Question

Evaluate: `int_1^4 (root(3)(x + 6))/(root(3)(x + 6) + root(3)(11 - x))*dx`

Answer 1

Let I = `int_1^4 (root(3)(x + 6))/(root(3)(x + 6) + root(3)(11 - x))*dx`   ...(1)

By properly, 

`int_a^bf(x)dx=int_a^bf(a+b-x)dx` we get

I = `int_1^4(root(3)(11-x))/(root(3)(11-x)+root(3)(x+6))dx`    ...(2)

Adding (1) and (2)

2I = `int_1^4(root(3)(x+6)+root(3)(11-x))/(root(3)(x+6)+root(3)(11-x))dx`

2I = `int_1^4 1*dx`

2I = `[x]_1^4`

2I = 4 – 1

∴ 2I = 3 

I = `3/2`