Question

Evaluate the following integrals : `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx`

Answer 1

Let I = `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx`        ...(i)

= `int_1^2 sqrt(1 + 2 - x)/(sqrt(3 - (1 + 2 - x)) + sqrt(1 + 2 - x))*dx       ...[because int_"a"^"b" f(x)*dx = int_"a"^"b" f("a" + "b" - x)*dx]`

∴ I = `int_1^2 sqrt(3 - x)/(sqrt(x) + sqrt(3 - x))*dx`           ...(ii)

Adding (i)  and (ii), we get

2I = `int_1^2 sqrt(x)/(sqrt(3 - x) + sqrt(x))*dx  + int_1^2 sqrt(3 - x)/(sqrt(x) + sqrt(3 - x))*dx`

= `int_1^2 (sqrt(x) + sqrt(3 - x))/(sqrt(x) + sqrt(3 - x))*dx`

= `int_1^2 1*dx`

= `[x]_1^2`
∴ 2I = 2 – 1 = 1
∴ I = `(1)/(2)`.