Question

Evaluate `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x))  "d"x`

Answer 1

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

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

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

Adding (i) and (ii), we get

2I = `int_1^2 (sqrt(x))/(sqrt(3 - x) + sqrt(x))  "d"x + int_1^2 (sqrt(3 - x))/(sqrt(x) + sqrt(3 - x))  "d"x`

= `int_1^2 (sqrt(x) + sqrt(3 - x))/(sqrt(x) + sqrt(3 - x))  "d"x`

= `int_1^2 1* "d"x`

= `[x]_1^2`

∴ 2I = 2 – 1 = 1

∴ I = `1/2`