# Basic Calculus (Integration)

I came across an integration in my notes, but I can't seem to make out the steps to getting the answer.

All I got was the equation started from $6(x-y)$ and ended as $3(x^2y - y^2x)$.

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What do you mean that it started from $6(x-y)$? –  Thomas Dec 1 '12 at 20:42
$\int\int6(x-y) dx dy = 3(x^2y - y^2x)$ if you ignore constants. But I don't understand what you are really asking. Please be more specific. –  cheepychappy Dec 1 '12 at 20:45
@cheepychappy: It seems like what you have written is what OP is asking. I suggest you make an answer... –  Thomas Dec 1 '12 at 21:01

So if you are integrating $$\iint 6(x-y)dxdy$$ Start like the insides like you were taught:\begin{align}\int 6(x-y)\quad dx\\ =3x^2-6xy\end{align} Dont forget to think of $y$ as a constant and we can ignore the $+C$ since we have to integrate again. We are now left with: $$\int3x^2-6xy\quad dy$$ Now integrate this with respect to $y$ viewing $x$ as a constant now. \begin{align}\int3x^2-6xy\quad dy\\ =3x^2y-3xy^2+C\\ =3(x^2y-xy^2)+C\end{align} You now have the steps of what is in your notes! Except you are missing the $+C$ in your notes. You could also do it in reverse: \begin{align}\iint 6(x-y) \quad dydx\\=\iint 6x-6y \quad dydx\\=\int 6xy-3y^2 \quad dx\\=3x^2y-3xy^2+C\\ =3(x^2y-3xy^2)+C \end{align}
Because by distributing $6(x-y)$ as you know we would have $6x-6y$ and by basic rules of multiple integration, we treat $y$ as a constant. You're half right in that it is indeed $6(\frac{x^2}{2})$ but you also need to integrate upon the $6y$ with respect to $x$, thus leaving you with $6xy$. See the part I added in my answer about how you could also do it in reverse by doing dy first, maybe that will make it a little more clear. –  TheHopefulActuary Dec 2 '12 at 8:07
$\int\int6(x-y) dx dy = 3(x^2y - y^2x)$ if you ignore constants.