The value of the integral ∫₀^π |sin 2x|dx is ___________.

Solution

Maths sir · 4 steps

The modulus makes every hump positive, so count the humps of sin 2x and add their areas.

1. Find the period. sin 2x has period π, so |sin 2x| has period π/2. The interval [0, π] holds exactly 2 of these periods.
2. Use periodicity. For a function with period T, the integral over nT is n times the integral over one period. Here n = 2 and T = π/2.

   ∫₀^π |sin 2x| dx = 2∫₀^(π/2) sin 2x dx = 2[-(cos 2x)/2]₀^(π/2)

3. Evaluate one hump. On [0, π/2], sin 2x ≥ 0, so the modulus drops. The bracket gives -(cos π)/2 + (cos 0)/2 = 1/2 + 1/2 = 1.
4. So. The integral is 2 × 1 = 2.

The common slip: 0. Dropping the modulus and integrating sin 2x from 0 to π gives 0, because the positive and negative humps cancel.

The idea to keep. For ∫ |sin kx| or ∫ |cos kx|, find the new period π/k, count how many fit in the limits, and multiply by the area of one hump.

Answer from the NTA final answer key. Solution by PrepWiser.