For any real number x, the value of `int_(0)^(x) [x]dx`, is

Options

A. x[x]
B. x[x]-[x]([x]+1)
C. `x[x]-(1)/(2)[x]([x]+1)`
D. none of these

Answer

C

Explanation

Let k be an integer such that `k le x lt k+1`. Then,
` rArroverset(x)underset(0)int [x]dx=overset(k)underset(0)int[x]dx+overset(x)underset(k)int[x]dx`
` rArr overset(x)underset(0)int[x]dx=overset(1)underset(0)int[x]dx+overset(2)underset(1)int [x]dx=+..........+overset(k)underset(k-1)int[x]dx+overset(x)underset(k)int[x]dx`
`rArr overset(x)underset(0)int[x]dx=overset(k)underset(r=1)sum overset(r )underset(r-1)int (r-1)dx+overset(x)underset(k)int k dx`
`rArr overset(x)underset(0)int[x]dx=overset(k)underset(r=1)sum (r-1) +k(x-k)`
` rArr overset(x)underset(0)int[x]dx=(k(k-1))/(2)+kx-k^(2)`
`rArr overset(x)underset(0)int[x]dx=kx-(k(k+1))/(2)=[x]x-([x]([x]+1))/(2)`