Greatest Integer Function Integrals

Integrating the greatest integer or floor function requires splitting the interval at the integer jump points where \lfloor x \rfloor changes value, then integrating its constant value on each piece separately. The reason is that the floor is a step function, constant between consecutive integers and discontinuous at each whole number, so direct antidifferentiation is impossible but piecewise summation is easy. On the subinterval [0,1) the floor equals 0; on [1,2) it equals 1; on [2,3) it equals 2. Therefore \int_0^3 \lfloor x \rfloor,dx = \int_0^1 0,dx + \int_1^2 1,dx + \int_2^3 2,dx = 0 + 1 + 2 = 3. Option 6 mistakenly integrates x itself rather than its floor. Option 9/2 equals \int_0^3 x,dx, again ignoring the floor and integrating the continuous line. Option 2 omits one of the three constant strips. This step-function strategy handles every integer-valued integrand uniformly. As a final plausibility check, the floor always lies at or below x, so the integral 3 must be less than \int_0^3 x,dx = 4.5, and indeed it is, confirming the underestimate consistent with the staircase sitting beneath the line y = x.