\sqrt{97}}{6}}$
Approximating the value of ${\sqrt{97}}$ up to 3 decimal places, we get c = 2.475 or -0.808, respectively.
Since c = 2.475 Ð„ [0, 4], the value of c is 2.475
Thus, the average value of f(x) = f(c) = 6, and the value of â€˜câ€™ is approximately 2.475 on [0, 4].