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Bear with my naivety, I wanted to ask if it is possible to cancel out $\int$ with a $\frac{\mathrm d}{\mathrm dx}$. I had $\frac{\partial}{\partial v}$ in a question and I took $\partial v$ to the other side and took integration on both sides. So now on left hand side only $\partial$ remains so will it cancel out with the integral?

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Saad Ibrahim Sherpao
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    seems a bit harsh that this received 2 downvotes and no comments, this is a new user. – operatorerror Jan 06 '17 at 04:05
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    Maybe this is OK, maybe not. Without seeing the derivation we can't tell. Probably the d should be part of the d(something) that you integrated. – Ross Millikan Jan 06 '17 at 04:08
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    @qbert Lots and lots of people on this site are lazy. They vote without contributing to the question in any other way, as if somehow their downvote will by itself magically fix whatever problem they thought they saw in the question. – layman Jan 06 '17 at 04:09
  • The equation is as follows dM/dx=V Now if I take dx to the other side and then integrate both sides, so I am anticipating that only M remains on the left hand side. Am I right. I am not a mathematics major, so don`t know much about this stuff. – Saad Ibrahim Sherpao Jan 06 '17 at 04:15

2 Answers2

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Usually when we take integration we remove d or $\partial$

You can say it is cancel out or it is integrated.

Example -

$\frac{d}{dv} x = v^2 + 3$

$dx = (v^2 + 3) dv$

On integrating both sides,

x = $\int(v^2 + 3)dv$

Kanwaljit Singh
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I wouldn't say that $\frac{\mathrm d}{\mathrm dx}$ and $\int$ cancel each other out. Consider the following examples $$\frac{\mathrm d}{\mathrm dx}\int e^t\ \mathrm dt=\frac{\mathrm d}{\mathrm dx}\left(e^t+C\right)=0$$ $$\frac{\mathrm d}{\mathrm dx}\int_0^2 e^x\ \mathrm dx=\frac{\mathrm d}{\mathrm dx}\left(e^2-1\right)=0$$ Study the fundamental theorem of calculus for a deeper understanding of what's going on.

I'll also address what you mentioned in the comments. Although $\frac{\mathrm dM}{\mathrm dx}$ look and act like a ratio, in modern mathematical analysis, it's not a ratio. It just so happens that $$\mathrm dM=\frac{\mathrm dM}{\mathrm dx}\ \mathrm dx$$ Therefore $$V=\frac{\mathrm dM}{\mathrm dx}$$ $$\int V\ \mathrm dx=\int\frac{\mathrm dM}{\mathrm dx}\ \mathrm dx=\int\ \mathrm dM=M+C$$

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