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Question
Evaluate the following integrals:
`int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x))*dx`
Sum
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Solution
Let I = `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x))*dx` ...(i)
= `int_1^3 (root(3)((1 + 3 - x) + 5))/(root(3)((1 + 3 - x) + 5) + root(3)(9 - (1 + 3 - x)))*dx ...[because int_"a"^"b" f(x)*dx = int_"a"^"b" f("a" + "b" - x)*dx]`
∴ I = `int_1^3 (root(3)(9 - x))/(root(3)(9 - x) + root(3)(5 + x))*dx` ...(ii)
Adding (i) and (ii), we get
2I = `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x))*dx + int_1^3 (root(3)(9 - x))/(root(3)(9 - x) + root(3)(5 + x))*dx`
= `int_1^3 (root(3)(x + 5) + root(3)(9 - x))/(root(3)(x + 5) - root(3)(9 - x))*dx`
= `int_1^3 1*dx`
= `[x]_1^3`
∴ 2I = 3 – 1 = 2
∴ I = 1
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