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

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

Related Calculators | |

Average Calculator | Average Acceleration Calculator |

Average Atomic Mass Calculator | Average Deviation Calculator |

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].

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

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

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:

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

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].

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

The average value of the function is given by,

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

$\bar{f}= f_{avg}$ = $\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

F_{avg} = $\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

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

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

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

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

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

c + 3 = 20

c = 17

Related Topics | |

Math Help Online | Online Math Tutor |