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I know this is a very simple question but why is this wrong?$$\int(xdy+ydx)=\int xdy+\int ydx=x\int dy+y\int dx=2xy$$

I saw a similar question on Stack Exchange, but it was too complicated for me to understand. I am in 11th Grade and I have just done basic differentiation and integration for physics. Any help would be appreciated!

J.G.
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Niescte
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  • Could you link to the similar question you saw elsewhere? – J.G. Jul 07 '20 at 06:06
  • https://math.stackexchange.com/questions/1300023/easy-question-int-xdyydx here – Niescte Jul 07 '20 at 06:07
  • generally everything is wrong if you integrate without limits. In first place you need to define the integral properly making it a definite integral, after you will need to parameterize the functions $x$ and $y$ to finally get an answer. take a look here – Masacroso Jul 07 '20 at 06:10
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    @Masacroso “everything is wrong if you integrate without limits” — Seriously? That goes against the entire concept of an indefinite integral. – gen-ℤ ready to perish Jul 07 '20 at 06:50
  • @gen-zreadytoperish an indefinite integral is, in general, not well defined, by example $\int x^{-1/2} \mathop{}!d x=\ln|x|+C$ doesn't make sense if you consider, by example $\int_{-1}^1 x^{-1}\mathop{}!d x$. There are plenty of examples with different behavior showing how bad behaved is the concept of indefinite integral. – Masacroso Jul 08 '20 at 02:48
  • @Masacroso Everyone knows that particular indefinite integral, like some others, has a restriction on the domain of integration. That doesn’t mean “everything is wrong.” – gen-ℤ ready to perish Jul 08 '20 at 02:50
  • @gen-zreadytoperish when something is not well-defined then "everything is wrong" is to me a fine and correct way to describe it. – Masacroso Jul 08 '20 at 02:51
  • @Masacroso My point is that you were saying one must define limits of integration before continuing with this problem, and as J.G. demonstrated, that’s simply not true. – gen-ℤ ready to perish Jul 08 '20 at 02:53

1 Answers1

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Since $x,\,y$ are in general not independent, you can't treat $x$ as a constant as in $\int xdy=x\int dy$. Your original problem would make this clearer if you wrote $x(y)dy+y(x)dx$. In fact,$$\int(xdy+ydx)=\int d(xy)=xy+C.$$

J.G.
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  • But if we move parallel to the x-axis first and then parallel to the y-axis, won't we get 2xy. – Niescte Jul 07 '20 at 06:10
  • @Niescte No, because you'll move at an angle. Let's consider the example $y=x$. As you try to vary $x$, you'll vary $y$ instead & vice versa. In this case, the problem reduces to $\int2xdx=x^2+C=xy+C$. – J.G. Jul 07 '20 at 06:13
  • Ok, i get it now. If we move parallel to the x axis first, then the value of x should change the next time when we integrate it moving parallel to y. – Niescte Jul 07 '20 at 06:21