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AIEEE 2002

JEE Main 2002 Mathematics question on Definite Integrals and Areas

The integral ∫−1/21/2[[x]+log⁡(1+x1−x)]dx\int_{-1/2}^{1/2} \left[ [x] + \log\left(\frac{1+x}{1-x}\right) \right] dx equals
  1. A.
    −12-\frac{1}{2}
  2. B.
    00
  3. C.
    11
  4. D.
    log⁡(12)\log\left(\frac{1}{2}\right)
Show answer and solution

Correct option: A

Solution

The integral splits into two parts. The term log((1+x)/(1-x)) is an odd function, so its integral over the symmetric interval [-1/2, 1/2] is 0.

For [x] (floor function): on [-1/2, 0), [x] = -1, and on [0, 1/2], [x] = 0. So the integral of [x] = (-1)(0-(-1/2)) + (0)(1/2-0) = -1/2.

Total = -1/2 + 0 = -1/2.

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