∫₀²(|2 x²-3 x|+[x-1/2]) d x, where [t] is the greatest integer function, is equal to :

Solution

Maths sir · 3 steps

Break the integral at every point where the modulus or the box changes form. Then each piece is easy.

1. Open the modulus. 2x² - 3x = x(2x - 3) is zero at x = 0 and x = 3/2. It is negative on (0, 3/2) and positive on (3/2, 2), so split the integral at 3/2.

   ∫₀^(3/2)(3x - 2x²) dx + ∫_(3/2)²(2x² - 3x) dx = 9/8 + 11/24 = 19/12

2. Greatest integer part. On [0, 1/2), [x - 1/2] = -1. On [1/2, 3/2) it is 0, and on [3/2, 2] it is 1. Its integral is -1 × 1/2 + 0 × 1 + 1 × 1/2 = 0.
3. So. The total is 19/12 + 0 = 19/12, which is option (b).

The idea to keep. For a modulus, split where the expression inside is zero. For [f(x)], split where f(x) crosses an integer. Integrate each piece separately and add.

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