How to compute $\int_{a}^{b}\sqrt{\mathrm dx}$? I do not think setting $\sqrt {\mathrm dx} = \mathrm du$ is the right idea. Is it even sensible to have $\mathrm dx$ square rooted?
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1What book/paper/homework did you see this in? – J. M. ain't a mathematician Mar 04 '19 at 06:24
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1Just my own curiosity, wondering what would happen if $dx$ gets other forms. For example, can we have $dx^2$? I do not know if that is meaningful. – Rob Mar 04 '19 at 06:31
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3I asked because it's not really sensible to apply a square root to a differential. One could perhaps interpret your $\mathrm dx^2$ as $\mathrm d(x^2)=2x\mathrm dx$ however. – J. M. ain't a mathematician Mar 04 '19 at 06:33
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A sensible way to define such an expression would be through the Reimann sums $$\int_a^b \sqrt{\mathrm dx}=\lim_{\Delta x \to 0} \sum_{x=a}^{x=b}\sqrt{\Delta x}$$ Taking $a=0$, $b=1$ and $\Delta x =\frac1n$ we get $$\int_0^1 \sqrt{\mathrm dx}=\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{n}}=\lim_{n \to \infty}\sqrt{n}$$ which clearly diverges. Its pretty easy to see that $a=0$ and $b=1$ are not special here.
mrtaurho
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