Introduction
The average change of
f
on
[x_1,x_2]
is
m=\frac{f(x_2)-f(x_1)}{x_2-x_1}
Thus, that is the average value of the derivative for that domain.
Therefore, the average value of
f
is equal to:
\frac{F(x_2)-F(x_1)}{x_2-x_1} = \frac{\int_{a}^{b} f(x) \, dx}{b-a}
Examples
Find the average value of
f(x) = 2x + 3
on
[1,5]
.
\begin{align*} \frac{1}{5-1}\int_{1}^{5} (2x + 3) \, dx &= \frac{1}{4}[x^2 + 3x]\Big|_{1}^{5} \\\\
&= \frac{1}{4}[(5^2 + 3 \cdot 5) - (1^2 + 3 \cdot 1)] \\\\
&= \frac{1}{4}[25 + 15 - 1 - 3] \\\\
&= 9
\end{align*}
Find the average value of
f(x) = x^2
over the interval
[0, 2]
.
\begin{align*}
\frac{1}{2-0}\int_{0}^{2} x^2 \, dx &= \frac{1}{2}[ \frac{x^3}{3} ]\Big|_{0}^{2} \\\\
&= \frac{1}{2} \left( \frac{8}{3} - 0 \right) \\\\
&= \frac{4}{3}
\end{align*}
Find the average value of
f(x) = \sin(x)
over the interval
[0, \pi]
.
\begin{align*}
\frac{1}{\pi - 0}\int_{0}^{\pi} \sin(x) \, dx &= \frac{1}{\pi}[ -\cos(x)]\Big|_{0}^{\pi} \\\\
&= \frac{1}{\pi} \left( -\cos(\pi) + \cos(0) \right) \\\\
&= \frac{1}{\pi} (1 + 1) \\\\
&= \frac{2}{\pi}
\end{align*}
Find the average value of
f(x) = e^x
over the interval
[0, 1]
.
\begin{align*}
\frac{1}{1-0}\int_{0}^{1} e^x \, dx &= [ e^x ]\Big|_{0}^{1} \\\\
&= e^1 - e^0 \\\\
&= e - 1
\end{align*}