Question

Evaluate the integral ${\int }_0^{50}\left(\right[x]-x)dx$

where $[.]$ is a greatest integer function

• A. $-25$
• B. $-20$
• C. $-15$
• D. $-10$

Answer 1

The correct option is A $-25$

Since $\left[x\right]-x$ is periodic function with period one unit.
Therefore
${\int }_0^{50}(\left[x\right]-x)dx=50{\int }_0^1(\left[x\right]-x)dx=50[{\int }_0^1\left[x\right]dx-{\int }_0^{1}xdx]=50[-{\left[\frac{x^2}{2}\right]}_0^1+0]=-25$