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Pattern – 8 : Definite integration

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Pattern – 8 :

$\int_{0}^{b-c}f\left(x+c\right)dx=\ \int_{c}^{b}f\left(x\right)dx$

Proof :

Let, x + c = t

∴ dx = dt

x = 0 ⇒ t = c

x = b-c ⇒ t = b – c + c = b

$\therefore\ \ \int_{0}^{b-c}\left(x+c\right)dx=\ \int_{c}^{b}f\left(t\right)dt\$

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