## Question on the functionality of dx

My question is based around when you integrate dx as if it were a function and when it just represents the quantity that is infinitely small.

Say we have the function y=3, a straight horizontal line. We want to find the area under this line from x = 0 to x = 10. I know, you wouldn't use calculus to solve this because it is just a rectangle, but humor me. You could set up the integral like so:

$\int$3dx with limits from 0 to 10. In this situation, 3 is a constant, so you can pull it out, leaving 3$\int$dx from 0 to 10. Integration of the dx leaves just x, and plugging this into the limits gives a value of 10, which you then multiply by 3. This gives 30, the correct answer as easily calculated also by geometry.

Now we have the function y=x. We want to again find the area under the line from 0 to 10, so the integral becomes $\int$xdx. Here when we integrate, we simply use the power rule to integrate, x becomes x^2/2, and we plug in our 10 value. We get 50, the correct answer, also easily calculated through geometry. The dx, effectively, was ignored during the integration process.

My question is: why in the first case did we actually use the dx and integrate it, but in the second example it was there just to show us what the differential is?

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 The $\mathrm{d}x$ in terms of definite integration is an infinitely small rectangle beyond feasible measurements. When you integrate a function such as $f(x) = x$ from $a$ to $b$, you're finding the height from the $x$-axis to the graph of $f(x) = x$. In theory, you're multiplying the height of the curve by an infinitely small rectangle of width $\mathrm{d}x$ which gives you an area. When you integrate, you're summing each area to find the total area beneath the curve. So to answer your question, you're using $\mathrm{d}x$ in both cases you presented.
 In your first case, you were integrating the invisible '1' between the integral sign and $\mathrm{d}x$

## Question on the functionality of dx

You are using $dx$ in the second one.
You have three parts in a simple integral:
$$\int_a^b f(x)dx$$
The function, $f(x)$ is the height of the function away from the $x$-axis and $dx$ is an infinitely small width. So, together, you have the area of an infinitely thin rectangle with height $f(x)$.
$$f(x)dx$$

Now all that's left is the integration symbol, $\int_a^b$. What this symbol says is to sum up these infinitely small rectangles from one point to another, from a to b.

An example would be the function $f(x) = x^2$.
Here is a graph of it:

Let's say you want the area from 1 to 2. Then you have this region:

Those blue lines at 1 and 2 represent the infinitely thin rectangles of width $dx$, each with different heights. You multiply the heights, $1^2$ and $2^2$ by this infinitely small width $dx$, and you get the areas of the rectangles at those points. Now, since we want all of the area, we keep adding these rectangles at all points from 1 to 2. That's where the integral symbol comes in. Since each point $x$ between 1 and 2 has a rectangle of height $x^2$, we notate this infinite sum of infinitely thin rectangles by writing
$$\int_1^2 x^2 dx$$

So as a summary of the above text, just to wrap it all up:
Each point $x$ between a and b has a rectangle of height $f(x)$, so you multiply that with the infinitely small width, $dx$, to get $f(x)dx$ and add them using integration going between a and b, $\int_a^bf(x)dx$.
So just imagine these blue lines, coming from the $x$-axis and extending up to the graph, filling up the area below the graph until it's just a solid block of blue. That is basically what is happening.

So your first case was the sum of rectangles with a constant height of 3, and the second case was a variable height of $x$.
 Recognitions: Gold Member Science Advisor Staff Emeritus The crucial point is what daldce said and it bears repeating. When you integrate $\int dx$, you are really integrating $\int f(x)dx$ with $f(x)= 1$. The function you are integrating is $f(x)= 1$ for all x, not "dx".

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