The value `int_(1)^(a)[x]f'(x)dx, a gt 1`,where [x] denotes the greatest integer not exceeding x, is

Options

A. `a f(a)-{f(1)+f(2)+......+f([a])}`
B. `[a] f(a)-{f(1)+f(2)+......+f([a])}`
C. `[a] f([a])-{f(1)+f(2)+......+f(a)}`
D. `a f([a])-{f(1)+f(2)+......+f(a)}`

Answer

B

Explanation

Suppose `n le a lt n+1`, where `n in N`. Then, [a]=n.
`:. I=overset(a)underset(1)int [x]f'(x)dx`
`rArrI=overset(n-1)underset(r=1)sum overset(r+1)underset(r=1)int[x]f'(x)dx+overset(a)underset(n)int [x]f'(x)dx`
`rArrI=overset(n-1)underset(r=1)sumoverset(r+1)underset(r=1)int r f'(x)dx+overset(a)underset(n)int nf'(x)dx`
`rArrI=overset(n-1)underset(r=1)sumr[f(x)]_(r )^(r+1)+n[f(x)]_(n)^(a)`
`rArr I=overset(n-1)underset(r=1)sum r{f(r+1)-f(r )}+n{f(a)-f(n)}`
`rArr I={f(2)-f(1)}+2{f(3)-f(2)}+3{f(4)-f(3)}+...........+(n-1){f(n)-f(n-1)}+n{f(a)-f(n)}`
`I=-f(1)-f(2)f-(3)-f(4)-.........-f(n-1)-f(n)+nf(a)`
`rArr I=[a]f(a)-{f(1)+f(2)+f(3)+.........+f[a]}`