I don't know how to solve this integral:
$$\int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3 + 1} dx dy$$ I tried substitution with $u = x^3$ , $\sqrt[3]{u} = x$ , $u = x^3 + 1$ and others, but it seems hopeless. I tried using Fubini's theorem then I get $$\int_{\sqrt{y}}^1 \int_0^1 \sqrt{x^3 + 1} dy dx = \int_{\sqrt{y}}^1 \sqrt{x^3 + 1} dy dx $$ but now I'm stuck again.
Thank you for your help.
On
As highlighted, you need to change the order of the integral to make the integration a bit easier (i.e. consider a direction parallel to the y-axis-sketching a graph of the region will definitely help here)
$$\int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3 + 1}\ dx\ dy = \int_0^1 \int_0^{x^2}\sqrt{x^3 + 1}\ dy\ dx $$
$$ \int_0^1 \left(y\sqrt{x^3 + 1}\right)_0^{x^2} \ dx = \int_0^1 x^2\sqrt{x^3 + 1} \ dx $$
Let $u = x^3$, $\frac{du}{dx} = 3x^2$
$$\int_0^1 \frac{(u+1)^{\frac{1}{2}}du}{3} = \left(\frac{2(u+1)^{\frac{3}{2}}}{9}\right)_0^1 $$
I'm sure you can finish the rest
Change the order of integration. If you draw a picture of the integration region, you may see that the integral is actually equal to
$$\int_0^1 dx \, \int_0^{x^2} dy \, \sqrt{x^3+1} =\int_0^1 dx\, \sqrt{x^3+1} \, \int_0^{x^2} dy $$
This should be much easier.