Differentials and integrals: Applications of integration
Average value of a function on an interval
The average value \(\overline{f}_{[a,b]}\) of a function \(f(x)\) on the interval \([a,b]\) is \[\overline{f}_{[a,b]}=\frac{1}{b-a}\int_a^b f(x)\,\dd x\] that is, the integral over the interval divided by the length of the interval. The integral is the (signed) area under the curve, and this is equal to the area of the rectangle with the height \(\overline{f}_{[a,b]}\). The following specific examples illustrate this application.
The calculation of the average value of \(f(t)=e^{-t}\) on the interval \([0,100]\).
\[\overline{f}_{[0,100]}=\frac{1}{100}\int_0^{100} e^{-t}\,\dd t=\frac{1}{100}\left[-e^{-t}\right]_0^{100}=\frac{1}{100}\left(1-e^{-100}\right)\]
Note that the average value is approximately equal to \(\frac{1}{100}\) because the term \(e^{-100}\) is negligibly small.
The calculation of the average value of \(\sin(x)^2\) on the interval \([0,2\pi]\) .
Via integration by parts we find: \[\begin{aligned}\int_0^{2\pi}\bigl(\sin(x)\bigr)^2\,\dd x&=\int_0^{2\pi}\sin(x)\cdot \bigl(-\cos(x)\bigr)'\,\dd x\\ \\ &=\bigl[-\sin(x)\cos(x)\bigr]_{0}^{2\pi}-\int_0^{2\pi}\cos(x)\cdot\bigl(-\cos(x)\bigr)\,\dd x\\ \\ &=\int_0^{2\pi}\bigl(\cos(x)^2\bigr)\,\dd x\end{aligned}\] But then also: \[\begin{aligned}2\int_0^{2\pi}\bigl(\sin(x)\bigr)^2\,\dd x &= \int_0^{2\pi}\bigl(\sin(x)\bigr)^2\,\dd x+\int_0^{2\pi}\bigl(\cos(x)^2\bigr)\,\dd x\\ \\ &= \int_0^{2\pi}\biggl(\bigl(\sin(x)\bigr)^2+\bigl(\cos(x)^2\bigr)\biggr)\,\dd x\\ \\ &=\int_0^{2\pi} 1\,\dd x=\bigl[x\bigr]_0^{2\pi}=2\pi.\end{aligned}\] So: \[\int_0^{2\pi}\bigl(\sin(x)\bigr)^2\,\dd x=\pi\] and the average value of \(\sin(x)^2\) on the interval \([0,2\pi]\) is equal to \(\frac{1}{2}\).
