0
$\begingroup$

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.

  • 0
    What do you mean that it started from $6(x-y)$?2012-12-01
  • 4
    $\int\int6(x-y) dx dy = 3(x^2y - y^2x)$ if you ignore constants. But I don't understand what you are really asking. Please be more specific.2012-12-01
  • 1
    @cheepychappy: It seems like what you have written is what OP is asking. I suggest you make an answer...2012-12-01

2 Answers 2

2

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: $$\begin{align}\iint 6(x-y) \quad dydx\\=\iint 6x-6y \quad dydx\\=\int 6xy-3y^2 \quad dx\\=3x^2y-3xy^2+C\\ =3(x^2y-3xy^2)+C \end{align}$$

  • 0
    Thank you so much @Kyle! =) May I get your kind advice on how the first part works out on how you got to 3(x^2)−6xy? This is how I thought it should work 6(x^2 / 2) - 6 (y), but couldn't seem to make sense of it. How did the 'x' creep up to becoming 6xy? Many thanks once again.2012-12-02
  • 0
    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.2012-12-02
  • 0
    Thank you so much @Kyle! =) May I get your kind advice on how the first part works out on how you got to 3(x^2)−6xy? This is how I thought it should work 6(x^2 / 2) - 6 (y), but couldn't seem to make sense of it. How did the 'x' creep up to becoming 6xy? Many thanks once again.2012-12-02
  • 0
    Thank you so much @Kyle . Really appreciate your advice. May I clarify if this {y[6x - 6(0.5)y]} is why it became 6xy - 3y^2 from the second to third line? In the final line, may I also clarify that it's 3[(x^2)(y) - (x)(y^2)] + C2012-12-02
  • 0
    Yep that would be correct!2012-12-02
  • 0
    Many thanks @Kyle2012-12-03
1

$\int\int6(x-y) dx dy = 3(x^2y - y^2x)$ if you ignore constants.