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Pattern – 10 : Definite integration

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⋆ Pattern – 10

$\int_{-a}^{a}f\left(x\right)dx$ = $\begin{cases} & \text{ x } \int_{0}^{a} f(x) dx : \text{f is even} \\ & \text{ 0 : f is odd } \end{cases}$

Proof :

Now,

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

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

where ,

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

Let, x = -t

∴ dx = – dt

x = -a ⇒ t = a

x = 0 ⇒ t = 0

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

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

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

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