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

JohnD
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gauss115
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    I haven't seen anything like this. Where did it come from? – Ross Millikan Feb 15 '13 at 04:15
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    What does $(dx)^2$ even mean? – JohnD Feb 15 '13 at 04:17
  • @RossMillikan, A friend asked.. I dont see how to deal with these definite integral or indefinite integral :( – gauss115 Feb 15 '13 at 04:17
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    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 04:17
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    @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 04:19
  • @RossMillikan: $(dx)^2$ in an integral and the $dx^2$ in the denominator of your expression involve the same letters and thus look similar. But unless someone declares what the symbol $(dx)^2$ means in a definite integral, I have no idea... – JohnD Feb 15 '13 at 04:25
  • Yea, I guess so..Cant see any meaning for $(dx)^2$ – gauss115 Feb 15 '13 at 04:25
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    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 04:28
  • @JohnD:I agree. I was speculating on where it came from. I didn't claim it had standard mathematical meaning. But I can see someone starting from the equation I gave and turning it into the integral in the question. – Ross Millikan Feb 15 '13 at 04:28
  • @RossMillikan: Gotcha. – JohnD Feb 15 '13 at 04:34

2 Answers2

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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.

Thomas Andrews
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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.

Ross Millikan
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