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Pattern – 11 : Definite integration

February 28, 2021
Category: Pattern – 11 : Definite integration

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⋆ Pattern – 11 :

$\int_{0}^{2a}f\left(x\right) dx$ = $\int_{0}^{a}$ [ f(x) + f(2a-x) ] dx deduce that, $\int_{0}^{2a}f\left(x\right)dx$ = $\begin{cases} & 2 \int^a_0 f(x) dx : f(2a-x)=f(x)\\ & x= 0 : f(2a-x) = -f(x) \end{cases}$

Proof :

Here, I = $\int_{0}^{2a}f\left(x\right)dx$

$=\ \int_{0}^{a}{f\left(x\right)\ dx}+\int_{a}^{2a}f\left(x\right)dx$

$=\int_{0}^{a}f\left(x\right)dx+I_1---\left(1\right)$

where , $I_1=\ \int_{a}^{2a}f\left(x\right)dx$

Let, x = 2a – t

∴ dx = – dt

x = a ⇒ t = a

x = 2a ⇒ t = 0

$\therefore\ \ I_1=\ -\int_{a}^{0}f\left(2a-t\right)dt$

$=\ \int_{0}^{a}f\left(2a-x\right)dx$

∴ from (1),

I= $\ \int_{0}^{a}\left[f\left(x\right)+f\left(2a-x\right)\right]dx$

$\therefore I = \left \{ 2 \int_{0 \ f(x)dx : f(2a-x)= f(x)}^{a \ f(x)dx : f(2a-x)= f(x)} \right.$

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Pattern – 11 : Definite integration

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