# Average Value of a Function

It is easy to compute average value of finitely many values y_1, y_2,...,y_n: y_(ave)=(y_1+y_2+...+y_n)/n.

But how to compute average of infinitely many values? In general, let's try to compute the average value of a function y=f(x), a<=x<=b.

As always we start by dividing the interval [a,b] into n equal subintervals, each with length Delta x=(b-a)/n. Then we choose points x_1^(**), . . . , x_n^(**) in successive subintervals and calculate the average of the numbers f(x_1^(**)), . . . , f(x_n^(**)): (f(x_1^(**))+f(x_2^(**))+...+f(x_n^(**)))/n.

Since Delta x=(b-a)/n then n=(b-a)/(Delta x) and the average value becomes (f(x_1^(**))+f(x_2^(**))+...+f(x_n^(**)))/((b-a)/(Delta x))=

=1/(b-a)(f(x_1^(**))+f(x_2^(**))+...+f(x_n^(**)))Delta x=

=1/(b-a)sum_(i=1)^nf(x_i^(**))Delta x .

If we let n increase, we would be computing the average value of a large number of closely spaced values. The limiting value is lim_(n->oo) 1/(b-a)sum_(i=1)^n f(x_i^(**))Delta x=1/(b-a)int_a^b f(x)dx by the definition of definite integral.

Average Value of a Function f on the interval [a,b] is f_(ave)=1/(b-a)int_a^b f(x)dx.

For a positive function, we can think of this definition as saying text(area)/text(width)=text(average height).

Example 1. Find the average value of function f(x)=4-x^2 on interval [0,2].

Here, a=0 and b=2, therefore f_(ave)=1/(2-0)int_0^2(4-x^2)dx=1/2 (4x-1/3x^3)|_0^2=1/2 (4*2-1/3*2^3)=8/3.

Now, recall that Mean Value Theorem for Integrals states that there exists such number c from [a,b] that f(c)=1/(b-a) int_a^b f(x)dx.

This means that there exists such number (or numbers) c that f(c)=f_(ave). In other words function takes its average value at least once.

Example 2. Find such c in interval [0,2] that f(c)=f_(ave) for y=4-x^2 on [0,2].

In Example 1 we've found that f_(ave)=8/3.

Therefore, we need to find such c that f(c)=4-c^2=8/3.

This gives c^2=4-8/3=4/3 or c=+-2/sqrt(3) , however c=-2/sqrt(3) doesn't belong to interval [0,2], therefore the only applicable number is c=2/sqrt(3).