Average Value of a Function

The average value of a function is the average height of the graph of a function.

The average value of a function f(x) over the interval [a,b] is given by the formula $$f _{ave}= \frac{1}{b-a}{\int}^b_a{f(x)dx}$$

Well, here is how to solve this problem:

We know that
$$f(x) = x^{2}$$ and $$a= 0, b = 2$$

So, $$f _{ave}= \frac{1}{2-(0)}{\int}^2_0{x^{2}dx}$$

=$$\frac{1}{2}\left.[\frac{x^3}{3}]\right|_0^2$$ = $$\frac{1}{2} [\frac{8}{3}-\frac{0}{3}]$$ = $$[\frac{1}{2}][\frac{8}{3}]$$ = $$[\frac{4}{3}]$$

The Mean Value Theorem for Integrals states that if $f$ is continuous on the closed interval $[a,b]$, then there exists a point $c$ in $[a,b]$ such that
$$f(c)= f_ {ave}= \frac{1}{b-a}{\int}^b_a{f(x)dx}$$

There exists some point $c$ where the instantaneous rate of change at that x value is the same as the average rate of change.

So, Find $c$ such that $f(c)= f _{ave}$ for the function $$f(x)=(x-3)^{2}$$ on the closed interval $[2,5]$

$$f _{ave} = \frac{1}{(5-2)}{\int}^5_2{(x-3)^{2}dx}$$ = $$\left.\frac{1}{9}(x-3)^{3} \right|_2^5$$ = $$1$$

$$f _{ave}= f(c)$$

$$1= (c-3)^{2}$$

$$(c-3-1)(c-3+1)= 0$$

The solutions are $$c= 4$$
or $$c =2$$

The average value of a function over a particular interval can be found with an expression involving an integral. Obviously you must know the algebraic function, $f(x)$. Also, the interval will be known. Let's say that interval is $[a, b]$. Then the average value of $f(x)$ over said interval is

$\frac{1}{b - a}{\int}^b_a{f(x)dx}$

Here is an example to demonstrate how this works.

What is the average value of the function $f(x) = 2\cos(2x)$ on the interval $[0,\pi]$?

In the above formula, $f(x) = 2\cos(2x)$, $a = 0$, and $b = \pi$, so the average value is

$$\begin {align*} \frac{1}{\pi - 0}{\int}^{\pi}_{0}{2\cos(2x)dx} &= \frac{1}{\pi}{[\sin(2x)]}^{\pi}_{0} \\&= \frac{1}{\pi}[\sin(2\pi) - \sin(0)] = \frac{1}{\pi}[0 - 0] \\&= 0. \end {align*}$$

Say we want to find the average value of the function $f(x) = 1 + x^2$ on the interval $[-1,2].$

The average value of a function, or the average height of a function, is the height of the rectangle that has area equivalent to the area under the curve of the function. It is defined generally by the equation

$\frac{1}{(b-a)}\int_a^{b}f(x)dx$

Now we replace $f(x)$ with our function from the above graph and get

$\frac{1}{(b-a)}\int_a^{b}(1+x^2)dx$

Then we switch the endpoints and get

$\frac{1}{(b-a)}\int_{-1}^{2}(1+x^2)dx$

And finally, in the equation before the integrand we replace $(b-a)$ with the given values of $b$ and $a$

$\frac{1}{(2-(-1))}\int_{-1}^{2}(1+x^2)dx$ = $\frac{1}{3}\int_{-1}^{2}(1+x^2)dx$

An integral is the area under a curve. A definite integral is the area under a curve from one point to another along the x-axis. There is a specific point on the y-axis where if you multiply the x-value and the y-value, you get the exact value of the area. Average Value of a function is how to solve for that y-value.

We know how to find the area under a curve through the anti-derivative method: $\int_0^3 x^2 dx$

We take the anti-derivative of $x^2$ and get $(x^3)/3$.

Then we plug $0$ and $3$ in for $x$ and subtract $0$ from $9$, which equals $9$. We can find the average value if we take our final answer (in this case, $9$) and multiply it by $1/(b-a)$. $b$ is the upper limit of integration and $a$ is the lower limit of integration. In this case, $a$ is $0$ and $b$ is $9$ so $1/3(9)=9/3=3$. The average value of this function is $3$.

$$\begin{align} \frac{1}{b-a}\int_a^b f(x) dx &= \frac{1}{2-0}\int_0^2 x^2 dx\\ &= \frac{1}{2}\int_0^2 x^2 dx\\ &=\frac{1}{2}\frac{x^3}{3}\bigg|_0^2\\ &=\frac{1}{2}\bigg(\frac{2^3}{3}-\frac{0^3}{3}\bigg)\\ &=\frac{1}{2}\cdot \frac{2^3}{3}\\ &=\frac{1}{2}\cdot\frac{8}{3}\\ &=\frac{4}{3}\\ \end{align}$$