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

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

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and by integral i mean the definite integral –  nuhrin Sep 28 '13 at 22:59
Try calculating the area of a rectangle with height=1 ( since $dx$=$1dx$, based on the Real line going from $a$ to $b$ –  DBFdalwayse Sep 28 '13 at 23:03
The derivative of $x$ isn't "dx", it's $1$. So $x$ is an antiderivative of $1$. –  littleO Sep 28 '13 at 23:08
@DBF, Just saw your comment, Sorry for repeating your answer. –  Arash Sep 28 '13 at 23:15
random follow up question...i have the equation force = mass*(dv(t)/dt), where v is velocity and t is time. the book says to multiply both sides by dt to get f(dt) = m(dv), then integrate the left equation from a=t(initial) b=t(final) and integrate the right equation from a=v(initial) to b=v(final). can you explain why these operations are legal? –  nuhrin Sep 28 '13 at 23:24

## 5 Answers

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 $x\cdot d$.

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random follow up question...i have the equation force = mass*(dv(t)/dt), where v is velocity and t is time. the book says to multiply both sides by dt to get f(dt) = m(dv), then integrate the left equation from a=t(initial) b=t(final) and integrate the right equation from a=v(initial) to b=v(final). can you explain why these operations are legal? –  nuhrin Sep 28 '13 at 23:17
@nuhrin: This should hopefully explain things: math.stackexchange.com/questions/47092/… –  Hovercouch Sep 28 '13 at 23:32
Of course, "you're integrating 'dx'" can be made to make sense, and it's a really useful notion, even at the introductory level. Unfortunately, AFAIK it's usually only explicitly discussed in differential or algebraic geometry. However, the ideas are still usually used often in calculus and physics, giving the student a chance to learn it through osmosis. –  Hurkyl Sep 29 '13 at 1:56

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 $b-a$.

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.

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

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random follow up question...i have the equation force = mass*(dv(t)/dt), where v is velocity and t is time. the book says to multiply both sides by dt to get f(dt) = m(dv), then integrate the left equation from a=t(initial) b=t(final) and integrate the right equation from a=v(initial) to b=v(final). can you explain why these operations are legal? –  nuhrin Sep 28 '13 at 23:25
It does not makes sense to have the bound variable in the limits. –  L. F. Sep 28 '13 at 23:31
@L.F., it was just a typo. –  Arash Sep 28 '13 at 23:33

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.

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One has a very precise definition of what an integral is, why would you want to change it into "...is to take the infinitesimal quantity..."? –  Pedro Tamaroff Sep 28 '13 at 23:18
random follow up question...i have the equation force = mass*(dv(t)/dt), where v is velocity and t is time. the book says to multiply both sides by dt to get f(dt) = m(dv), then integrate the left equation from a=t(initial) b=t(final) and integrate the right equation from a=v(initial) to b=v(final). can you explain why these operations are legal? –  nuhrin Sep 28 '13 at 23:21
@nuhrin This follows from the chain rule. On definite integrals, the change of variables requires a change of bounds of integration, but $v$ being a function of $t$ in the first place, the appropriate bounds are the values of $v$ at the initial and final time; abbreviated to just the initial and final $v$. –  Loki Clock Sep 28 '13 at 23:31
@LokiClock: Infinitesimals are physicists' and engineers' domain — nothing to do with differential forms, Kähler differentials, or cohomology. The math is made rigorous by model theory and non-standard analysis. Still nothing to do with what you cite. –  Ted Shifrin Sep 29 '13 at 4:21
@TedShifrin Sure it has to do with those things. The dual space to the tangent space is the cotangent space. Viewing the cotangent space as the cochains and the tangent space as the chains, you can formally integrate a cochain - a formal infinitesimal - over a set of chains by summing over the values of the evaluation map as the chain varies. –  Loki Clock Sep 29 '13 at 11:00

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

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