NDA Mathematics Question (2025) — Solution

Step-by-step solution

The given integral is I = _n^ n+1 (x - [x])\, dx. We know that x - [x] = \ x\ , which is the fractional part function. The fractional part function is periodic with a period of 1. Therefore, the integral over any interval of length 1 is equal to the integral over [0, 1]. I = _0^1 (x - [x])\, dx In the interval [0, 1), [x] = 0. I = _0^1 x\, dx = [ x^2 2 ]_0^1 = 1 2 Alternatively, in the interval [n, n+1), [x] = n. I = _n^ n+1 (x - n)\, dx = [ (x-n)^2 2 ]_n^ n+1 = 1^2 2 - 0 = 1 2 Answer: 1 2