Question

Evaluate: `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x)) dx`

Solution:

Let I = `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x)) dx`    ...(i)

By property `int_a^b f(x) dx = int_a^b f(a + b - x)dx`,

I = `int_1^3 square/(root(3)(9 - x) + square)`    ...(ii)

Adding (i) and (ii)

∴ I + 1 = `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x)) dx + int_1^3 square/(root(3)(9 - x) + square)`

= `int_1^3 (root(3)(x + 5) + root(3)(9 - x))/(root(3)(x + 5) + root(3)(9 - x)) dx`

∴ 2I = `int_1^3 square` dx

= `[square]_1^3`

∴ 2I = `square`

∴ I = 1

Answer 1

Let I = `int_1^3 (root(3)(x + 5))/(root(3)(x + 5) + root(3)(9 - x)) dx`    ...(i)

By property `int_a^b f(x) dx = int_a^b f(a + b - x)dx`,

\[\text{I} = \int_1^3\frac{\boxed{\sqrt[3]{9-x}}}{\sqrt[3]{9-x} + \boxed{\sqrt[3]{x+5}}}dx\]    ...(ii)

Adding (i) and (ii)

\[\text{I} + 1 = \int_1^3\frac{\sqrt[3]{x + 5}}{\sqrt[3]{x + 5} + \sqrt[3]{9 - x}}dx + \int_1^3 \frac{\boxed{\sqrt[3]{9 - x}}}{\sqrt[3]{9 - x} + \boxed{\sqrt[3]{x + 5}}}\]

= `int_1^3 (root(3)(x + 5) + root(3)(9 - x))/(root(3)(x + 5) + root(3)(9 - x)) dx`

∴ \[\text{2 I} = \int_1^3 \boxed{1}dx\]

= \[[\boxed{x}]^3_1\]

∴ 2I = \[\boxed{2}\]

∴ I = 1