If [x] is the greatest integer ≤ x, then
π ²∫₀² ( sin (π x)/2 )(x - [x] )^([x])dx is equal to :

Solution

Maths sir · 4 steps

Break the integral at x = 1, where [x] changes. Then each piece is an ordinary integral.

1. Split at integers. On [0, 1), [x] = 0, so (x - [x])^([x]) = 1. On [1, 2), [x] = 1, so (x - [x])^([x]) = x - 1.

   ∫₀² sin(π x)/2 (x-[x])^([x]) dx = ∫₀¹ sin(π x)/2 dx + ∫₁² (x-1)sin(π x)/2 dx

2. First piece. ∫₀¹ sin(π x)/2 dx = [-2/π cos(π x)/2]₀¹ = 2/π.
3. Second piece. Put t = x - 1. Then sin π(t+1)/2 = cos(π t)/2, and the integral becomes ∫₀¹ t cos(π t)/2 dt. Integration by parts gives [2t/π sin(π t)/2]₀¹ - 2/π∫₀¹ sin(π t)/2 dt = 2/π - 4/π².
4. So. The integral is 2/π + 2/π - 4/π² = 4/π - 4/π². Multiplying by π² gives 4π - 4 = 4(π - 1), option (b).

The idea to keep. Whenever [x] appears inside an integral, cut the range at every integer. On each piece [x] is a constant, and the integrand becomes something you already know how to integrate.

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