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