# Troubles with double integral

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.

• You didn't switch the order of integration correctly. When you do it right, you'll be a happy camper. – zhw. Jul 21 '15 at 20:37
• Be careful: changing the order of integration is not just switching the integral symbols. You have to change also the integration limits (that's the most difficult part). – A.Γ. Jul 21 '15 at 20:43

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.

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