For any natural number n, theb value of `rArr int_(0)^(n^(2))[ sqrt(x)]dx,` is

Options

A. `(n(n+1)(4n+1))/(6)`
B. `(n(n-1)(4n+1))/(6)`
C. `(n(n-1)(4n-1))/(6)`
D. none of these

Answer

B

Explanation

We have,
`I=underset(0)overset(n^(2))int [ sqrt(x)]dx`
`rArr I=underset(r=1)overset(n)sum underset((r-1)^(2))overset(r^(2))int [ sqrt(x)]dx`
`rArr I=underset(r=1)overset(n)sum underset((r-1)^(2))overset(r^(2))int (r-1)dx`
`rArr I=underset(r=1)overset(n)sum (r-1)(2r-1)`
`rArr I=underset(r=1)overset(n)sum (2r^(2)-3r+1)`
`rArr I=2 xx(n(n+1)(2n+1))/(6)-3xx(n(n+1))/(2)+n`
`rArr I=(n)/(6){2(2n^(2)+3n+1)-9n-9+6}`
`rArr L=(n)/(6) (n-1)(4n+1)=(n(n-1)(4n+1))/(6)`