JEE Main 2020 · 2 September · evening · Numerical answer · +4 / −0

Let [t] denote the greatest integer less than or equal to t.
Then the value of ∫₁² | 2x - [ 3x ] |dx is ______.

Answer and solution

Answer 1

Solution

Maths sir · 3 steps

Split [1, 2] where 3x hits an integer; on each piece [3x] is constant and the modulus opens.

  1. Find the breaks. For x from 1 to 2, 3x runs from 3 to 6. So [3x] jumps at 3x = 4 and 3x = 5, that is at x = 4/3 and x = 5/3.
  2. Open the modulus. On each piece 2x stays below [3x]: 2x < 8/3 < 3 on the first, 2x < 10/3 < 4 on the second, and 2x ≤ 4 < 5 on the third. So 2x - [3x] < 0 throughout, and the integrand is [3x] - 2x.
    Interval[3x]IntegrandIntegral
    [1, 4/3)33 - 2x1 - 7/9 = 2/9
    [4/3, 5/3)44 - 2x4/3 - 1 = 3/9
    [5/3, 2]55 - 2x5/3 - 11/9 = 4/9

    ∫₁² |2x - [3x]| dx = 2/9 + 3/9 + 4/9 = 1

  3. So. The value of the integral is 1.

The common slip: -1. Integrating 2x - [3x] without the modulus gives -1. That expression is negative on all of [1, 2], so take [3x] - 2x instead.

The idea to keep. With [kx] inside an integral, split the interval at the points where kx is an integer. On each piece the greatest integer is a fixed number, and the rest is routine.

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 2020.