Why is the integral of $dx$ equal to $x$?

Question

Please give a thorough explanation, not just "$dx$ is the derivative of $x$, so the antiderivative of $dx$ is $x$, duh".

Answer 1

It's not. It's equal to x+C.

More seriously, You're not integrating "dx". When we write $\int dx$,we mean to solve the problem $\int 1dx$. "dx" has no meaning outside "I'm trying to integrate by x" or "I'm taking the x derivative", unless you possibly mean $x\cdot d$.

Answer 2

Hint: the integral $\int_0^xdt$ is the area of a rectangle by length of $x$ and width of 1.

Answer 3

To take a definite integral from $a$ to $b$ of $dx$ is to take the infinitesimal quantity $dx$ and sum the infinite number of occurrences of it between $a$ and $b$. $dx$ is defined so that the sum from $a$ to $a+1$ of $dx$ is always 1. Or, if you view $dx$ as the mesh of a partition, as in the Riemann integral's definition, then you're approximating a set of quantities whose sum is by definition the length of the interval of integration.

Answer 4

IMO, a better viewpoint on the problem is:

Okay, definite integrals are designed so that $\int_P \, df = f(b) - f(a)$, where $a,b$ are the endpoints of the path $P$ (think of a path as being a line segment). Why can we compute this as a limit of Riemann sums?

The answer is analysis -- we take the problem and break it into smaller pieces that are hopefully easier to solve. In this case, we split the path $P$ into a sequence of many smaller paths (e.g. split $[0,5]$ into $[0,1]$ then $[1,2]$ then $[2,3]$ then $[3,4]$ then $[4,5]$).

If $f$ is a continuously differentiable function, then differential approximation tells us that over a small interval $[x, x+\epsilon]$, the definite integral of $df$ -- that is, the value $f(x+\epsilon) - f(x)$ -- is approximately $f'(x) \epsilon$ (we could replace $x$ with any point in $[x, x+\epsilon]$ if we like).

If we add this up over all the small paths, the result is a Riemann sum, and if we take a suitable limit, the approximation error goes to zero.

For $f(x) = x$ specifically, we're adding up $\epsilon$ for each of the paths: no matter how we split $[a,b]$ into paths, the resulting sum is going to be $b-a$.

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As usual, non-standard analysis gives us a cleaner description: we split the path $P$ into a hyperfinite number of paths $P_n$, each one of infinitesimal length. Adding up $f'(x) \epsilon$ for each one gives us a value infintiesimally close to the value of the integral.

Answer 5

From $0$ t0 $x > 0$ and $N = 1,2,3,\ldots$: $$ \overbrace{\quad% {x \over N}\times 1\quad +\quad {x \over N}\times 1 \quad + \quad \cdots\quad +\quad {x \over N}\times 1\quad}^{N\,\,\,\,\, \mbox{times}} = N\times{x \over N} = \color{#ff0000}{\large x} $$