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# Average Value of a Function

The integral of a function over an integral [a, b] is the area formed by the curve of the function and rectangular axes. The average value of a function is a function f on the interval [a, b] is that y-value which specifies the height of the rectangle which has an area exactly equal to the define integral. Consider f is a function which is continuous on the closed interval [a, b]. The average value of the function from x = a to x = b is the integral

$\bar{f} = f_{avg} = \frac{1}{b-a} \int_{a}^{b}f(x)dx$

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## Definition

For n numbers, $y_{1}, y_{2}.., y_{n}$ the average value can be written as,

$y_{avg}$ = $\frac{y_{1} + y_{2} + .. + y_{n}}{n}$

For a function f(x), the average value of the function is a constant $f_{avg}$ whose integral over [a, b] gives same value as integral of f(x) over [a, b].

$\int_{a}^{b}f(x)dx$ = $\int_{a}^{b}f_{avg}dx$ = $f_{avg}\int_{a}^{b}dx$

$\Rightarrow$ $\int_{a}^{b}f(x)dx$ = $f_{avg}(b - a)$

Average value of a function: $f_{avg}$ = $\frac{1}{b-a}$ $\int_{a}^{b}f(x)dx$

## Proof of Average Value of a Function

If we break a continuous function f(x) over [a, b] into n equal sub-intervals with length x = $\frac{b - a}{n}$, then we can calculate the average of the function by:

$\frac{f(x_1) + f(x_2) + ....... + f(x_n)}{n}$

But by saying that n = $\frac{b - a}{x}$, we show that

$\frac{x}{b-a}$ $(f(x_1) + f(x_2)+ ..... + f(x_n))$

= $\frac{1}{b-a}$ $(f(x_1)x + f(x_2)x+ ..... + f(x_n)x)$

Hence, the average value of a continuous function f(x) over [a, b] is defined as:

$\bar{f}$ = $f_{avg}$ = $\frac{1}{b - a}$ $\int_{a}^{b}f(x)dx$

## Mean Value theorem of integration

The mean value theorem of integration states that if $f$ is a function continuous over an interval [a, b] then there exists at least one number c, a<c<b such that,

$f(c)(b-a)$ = $\int_{a}^{b}f(x)dx$

This implies that $f(c)$ = $f_{avg}$.

We can say that there exists a point in the interval [a, b] where the value of the function equals the average value of the function if the function is continuous over [a, b].

## Average Value of a Function Examples

Given below are some of the examples on average value of a function:

### Solved Examples

Question 1: Find the average value of f(x) = x4 on the interval [1, 6].
Solution:
The average value of the function is given by,
$\bar{f}= f_{avg}$ = $\frac{1}{b - a}$$\int_{a}^{b}f(x)dx Then, \bar{f} = f_{avg} = \frac{1}{6 - 1}$$\int_{1}^{6}x^{4}dx$
= $\frac{1}{5}[\frac{x^{5}}{5}]_{1}^{6}$

= $\frac{1}{25}$ $(7775)$

= 311.04

Question 2: Find the average value of f(x) = x2 on [1, 4]. Show that the average value is the rectangle height with base width equal to 4 - 1 = 3 and area equal to the definite integral of f over [1, 4].
Solution:

Favg = $\frac{1}{4 - 1}$$\int_{1}^{4}x^{2}dx = \frac{1}{4}$$\left ( x^{3} \right )_{1}^{4}$

= $\frac{1}{3} (\frac{4^3}{3} - \frac{1^3}{3})$

= 7

Question 3: Determine the average value of a function f(x) = 4x3 - 3x2 + 6 on [0,2].
Solution:

There’s really not a whole lot to do in this problem other than just use the formula.

$\bar{f} = f_{avg}$ = $\frac{1}{b - a}$$\int_{a}^{b}f(x)dx = \frac{1}{2 - 0}$$\int_{0}^{2}\left (4x^{3} - 3x^{2} + 6 \right ) dx$

= $\frac{1}{2}$.
$\left [ x^{4}- x^{3} + 6x \right ]_{0}^{2}$

=
$\frac{1}{2}$$(16 - 8 + 12) = 10 Question 4: If the average value of a function f(x) = x on [3, c] is 10. Then, find the value of c Solution: Since we know that the average value of the function is given by \bar{f} = f_{avg} = \frac{1}{b - a}\int_{a}^{b}f(x)dx Then we have, \bar{f} = f_{avg} = \frac{1}{c -3}$$\int_{3}^{c} x dx$ = 10(given)

Hence
$\frac{1}{c - 3}$$\int_{3}^{c}x dx = 10 \frac{1}{2(c - 3)}$$\left [ x^{2}\right ]_{3}^{c}$ = 10

$\frac{1}{2(c - 3)}$ ( c2 - 9 ) = 10

$\frac{1}{2(c - 3)}$ $(c - 3 )(c + 3)$ = 10

$\frac{1}{2}$ $(c + 3)$ = 10

c + 3 = 20

c = 17

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