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Sometimes I come up with an integral with a $dx^2$ term. Whenever I have this, I omit the integral with the $dx^2$ with the idea that it's negligible relative to other integrals with only a $dx$ term, the same way I would when I'm using Newton's method for approximations. I've never read that this is what should be done, but when I do it I get the correct answer. here's an example:

I've been trying to figure out how to find the centre of mass of a lamina. This is what I worked out for the vertical pivot line. If the line is $X_c$ and the total mass is M, then:

$$X_c = \frac{1}{M}\int_a^b f(x)\;dx\left(x+\frac{dx}{2}\right)$$

$$=\frac{1}{M}\int_a^b f(x)x\;dx + \frac{1}{2}\int_a^b f(x)\;dx^2$$

If I then omit the 2nd term I get the answer that I see in textbooks:

$$=\frac{1}{M}\int_a^b f(x)x\;dx$$

This is the way I've figured out on my own and it seems to work but it doesn't seem legit. What happens to that 2nd term really? It may be negligible but it still exists, so...how do you deal with it?

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    how do you get dx/2 there ?2012-02-08
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    The moment arm from the y-axis is $x+\frac{dx}{2}$.2012-02-08
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    This doesn't make sense. As there's a Physics flavour to it, and I am an 'extraordinarily' bad in Physics, I didn't read the problem but, mathematically, the integrand does not make sense. But, nature is very careful not to contradict Mathematics, (and the converse being true, as well), I suggest you revisit the concept!2012-02-08
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    On the one hand, if you get an integrand with a $dx^2$ term, the only reasonable way to interpret it (particularly for a definite integral) is that $dx^2 = 0$. On the other hand, if you get an integrand with a $dx^2$ term, you are probably doing something wrong somewhere--at least from the point of view of a mathematician. Perhaps you'd have better luck asking this question on some sort of physics forum.2012-02-08

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