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 :




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.



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