Prove that, \(\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx\), if \(f(x)\) is even\(= 0\), if \(f(x)\) is odd

Answer: Proof (split at 0 and substitute \(x = -t\)).

\(\int_{-a}^{a}f(x)\,dx = \int_{-a}^{0}f(x)\,dx + \int_{0}^{a}f(x)\,dx\) ... (1)

In \(I_1 = \int_{-a}^{0}f(x)\,dx\) put \(x = -t\), \(dx = -dt\); when \(x = -a\), \(t = a\) and when \(x = 0\), \(t = 0\):

\(I_1 = \int_{a}^{0}f(-t)(-dt) = \int_{0}^{a}f(-t)\,dt = \int_{0}^{a}f(-x)\,dx\).

So from (1): \(\int_{-a}^{a}f(x)\,dx = \int_{0}^{a}f(-x)\,dx + \int_{0}^{a}f(x)\,dx = \int_{0}^{a}[f(-x) + f(x)]\,dx\) ... (2)

If f is even, \(f(-x) = f(x)\), and (2) gives \(\int_{-a}^{a}f(x)\,dx = \int_0^a 2f(x)\,dx = 2\int_0^a f(x)\,dx\).

If f is odd, \(f(-x) = -f(x)\), and (2) gives \(\int_{-a}^{a}f(x)\,dx = \int_0^a 0\,dx = 0\).

Hence proved.