Definite Integration — Maths STD 12 Science — Question

L.H.S becomes
$\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x$...(i)
Consider $\int_{-a}^0 f(x) d x$
Put x = – t
∴ dx = – dt
When x = – a, t = a
and when x = 0, t = 0
$\therefore \int_{-a}^0 f(x) d x=\int_a^0 f(-t)(-d t)$
$=-\int_a^0 f(-t) d t$
$\begin{aligned} & =\int_0^a f(-t) d t \quad \ldots . . .\left[\because \int_a^b f(x) d x=-\int_b^a f(x) d x\right] \\ & =\int_0^a f(-x) d x \quad \ldots . .\left[\because \int_a^b f(x) d x=\int_a^b f(t) d t\right]\end{aligned}$
Equation (i) becomes
$\begin{aligned} & \int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x \\ & =\int_0^a[f(-x)+f(x)] d x \ldots . . \text { (ii) }\end{aligned}$
Case I:
If f(x) is an even function, then f(– x) = f(x),
Equation (ii) becomes
$\int_{-a}^a f(x) d x=2 \cdot \int_0^a f(x) d x$
Case II:
If f(x) is an odd function, then f(– x) = – f(x),
Equation (ii) becomes
$\begin{array}{rlrl}\int_{-a}^a f(x) d x & =0 & \\ \int_{-a}^a f(x) d x & =2 \cdot \int_0^a f(x) d x, & & \text { If } f(x) \text { is an even function } \\ & =0, & & \text { if } f(x) \text { is an odd function }\end{array}$