The value of the integral `int_3^6 sqrtx/(sqrt(9-x)+sqrtx)dx` is

Options

A. `3//2`
B. 2
C. 3
D. 6

Answer

A

Explanation

Let `I=int_(3)^(6)(sqrt(x))/(sqrt(9-x)+sqrt(x))dx " " `…(i)
`=int_(3)^(6)(sqrt(9-x))/(sqrt(9-9+x)+sqrt(9+x))dx`
`rArr I=int_(3)^(6)(sqrt(9-x))/(sqrt(x)+sqrt(9-x))dx " " ` …(ii)
On adding Eqs. (i) and (ii), we get
`2I=int_(3)^(6)(sqrt(x)+sqrt(9-x))/(sqrt(x)+sqrt(9-x))dx=int_(3)^(6)1 dx`
`=[x]_(3)^(6)=(6-3)=3 rArr I=(3)/(2)`