# Is how I treat $dx^2$ how you do it?

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 ? – Geralt of Rivia Feb 8 '12 at 17:53
The moment arm from the y-axis is $x+\frac{dx}{2}$. – Korgan Rivera Feb 8 '12 at 17:54
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! – user21436 Feb 8 '12 at 18:03
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. – Charles Staats Feb 8 '12 at 19:12

Your expression, with a $(dx)^2$ in the integral, makes me uncomfortable. My comfort is not of overwhelming importance. But your expression is also highly non-standard, and may contribute to confusion later.

It is not how I do it when thinking informally. Imagine the vertical strip from $x$ to $x+h$ where $h$ is very very small. If you like, you can write $dx$ instead of $h$, and even think of $dx$ as being "infinitesimal." I will use $h$, but there are good arguments for $dx$, including a centuries-old tradition that still survives in some applied areas.

We find an approximate expression for the moment of our strip about the $y$-axis. If you had been using $h$, your expression would have been $hf(x)(x+h/2)$. That's a somewhat better approximation than $(hf(x))(x)$. The intuition you had is that it is more accurate to think of the mass of the strip as concentrated at a distance $x+h/2$ from the $y$-axis. For reasonably-behaved functions, this is true.

Note that your expression differs from the simpler expression $(hf(x))(x)$ by $(h^2/2)(xf(x))$. If $h$ is very small, and $f$ is at all respectable, this difference is tiny in comparison with $(hf(x))(x)$.

Actually, your small change, from $x$ to $x+h/2$, could have been improved on. For the area of the strip is better approximated by $hf(x+h/2)$. If our function is differentiable, then by the tangent line approximation, $f(x+h/2)\approx f(x)+(h/2)f'(x)$. If you take this correction into account, the estimate for the moment becomes $$h(x+h/2)(f(x)+(h/2)f'(x)).$$ So it would have been more accurate to use $hxf(x)[1+h/2)(f(x)+xf'(x)]$. But still, here we also have a main term $hxf(x)$, plus a (partial) correction term that is very small compared to the main term.

However, we are getting too fancy. A standard way to think informally is that the moment of the strip is $hxf(x)$ plus a term that approaches $0$ faster than any constant multiple of $h$, so that the error in our approximation is negligible compared to $hxf(x)$. For nice functions $f$, the error in our approximation behaves like a constant times $h^2$, and for $h$ small enough, $h^2$ is negligible compared to $h$.

Even more informally, we throw away terms in $h^2$ or higher powers of $h$!

So we have divided our interval $[a,b]$ into strips of tiny width. We want to "add up" the moments of these strips. There is a mnemonic advantage to using $dx$ instead of $h$. We are "adding up" moments $xf(x)dx$. The "sum" is $\int_a^b xf(x)dx$. Recall that the integral symbol $\int$ was originally an elongated $S$, for sum (in Latin).

We have discarded the terms in powers of $h$ that are $2$ or higher before writing down the integral. They leave no trace in the integral.

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