JEE Main 2026 — Jan 22, Shift 1 — Mathematics solution

JEE Main2026Jan 22, Shift 1Mathematics

The value of $\int_{- π /2}^{π /2} (\frac{1}{[ x ] + 4}) d x$ , where $[\cdot]$ denotes the greatest integer function, is

Solution

Split the interval where $[x]$ is constant.

$[x] = - 2$ on $[- π /2, - 1)$ (length $π /2 - 1$ ): integrand $= 1/2$ .

$[x] = - 1$ on $[- 1, 0)$ (length 1): integrand $= 1/3$ .

$[x] = 0$ on $[0, 1)$ (length 1): integrand $= 1/4$ .

$[x] = 1$ on $[1, π /2)$ (length $π /2 - 1$ ): integrand $= 1/5$ .

Total $= \frac{1}{2} (π /2 - 1) + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} (π /2 - 1)$ .

Group $π$ -terms: $\frac{π}{4} + \frac{π}{10} = \frac{7 π}{20}$ .

Group constants: $- \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5} = \frac{- 30 + 20 + 15 - 12}{60} = - \frac{7}{60}$ .

Total $= \frac{7 π}{20} - \frac{7}{60} = \frac{21 π - 7}{60} = \frac{7 ( 3 π - 1 )}{60}$ .

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