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Pattern – 14 : Definite integration

February 28, 2021
Category: Pattern – 14 : Definite integration

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⋆ Pattern – 14 :

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

; Where, m, n ∈ I & n > m.

Proof :

Let, I = $\ \int_{mT}^{nT}f\left(x\right)dx$

Let, (m + k + 1 )T

$I_k=\ \int_{\left(m+k\right)T}^{\left(m+k+1\right)T}f\left(x\right)\ dx\ :k\in I$

Let, x – (m+k)T = y

∴ dx = dy

x = (m+k)T ⇒ y = 0

x = (m+k)T + 1 ⇒ y = T

$=\ \int_{0}^{T}f\left[y+\left(m+k\right)T\ \right]dy$

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

Let, $\ I=\ \int_{mT}^{nT}f\left(x\right)dx$ &

$\int_{\left(m+k\right)T}^{\left(m+k+2\right)T}f\left(x\right)\ dx$ = $\ \int_{0}^{T}f\left(y\right)dy$

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

Now, (m+1)T (m+2)T (m+3)T

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