Integrating a Periodic Function

 (13) If \(f(x)\) is a periodic function with period T, then the area under \(f(x)\) for n periods would be n times the area under \(f(x)\) for one period, i.e.

\[\int\limits_0^{nT} {f(x)\,dx = n\int\limits_0^T {f(x)\,dx} } \]

Now, consider the periodic function \(f(x) = \sin x\) as an example. The period of \(\sin x\,\,{\rm{is }}2\pi \) .

Suppose we intend to calculate \(\int\limits_a^{a + 2\pi } {\sin x\,dx} \) as depicted above. Notice that the darkly shaded area in the interval \([2\pi ,\,\,a + 2\pi ]\) can precisely cover the area marked as A. Thus,

\[\int\limits_a^{a + 2\pi } {\sin x\,dx} = \int\limits_0^{2\pi } {\sin x\,\,dx} \]

This will hold true for every periodic function, i.e.

\(\int\limits_a^{a + T} {f(x)\,dx} = \int\limits_0^T {f(x)\,dx} \) (where T is the period of f (x))

This also implies that

\[\begin{align}& \quad\qquad\;\int\limits_{a}^{a+nT}{f(x)dx=\int\limits_{0}^{nT}{f(x)dx=n}\int\limits_{0}^{T}{f(x)dx}} \\ 
 & \text{and}\quad \int\limits_{a+nT}^{b+nT}{f(x)dx=\int\limits_{a}^{b}{f(x)dx}} \\ & \text{and}\quad\text{ }\!\!~\!\!\text{ }\int\limits_{a}^{b+nT}{f(x)dx=\int\limits_{a}^{b}{f(x)dx+n\int\limits_{0}^{T}{f(x)dx}}}\text{ } \\ 
\end{align}\]

Example –18

Show that \(\int\limits_0^{n\pi + V} {|\sin x|dx = (2n + 1) - \cos V,} \) where n is a positive integer and \(V=[0,\pi )\)

Solution: \(f(x) = |\sin x|\) is periodic with period p .

Therefore, as described in property-13,

\[\begin{align}&\int\limits_0^{n\pi + V} {|\sin x|dx} = \int\limits_0^V {|\sin x|dx + n\int\limits_0^\pi {|\sin x|dx} } \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= \int\limits_0^V {\sin xdx} + n\int\limits_0^\pi {\sin xdx} & & & \left[ \begin{gathered}\because \ \,\,\,\sin x \ge 0 \\\,\,\,\,\,\,{\text{for}}\,x \in [0,\pi ) \\\end{gathered} \right] \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= \left. { - \cos x} \right|_0^V + \left. {n( - \cos x)} \right|_0^\pi\\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= 1 - \cos V + n(2)\\ \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad= (2n + 1) - \cos V \\\end{align} \]

Example –19

Evaluate \(\int\limits_0^{n\pi } {\sqrt {1 - \cos 2x} \,dx,} \) where n is a positive integer.

Solution:

\[\begin{array}{l}I = \int\limits_0^{n\pi } {\sqrt {1 - \cos 2x} \,dx} \\\,\,\,\, = \int\limits_0^{n\pi } {\sqrt {2{{\sin }^2}x} \,dx} \,\,\,\,\\\,\,\, = \sqrt 2 \int\limits_0^{n\pi } {|\sin x|dx} \\\,\,\,\, = \sqrt 2 n\int\limits_0^\pi {|\sin x|dx} & & & ({\rm{Property - }}13)\\\,\,\,\, = 2\sqrt 2 n\end{array}\]