Evaluate the integral by reversing the order of integration

Evaluate the integral by reversing the order of integration. integral_0^1 integral_squareroot x^1 2/y^3 + 1 dy dx_________

Solution

Solution: Given

I =\\int_{0}^{1}\\int_{\\sqrt{x}}^{1} \\frac{2}{1+y^{3}}dydx ......(1)

Here outer limit is 0 \\leq x \\leq 1 and inner limit is \\sqrt{x} \\leq y \\leq 1.

By reversing the order of integration, we have

0 \\leq y \\leq 1 and 0 \\leq x \\leq y^{2}.

So I = \\int_{0}^{1}\\int_{0}^{y^{2}} \\frac{2}{1+y^{3}}dxdy

= \\int_{0}^{1} [\\frac{2x}{1+y^{3}]_{0}^{y^{2}}dy

= \\int_{0}^{1} [\\frac{2y^{2}}{1+y^{3}]dy

=(2/3)[log(1+y^{3})_{0}^{1} = (2/3)[log(1+1) -0

=(2/3)[log(2).(answer)