Question

Evaluate `int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx`

Answer 1

Let I = `int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx`

We use the property,`int_0^af(x)dx = int_0^af(a-x)dx`

Hence in I, we replace x by 3 - x

I = `int_0^3root3(3-x+4)/(root3(3-x+4)+root3(7-3+x))  dx`

I = `int_0^3root3(7-x)/(root3(7-x)+root3(x+4))  dx`

Adding (1) and (2), we get

`I =int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx +int_0^3root3(7-x)/(root3(7-x)+root3(x+4))  dx`

`=int_0^3 (root3(x+4) + root3(7-x))/(root3(x+4) + root3(7-x))  dx`

`int_0^31 dx = [x]_0^3 = 3-0 =3`

∴ `I = 3/2`

Hence, `int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx =3/2`