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Pattern – 18 : Definite integration

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⋆ Pattern – 18(Even – off function) :

If f is a periodic function with principal period T, then

$\int_{0}^{T}f\left(t\right)dt$ = $\begin{cases} & 2 \int^{T/2}_0 \text{ : f is even } \\ & 0 \text{ : f is odd } \end{cases}$

Proof :

Let, h(x) = $\int_{0}^{x}f\left(t\right)$ dt —–(1)

Now,

h(x+T) = $\int_{0}^{x+T}f\left(t\right)\ dt$

$=\ \int_{0}^{T}f\left(t\right)dt+\ \int_{T}^{x+T}f\left(t\right)dt$

$=h\left(T\right)+\ \int_{0}^{x}f\left(t\right)dt$

= h(T) + h(x) ——(2)

∴ h(T) = h(x+T) – h(x)

Let, x= $\frac{-T}{2}$

$\therefore\ \ h\left(T\right)$ = $h\left(\frac{-T}{2}+T\right)-h\left(\frac{-T}{2}\right)$

$=h\left(\frac{T}{2}\right)-h\left(\frac{-T}{2}\right)$

$=\ \int_{0}^{\frac{T}{2}}f\left(t\right)dt-\ \int_{0}^{\frac{-T}{2}}f\left(t\right)dt$

$=\ \int_{0}^{\frac{T}{2}}f\left(t\right)dt+\ \int_{\frac{-T}{2}}^{0}f\left(t\right)dt$

$\therefore\ \ \int_{0}^{T}f\left(t\right)dt=\ \int_{\frac{-T}{2}}^{\frac{T}{2}}f\left(t\right)dt$

$= \begin{cases} & 2 \int^{T/2 }_{0} \text{f(t) dt : f is even} \\ & \text{0 : f is odd} \end{cases}$