Question

If n is a positive integer and [x] is the greatest integer not exceeding x, then $\int _ { 0 } ^ { n } \{ x - [ x ] \} d x$ equals

• A. $n ^ { 2 } / 2$
• B. $n ( n - 1 ) / 2$
• C. $n / 2$
• D. $\frac { n ^ { 2 } } { 2 } - n$

Answer 1

Answer: (C)

$n / 2$

Answer 2

$x - [ x ]$ is a periodic function with period 1.

$\therefore \int _ { 0 } ^ { n } \{ x - [ x ] \} d x = n \int _ { 0 } ^ { 1 } ( x - [ x ] ) d x$

$= n \left[ \int _ { 0 } ^ { 1 } x d x - \int _ { 0 } ^ { 1 } [ x ] d x \right]$ .