Why I can't reduce an integral divided by another integral? For example,
$$
\frac{\int_{0}^{1}f(x)dx}{\int_{0}^{1}dx} = \int_{0}^{1}f(x),  \space\space\space  (1)
$$
The dx disappears since
$$
\frac{\int_{0}^{1}f(x)dx}{\int_{0}^{1}dx} = \lim_{N\to\infty}\left(\sum_{k=0}^{k\to N}{\frac{f(\frac{k}{N})\frac{1}{N}}{1\times \frac{1}{N}}}\right) = \lim_{N\to\infty}\left(\sum_{k=0}^{k\to N}{\frac{f(\frac{k}{N})}{1}}\right), dx = \Delta x = \frac{1}{N}  \space\space\space  (2)
$$
Above (2) is from the definition of the integral,
Can I do this reduction? If NOT, why?
 A: You cannot, since you're basically assuming
$$\frac{ \sum_i f(i) }{ \sum_i g(i) } = \sum_i \frac{f(i)}{g(i)}$$
This is not true in general. You're assuming this because each integral is basically its own, individual (limit of a) sum. For instance, it is generally not true that
$$\frac{a+b}{c+d} = \frac a c + \frac b d$$
It is also generally untrue that $\int f / \int g = \int (f/g)$ as a result.
A: No, and for several reasons:


*

*$\int_0^1 f(x)$ is not a thing. The operation is $\int_a^b(\bullet)\,dx$: it takes functions and it returns numbers.

*$\frac{\int_0^1 f(x)\,dx}{\int_0^1 g(x)\,dx}=\frac{\lim_{n\to\infty}\sum_{k=0}^{n-1}f(k/n)\frac1n}{\lim_{n\to\infty}\sum_{k=0}^{n-1}g(k/n)\frac1n}$. If both terms converge and the denominator converges to a non-zero number, then this limit is equal to $\lim_{n\to\infty}\frac{\sum_{k=0}^{n-1}f(k/n)\frac1n}{\sum_{k=0}^{n-1}g(k/n)\frac1n}$. Claiming this limit to be equal to $\lim_{n\to\infty}\sum_{k=0}^{n-1}\frac{f(k/n)}{g(k/n)}$ is basically the same fallacy as claiming that $\frac{x+y}{z+w}=\frac xy+\frac zw$, which is of course false.
