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Parin Sir > Mathematical Proofs > Definite Integration > Pattern – 13 : Definite integration > Pattern – 13 : Definite integration

Pattern – 13 : Definite integration

February 28, 2021
Category: Pattern – 13 : Definite integration

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⋆ Pattern – 13

If f is a periodic function with principal period T, then $\int_{0}^{nT}f\left(x\right)\ dx = n$ $\int_{0}^{T}f\left(x\right)\ dx$

; Where, n I

Proof :

Let,

$I_k=\ \int_{KT}^{\left(k+2\right)T}f\left(x\right)\ dx\ :k\in I$

Let, x- KT = y

∴ x = y + KT

∴ dx = dy

x KT ⇒ y = 0

x = KT + T ⇒ y = T

$\therefore\ \ I_k$ = $\ \int_{0}^{T}f\left(y+KT\right)dy$

$=\ \int_{0}^{T}f\left(y\right)\ dy$

$=\ \int_{0}^{T}f\left(x\right)dx$

$\int_{KT}^{\left(K+2\right)T}f\left(x\right)dx$ = $\ \int_{0}^{T}f\left(x\right)dx\ :K\in I$

Now,

$\int_{0}^{nT}f\left(x\right)dx$ = $\ \int_{0}^{T}f\left(x\right)dx$ + $\ \int_{T}^{2T}{f\left(x\right)\ dx}$ + $\ \int_{2T}^{3T}{f\left(5\right)\ dx}$ + … + $\int_{\left(n-1\right)T}^{nT}f \left(x\right)dx$

∴ from (1)

$\int_{0}^{nT}f\left(x\right)dx$ = $\ \int_{0}^{T}f\left(x\right)dx+\$ $\int_{0}^{T}f\left(x\right)dx$ + $\ \int_{0}^{T}f\left(x\right)dx+$ … $+\ \int_{0}^{T}f\left(x\right)dx$

$=n\ \int_{0}^{T}f\left(x\right)dx$

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