JEE Main 2021 · 31 August · morning · Numerical answer · +4 / −0

Let [t] denote the greatest integer ≤ t. Then the value of
8.∫_(- 1/2)¹ ([2x] + |x|)dx is ___________.

Answer and solution

Answer 5

Solution

Maths sir · 3 steps

Split the interval where [2x] jumps, at x = 0 and x = 1/2. Then each piece is a constant.

  1. Find the jumps. As x goes from -1/2 to 1, 2x goes from -1 to 2. So [2x] changes value at 2x = 0 and 2x = 1, that is, at x = 0 and x = 1/2.
    Interval[2x]∫ [2x] dx∫ |x| dx
    [-1/2, 0)-1-1 × 1/2 = -1/2∫_(-1/2)⁰ (-x) dx = 1/8
    [0, 1/2)00∫₀^(1/2) x dx = 1/8
    [1/2, 1]11 × 1/2 = 1/2∫_(1/2)¹ x dx = 3/8
  2. Add the pieces. The [2x] parts give -1/2 + 0 + 1/2 = 0. The |x| parts give 1/8 + 1/8 + 3/8 = 5/8.

    8∫_(-1/2)¹ ([2x] + |x|)dx = 8(0 + 5/8) = 5

  3. So. The value is 5.

The common slip: 3. Using x instead of -x for |x| on [-1/2, 0] gives ∫ |x| dx = 3/8, and the answer comes out as 3, not 5.

The idea to keep. For [kx], find where kx crosses an integer and break the integral there. On each piece the greatest integer is a constant, so its integral is just height times width.

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

Study Definite Integration

Asked about 230 times in JEE Main since 2020. This question is from 2021.