The integral ∫₋₁^(3/2) (| π² x sin(π x) |) dx is equal to:

Solution

Maths sir · 5 steps

Find where x sin(π x) changes sign, split the integral there, and the modulus comes off cleanly.

1. Sign of integrand. On (0,1) both x and sin(π x) are positive. On (-1,0) both are negative, so the product is positive again. On (1,3/2), x > 0 but sin(π x) < 0, so the product is negative there.
2. One antiderivative. Integration by parts gives ∫ π² x sin(π x) dx = F(x), where F(x) = sin(π x) - π x cos(π x). Useful values are F(0) = 0, F(1) = π and F(3/2) = -1.

   I = ∫₋₁¹ π² x sin(π x) dx - ∫₁^(3/2) π² x sin(π x) dx

3. First part. x sin(π x) is even, so the first integral is 2[F(1) - F(0)] = 2π.
4. Second part. -[F(3/2) - F(1)] = -(-1 - π) = 1 + π.
5. So. I = 2π + 1 + π = 1 + 3π, which is option (c).

The idea to keep. With a modulus inside an integral, first find where the expression changes sign. Split the limits at those points, and use evenness to halve the work on a symmetric interval.

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