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