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Pattern – 6 : Definite integration

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⋆ Pattern – 6 :

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

Proof :

Let, a + b – x = t

∴ -dx = dt

∴ dx = -dt

x = a ⇒ k = a + b – a = b

x = b ⇒ t = a + b – b =a

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

$=\ \int_{a}^{b}f\left(t\right)\ dt$

$=\int_{a}^{b}f\left(x\right)dx$