If {x} denotes the fractional part of x, then `int_(0)^(x)({x}-(1)/(2)) dx` is equal to

Options

A. `(1)/(2){x}({x}+1)`
B. `(1)/(2){x}({x}-1)`
C. `{x}({x}-1)`
D. `{x}({x}+1)`

Answer

B

Explanation

We have, {x}=x-[x]
Let `kle x lt k+1`, where `k in N`. Then,
`I=overset(x)underset(0)int({x}-(1)/(2))dx=overset(x)underset(0)int (x-[x]-(1)/(2))dx`
`rArr I=overset(x)underset(0)int (x-[x])dx-overset(x)underset(0)int(1)/(2)dx`
`rArr I=overset(k)underset(0)(x-[x])dx+overset(x)underset(k)int (x-[x])dx-(x)/(2)`
`I=(k)/(2)+overset(x)underset(k)int(x-k)dx-(x)/(2)`
`rArr I=(k)/(2)+((x^(2))/(2)-kx)-((k^(2))/(2)-k^(2))-(x)/(2)`
`rArr I=(x^(2))/(2)-kx+(k(k+1))/(2)-(x)/(2)`
`rArr I=(1)/(2)(x^(2)-2kx+k^(2))+(1)/(2)(k-x)`
`rArr I=(1)/(2)(x-k)^(2)-(1)/(2)(x-k)`
`rArr I=(1)/(2)(x-[x])^(2)-(1)/(2)(x-[x])`
`rArr I=(1)/(2){x}^(2)-(1)/(2){x}=(1)/(2){x}({x}-1)`