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 $3(x^2y - y^2x)$.
Appreciate any advice please.
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 $3(x^2y - y^2x)$.
Appreciate any advice please.
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: $\begin{align}\iint 6(x-y) \quad dydx\\=\iint 6x-6y \quad dydx\\=\int 6xy-3y^2 \quad dx\\=3x^2y-3xy^2+C\\ =3(x^2y-3xy^2)+C \end{align}$
$\int\int6(x-y) dx dy = 3(x^2y - y^2x)$ if you ignore constants.