Can anyone help me with a double integral question?
\[\int\limits_{?}^{?}\int\limits_{?}^{?}(5x^4y^3)/(x^5+1)dA\]
0<=x<=1 0<=y<=1
\[\int\int_D~\frac{5x^4y^3}{x^5+1}~dA\] \[\int_{0}^{1}\int_{0}^{1}~\frac{5x^4y^3}{x^5+1}~dx~dy\]
i believe all that interprets to something along those lines
yep thats how it looks on my paper
since there is only one y part; lets flop the dxdy into dydx since the bounds are the same this only changes which one we do first and has no other effect
and lets add some brackets to divvy this up into more doable pieces \[\int_{0}^{1}\left(\int_{0}^{1}~\frac{5x^4y^3}{x^5+1}~dy\right)~dx\] \[\int_{0}^{1}\frac{5x^4}{x^5+1}\left(\int_{0}^{1}~y^3~dy\right)~dx\] \[5\int_{0}^{1}\frac{x^4}{x^5+1}\left(\int_{0}^{1}~y^3~dy\right)~dx\]
ahhh that looks better already lol
do you see why i did that? or know why we can do that?
divide it up? because we have to do the parts separate right?
yes; and also, the "other" bits are constant when seen from the point of view of the part we are integrating with respect to; and constants can pull out.
and depending on which you do the other will act as constants
excellent :)
yep lol dident submit quick enough :)
ok so question
when can we take the y completly out without the x^3+1 denominator?
and, once we calculate the y part, its whole value is a constant so we can even rewrite this as \[5~*~\int_{0}^{1}\frac{x^4}{x^5+1}dx~*~\int_{0}^{1}~y^3~dy\]
its a factor of the whole thing, imagine\[\frac{3*4}{2+7}=\frac{3}{2+7}*\frac{4}{1}\]
alright makes sense
so i did the y part and i got 1/4
1/4 is good for that
so would the integral now be \[\frac{ 5 }{ 4 }\int\limits_{0}^{1}\frac{ x^4 }{ x^5+1 }\]
yes, and as i stare at the, the 5 would be more useful inside the integral; that way the top is the derivative of the bottom
makes sense
AH YES I GOT IT!
lol thank you very much
:) youre welcome
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