# Calculate this integral $\int_0^3\int_0^1 x(x^2+y)^{0.5} dxdy$

I need to calculate the following integral:

$$\int_0^3\int_0^1 x(x^2+y)^{0.5} dxdy$$

I have seen that in wolframalpha solution they used this substitution :

$u=x^2+y$

$du=2xdx$

$\frac{1}{2}*du=xdx$

I don't understand how can they reach to new limits of the integral with a variable $y$, this is what I done.

$$\int_0^3\int_0^1 x(x^2+y)^{0.5} dxdy= \int_0^3(\frac{1}{2}\int_y^{y+1} u^{0.5}du)= \int_0^3(\frac{1}{2}\int_y^{y+1} \frac{u^{1.5}}{1.5} du)dy = \int_0^3(\int_y^{y+1} \frac{(x^2+y)^{1.5}}{1.5} xdx)dy$$

I think it not true, could anybody show/explain the method in this integral?

Once you integrate $$\int_{y}^{y+1}u^{1/2}du$$ you obtain $$\left.\frac{3}{2}u^{3/2}\right|_{y}^{y+1}$$
Replace your limits of integration, and integrate with respect to $y$ and you are done, ie.
$$\int\limits_0^3\int\limits_{y}^{y+1}\frac{u^{1/2}}{2}dudy=\int\limits_0^3\left.\frac{3}{4}u^{3/2}\right|_{y}^{y+1}dy=\frac{3}{4}\int\limits_0^3[(y+1)^{3/2}-y^{3/2}]dy$$