`int_(0)^(a)(sqrtx)/(sqrtx+sqrt(a-x))dx`

Explanation

माना `int_(0)^(a)(sqrtx)/(sqrtx+sqrt(a-x))dx" …(1)"`
`rArr" "I=int_(0)^(a)(sqrt(a-x))/(sqrt(a-x)+sqrt(a-(a-x)))dx`
`[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`=int_(0)^(a)(sqrt(a-x))/(sqrt(a-x)+sqrtx)dx" ...(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर
`2I=int_(0)^(a)(sqrtx+sqrt(a-x))/(sqrtx+sqrt(a-x))dx=int_(0)^(a)1dx =[x]_(0)^(a)=a`
`rArr" "I=(a)/(2)`