How to evaluate a definite integral that involves $(dx)^2$? For example: $$\int_0^1(15-x)^2(\text{d}x)^2$$
 A: 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.
A: 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.
