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Pattern – 9 : Definite integration

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⋆ Pattern – 9

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

Proof :

Obviously, $\frac{a}{2}\in\left[0,\ a\right]\ 0 < \frac{a}{2} < a$

Let, a

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

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

$=\ \int_{0}^{\frac{a}{2}}f\left(x\right)dx+I_1---\left(1\right)$

where,

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

Let, x = a -t

∴ dx = -dt

x = $\frac{a}{2}$ ⇒ t = a – $\frac{a}{2}$ = $\frac{a}{2}$

x = a ⇒ t = 0

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

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

we have,

$I=\ \int_{0}^{\frac{a}{2}}f\left(x\right)dx+I_1----\left(1\right)$

where,

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

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

∴ from (1)

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