A calculus problem by Rahil Sehgal

$\displaystyle\int_{0}^{4} \displaystyle\int_{\sqrt{x}}^{2} \dfrac{1}{1+y^3} dy \; dx$

Consider the limits $x=0 \to 4$ and $y=\sqrt{x} \to 2$

Changing the order of the integration we get $y= 0 \to 2$ and $x= 0 \to y^2$

So, the integral becomes $\displaystyle \int_{0}^{2} \int_{0}^{y^2} \dfrac{1}{1+y^3} dy \; dx \\ =\displaystyle \int_{0}^{2} \dfrac{1}{1+y^3} \int_{0}^{y^2} dx \,\, dy \\= \displaystyle \int_{0}^{2} \dfrac{1}{1+y^3} .y^2 \,\, dy \\=\dfrac{1}{3} [\ln (1+y^3)]_{0}^{2} \\ =\dfrac{2}{3} \ln (3)$