Would I be right to think that $$\int dx \,\,\,\frac{\partial}{\partial x} f(x,y)=f(x,y)$$ Or are there pathological cases?
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It seems correct to me. – Anastasiya-Romanova 秀 Apr 15 '14 at 12:31
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1Up to a constant, yes. Note that to be differentiable, a function must be continuous, which guarantees it is Riemann integrable. – CunningTF Apr 15 '14 at 12:36
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1Up to the fact that the result might be $f(x,y)+C$, it looks reasonable. – Siminore Apr 15 '14 at 12:36
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It depends on what you mean with $\int dx ,,,\frac{\partial}{\partial x} f(x,y)=f(x,y)$. If you mean that that equality is shorthand for "differentiating with respect to $x$ on both sides gives an equality", then it is correct. But there's another reasonable interpretation and accordidng to that one, it is wrong. – Git Gud Apr 15 '14 at 12:38
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1Related. – Git Gud Apr 15 '14 at 12:44
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If $x$ and $y$ are independent variables (and thus the $y$ is held constant during integration), then it is true that
$$ \int \frac{\partial f}{\partial x} dx = f(x,y) + C(y) $$ where $C(y)$ is equivalent to the integration constant for the univariate case. As such, up to the "constant", you are right.
If $y=y(x)$, then it is not that simple. For instance, if $f(x,y)=x^2-xy+y^2$ and you integrate along the line $y=2x$, then you are actually integrating $$ \int \frac{\partial f}{\partial x} dx = \int (2x-y) dx = \int 0 dx = 0 $$
Glen O
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