The value of the integral, ∫₁³ [x² - 2x - 2]dx, where [x] denotes the greatest integer less than or equal to x, is :

Solution

Maths sir · 4 steps

Complete the square, pull the integer -3 out of the bracket, and only [t²] is left to split.

1. Complete the square. x² - 2x - 2 = (x-1)² - 3. An integer can come out of the greatest integer function, so [(x-1)² - 3] = [(x-1)²] - 3.
2. Shift the variable. Put t = x - 1. As x goes from 1 to 3, t goes from 0 to 2. The -3 over an interval of length 2 contributes -6.

   ∫₁³ [x² - 2x - 2] dx = ∫₀² [t²] dt - 6

   Interval of t	[t²]	Length	Contribution
   0 ≤ t < 1	0	1	0
   1 ≤ t < √2	1	√2 - 1	√2 - 1
   √2 ≤ t < √3	2	√3 - √2	2√3 - 2√2
   √3 ≤ t < 2	3	2 - √3	6 - 3√3

3. Add the pieces. The contributions add to ∫₀² [t²] dt = 5 - √2 - √3.
4. So. The integral is 5 - √2 - √3 - 6 = -√2 - √3 - 1, which is option (d).

The idea to keep. For [f(x)], first take any integer out of the bracket. Then break the interval wherever f(x) crosses an integer and add the value of the bracket times the length of each piece.

Answer from the NTA final answer key. Solution by PrepWiser.