NDA2025MathematicsDefinite IntegrationActual
What is _n^ n+1 (x - [x]) , dx , where [ ] is the greatest integer function and n is a natural number?
Options
- A4n+1 2
- B2n+1 2
- C1 2
- D1
Correct answer
C. 1 2
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