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Pattern – 12 : Definite integration

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⋆ Pattern – 12 :

$\int_{a}^{b}f\left(x\right)dx$ = (b-a) $\int_{0}^{1}f[ (b-a) x + a ] dx$

Proof :

Let, x = (b-a)t + a

∴ dx = (b-a) dt

x = a ⇒ a = (b-a)t +a ⇒ 0 = t

x = b ⇒ (b-a) = (b-a)t ⇒ t = 1

$\therefore\ \ \int_{a}^{b}f\left(x\right)dx$ = $\ \left(b-a\right)\int_{0}^{1}f\left[\left(b-a\right)+a\right]dt$

$=\ \left(b-a\right)\int_{0}^{1}f\left[\left(b-a\right)x+a\right]dx$