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NDA2025MathematicsDefinite IntegrationActual

What is _n^ n+1 (x - [x]) , dx , where [ ] is the greatest integer function and n is a natural number?

Options

  1. A4n+1 2
  2. B2n+1 2
  3. C1 2
  4. 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

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