Average value of a function

What does the average value of a function over an interval represent?

Average value of a function

For an integrable function over an interval with positive width, the average value is the signed definite integral divided by that width. It is the constant function value with the same signed integral over that interval. A negative average lies below the horizontal axis; it is not a negative geometric area

1. 1Find a primitive Choose an anti-derivative before evaluating the definite integral \(\displaystyle f(x)=x^2\Rightarrow F(x)=\frac{x^3}{3}\)
2. 2Evaluate the definite integral Subtract the primitive at the lower endpoint from the primitive at the upper endpoint \(\displaystyle \int_0^3x^2\,dx=F(3)-F(0)=9\)
3. 3Calculate the interval width Subtract the lower endpoint from the upper endpoint \(\displaystyle b-a=3-0=3\)
4. 4Divide the signed integral by the width The quotient is the average value. For this nonnegative function it also equals the height of an equal-area rectangle \(\displaystyle \overline{f}=\frac{9}{3}=3,\quad 3\times3=9\)

Evaluate the definite integral for the function on the interval

Divide the definite integral by the interval width to find the average value

Function and interval

\(\displaystyle f(x)=x^2,\quad 0\le x\le3\)

Which line correctly gives the average value of the function on the interval?

A negative average is a function value

Reflect the example in the horizontal axis. Its signed integral is −9, so its average value is −3. The constant function with value negative three has the same signed integral. The geometric area between either graph and the horizontal axis is positive, equal to 9

Function and signed integral

\(\displaystyle f(x)=-x^2,\quad 0\le x\le3,\quad \int_0^3f(x)\,dx=-9\)

Use the signed integral to find the average value. Do not replace it by geometric area