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Mathematics 27 Online
OpenStudy (anonymous):

Can anyone help me with a double integral question?

OpenStudy (anonymous):

\[\int\limits_{?}^{?}\int\limits_{?}^{?}(5x^4y^3)/(x^5+1)dA\]

OpenStudy (anonymous):

0<=x<=1 0<=y<=1

OpenStudy (amistre64):

\[\int\int_D~\frac{5x^4y^3}{x^5+1}~dA\] \[\int_{0}^{1}\int_{0}^{1}~\frac{5x^4y^3}{x^5+1}~dx~dy\]

OpenStudy (amistre64):

i believe all that interprets to something along those lines

OpenStudy (anonymous):

yep thats how it looks on my paper

OpenStudy (amistre64):

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

OpenStudy (amistre64):

and lets add some brackets to divvy this up into more doable pieces \[\int_{0}^{1}\left(\int_{0}^{1}~\frac{5x^4y^3}{x^5+1}~dy\right)~dx\] \[\int_{0}^{1}\frac{5x^4}{x^5+1}\left(\int_{0}^{1}~y^3~dy\right)~dx\] \[5\int_{0}^{1}\frac{x^4}{x^5+1}\left(\int_{0}^{1}~y^3~dy\right)~dx\]

OpenStudy (anonymous):

ahhh that looks better already lol

OpenStudy (amistre64):

do you see why i did that? or know why we can do that?

OpenStudy (anonymous):

divide it up? because we have to do the parts separate right?

OpenStudy (amistre64):

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.

OpenStudy (anonymous):

and depending on which you do the other will act as constants

OpenStudy (amistre64):

excellent :)

OpenStudy (anonymous):

yep lol dident submit quick enough :)

OpenStudy (anonymous):

ok so question

OpenStudy (anonymous):

when can we take the y completly out without the x^3+1 denominator?

OpenStudy (amistre64):

and, once we calculate the y part, its whole value is a constant so we can even rewrite this as \[5~*~\int_{0}^{1}\frac{x^4}{x^5+1}dx~*~\int_{0}^{1}~y^3~dy\]

OpenStudy (amistre64):

its a factor of the whole thing, imagine\[\frac{3*4}{2+7}=\frac{3}{2+7}*\frac{4}{1}\]

OpenStudy (anonymous):

alright makes sense

OpenStudy (anonymous):

so i did the y part and i got 1/4

OpenStudy (amistre64):

1/4 is good for that

OpenStudy (anonymous):

so would the integral now be \[\frac{ 5 }{ 4 }\int\limits_{0}^{1}\frac{ x^4 }{ x^5+1 }\]

OpenStudy (amistre64):

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

OpenStudy (anonymous):

makes sense

OpenStudy (anonymous):

AH YES I GOT IT!

OpenStudy (anonymous):

lol thank you very much

OpenStudy (amistre64):

:) youre welcome

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