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I know dy/dx for example means "derivative of y with respect to x," but there's another context that confuses me. You will generally just see a dx term sitting at the end of an integral equation and I just don't know exactly what it means or why it's there.

For instance, if I put into Wolfram Alpha "integral of 2x", it writes out:

enter image description here

That dx in there! I guess it's saying "the integral of 2x with respect to x" ? But why is that even necessary? Isn't it implied? It's annoying because it looks like multiplication, like it means "2 * x * dx" and it's just kind of misleading notation.

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    $\begingroup$ See Reimann Stieltjes Integral .. $\endgroup$ – r9m Jun 17 '15 at 16:54
  • $\begingroup$ This is going to be one of those viral questions - guys this isn't well researched. Any book with "real analysis" in the title will cover this. $\endgroup$ – Alec Teal Jun 17 '15 at 17:06
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    $\begingroup$ This has surely been asked in this site before. Did you take the time to search? $\endgroup$ – Mariano Suárez-Álvarez Jun 17 '15 at 17:09
  • $\begingroup$ I hate how these questions, the easy and already asked ones, get about 10 different answers in about an hour, while actual legitimate non-duplicate questions are lucky if they get 1. $\endgroup$ – Zach466920 Jun 17 '15 at 17:15
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What is $$ \int st^2 ? $$ Is $s$ the variable or is $t$ the variable? Sure, sometimes it is "obvious" what the variable is, but this is not always the case.

If you want to know about what the $dx$ "actually means", this question has been asked many times here. See for example What does $dx$ mean?

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    $\begingroup$ This answer shouldn't be upvoted like that. It is almost empty and not informative. It simply directs to another page for answer. Answers should be at least informative above all. I can understand what you mean, but some people may not. $\endgroup$ – Yannis Dran Apr 4 at 22:31
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It's because an integral means you are summing over a lot of very thin rectangles under a curve. The height of the rectangle is f(x) and the width is called $\delta x$ (These two symbols should be read as a single symbol, it doesn't mean $\delta \times x$). So you would write it like:

$\sum f(x) \delta x$

The $\Sigma$ sign is a sigma and stands for "sum". In an integral you take the limit as $\delta x$ goes to zero. So we replace the sigma with another type of s: $\int$. And the $\delta$ gets changed to a d. So it is now written:

$\int f(x) dx $

and it is the "integral of f(x) with respect to x". But the dx doesn't mean anything on it's own.

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I personally like to think of the $\int$ sign and the "d" of $dx$ like left- and right-parentheses: they delimit the extent of the integrand. The "x" in $dx$ is there to tell you the variable of integration.

There's an intuitive notion as well, which is that you're summing up values of the integrand as $x$ varies, and multiplying each by a tiny "width" to get the area of a rectangle, and summing these areas gets the integral...if you take a limit as the "tiny width $\Delta x$ gets smaller and smaller." The "$dx$" lets people think informally that you're multiplying a height, $f(x)$, by an "infinitesimal width", $dx$, and then taking an infinite sum.

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The way I see it the $dx$ represents the limit as $\Delta x_k$ goes to zero in the Riemann integral definition, which is $$ \int_{a}^{b}f(x)\,dx = \lim \limits_{\Delta x_k \to 0} \sum_{k=1}^n f(x_k)\Delta x_k$$

Which makes sense if you are familiar with approximating the area under the curve with a figure like a rectangle (Riemann sums). You are finding sum of the the areas of rectangles whose length gets smaller. When you take the limit as the number of rectangles increases to infinity, you get the area under the curve. Here is a helpful animation that shows the process of $\Delta x_k$ going to 0.

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In some sense, $dx$ just means an arbitrarily small change in $x$. To understand this better, let's look at your example $$\int2x\,dx,$$ and for simplicity say we're considering the definite integral $$\int_{0}^{1}2x\,dx.$$ If we use the Riemann integral, we have $$\int_{0}^{1}2x\,dx =\lim\limits_{k\rightarrow\infty} \sum\limits_{i=0}^{k-1}\left(2\frac{i}{k}\right)\left(\frac{i+1}{k}-\frac{i}{k}\right).$$

We typically need to be more careful here and prove that the integral exists and everything, but for our purposes let's assume that this equality holds.

So how I like to think of it is that "$\int_{0}^{1}$" means the same thing as "$\lim\limits_{k\rightarrow\infty}\sum\limits_{i=0}^{k-1}$", "$2x$" means the same thing as "$\left(2\frac{i}{k}\right)$", and $dx$ means the same thing as "$\left(\frac{i+1}{k}-\frac{i}{k}\right)$".

This line of thinking is of course not completely correct since $\lim\limits_{k\rightarrow\infty}\left(\frac{i+1}{k}-\frac{i}{k}\right)=0$, but it conveys the idea that $dx$ just refers to an arbitrarily small change in $x$.

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Suppose $x$ is a constant,then

$$\int (x^2+1)\,dy=(x^2+1)y+C$$.

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