Question

Prove that: \[ \int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_0^a f(2a - x) dx. \] Hence show that: \[ \int_0^\pi \sin x \,dx = 2 \int_0^{\pi/2} \sin x \,dx. \]

Hint:

For integrals of symmetric functions, use the property: \[ \int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_0^a f(2a - x) dx. \]

Answer 1

Step 1: Proof of Integral Identity
Using the substitution \( u = 2a - x \), we get: \[ du = -dx. \] Changing limits: \[ \int_0^a f(2a - x) dx = \int_{2a}^{a} f(u) (-du) = \int_a^{2a} f(u) du. \] Since: \[ \int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_a^{2a} f(x) dx, \] we conclude: \[ \int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_0^a f(2a - x) dx. \] Step 2: Apply to Sine Function
For \( f(x) = \sin x \), let \( a = \frac{\pi}{2} \), so \( 2a = \pi \): \[ \int_0^\pi \sin x dx = \int_0^{\pi/2} \sin x dx + \int_0^{\pi/2} \sin (\pi - x) dx. \] Since \( \sin (\pi - x) = \sin x \): \[ \int_0^\pi \sin x dx = 2 \int_0^{\pi/2} \sin x dx. \] Final Verification: \[ \int_0^\pi \sin x dx = [-\cos x]_0^\pi = -\cos \pi + \cos 0 = 1 + 1 = 2. \] \[ \int_0^{\pi/2} \sin x dx = [-\cos x]_0^{\pi/2} = -\cos (\pi/2) + \cos 0 = 0 + 1 = 1. \] \[ 2 \times 1 = 2. \]