# How to evaluate a definite integral that involves $(dx)^2$?

For example: $$\int_0^1(15-x)^2(\text{d}x)^2$$

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I haven't seen anything like this. Where did it come from? – Ross Millikan Feb 15 '13 at 4:15
What does $(dx)^2$ even mean? – JohnD Feb 15 '13 at 4:17
It's possible this is a typo in whatever book you are using; you might check if there is an errata for your textbook. – Clayton Feb 15 '13 at 4:17
@JohnD: the symbol comes up in second derivatives. Maybe it comes from $\frac {d^2y}{dx^2}=(15-x)^2$ but I am guessing. – Ross Millikan Feb 15 '13 at 4:19
That's meaningless. While "dx" looks like something numerical, it is not, and it cannot be squared this way. It's a bit like asking what an apple squared is. – Thomas Andrews Feb 15 '13 at 4:28

There's an old joke. A mathematician, a physicist and a engineer are asked by a student what the meaning of $$\int \frac{1}{dx}$$ is.

The mathematician says it is meaningless.

The physicist ponders it for a moment and wonders if there is some way to give it meaning.

The engineer says, "Hmmmm, I used to know how to do that."

This is a misuse of notation - $(dx)^2$ is essentially meaningless, because $dx$ is not something numeric, it is rather an indication of how we are measuring "area" in the integral.

If you replaced $(dx)^2$ with $d(x^2)$, there would be a meaning we could apply.

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and what is $\int \frac {setting curtain on fire}{dx}$? – Ross Millikan Feb 15 '13 at 4:41
+1 for the joke. (trust me, I´m engineer) – dwarandae Feb 15 '13 at 4:50

Just guessing, but maybe this came from $\frac {d^2y}{dx^2}=(15-x)^2$ The right way to see this is $\frac d{dx}\frac {dy}{dx}=(15-x)^2$. Then we can integrate both sides with respect to $x$, getting $\frac {dy}{dx}=\int (15-x)^2 dx=\int (225-30x+x^2)dx=C_1+225x-15x^2+\frac 13x^3$ and can integrate again to get $y=C_2+C_1x+\frac 12 225x^2-5x^3+\frac 1{12}x^4$ which can be evaluated at $0$ and $1$, but we need a value for $C_1$ to get a specific answer.

As I typed this I got haunted by the squares on both sides and worry that somehow it involves $\frac {dy}{dx}=15-x$, which is easy to solve.

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