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Question on the functionality of dx

  1. Jul 5, 2011 #1
    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:

    [itex]\int[/itex]3dx with limits from 0 to 10. In this situation, 3 is a constant, so you can pull it out, leaving 3[itex]\int[/itex]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 [itex]\int[/itex]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?

    Thank you for your assitance.
     
  2. jcsd
  3. Jul 5, 2011 #2
    The [itex]\mathrm{d}x[/itex] in terms of definite integration is an infinitely small rectangle beyond feasible measurements. When you integrate a function such as [itex]f(x) = x[/itex] from [itex]a[/itex] to [itex]b[/itex], you're finding the height from the [itex]x[/itex]-axis to the graph of [itex]f(x) = x[/itex]. In theory, you're multiplying the height of the curve by an infinitely small rectangle of width [itex]\mathrm{d}x[/itex] 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 [itex]\mathrm{d}x[/itex] in both cases you presented.
     
  4. Jul 6, 2011 #3
    In your first case, you were integrating the invisible '1' between the integral sign and [itex]\mathrm{d}x[/itex]
     
  5. Jul 7, 2011 #4
    You are using [itex]dx[/itex] in the second one.
    You have three parts in a simple integral:
    [tex]\int_a^b f(x)dx[/tex]
    The function, [itex]f(x)[/itex] is the height of the function away from the [itex]x[/itex]-axis and [itex]dx[/itex] is an infinitely small width. So, together, you have the area of an infinitely thin rectangle with height [itex]f(x)[/itex].
    [tex]f(x)dx[/tex]

    Now all that's left is the integration symbol, [itex]\int_a^b[/itex]. 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 [itex]f(x) = x^2[/itex].
    Here is a graph of it:
    2u4odty.jpg

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

    zmmil0.png

    Those blue lines at 1 and 2 represent the infinitely thin rectangles of width [itex]dx[/itex], each with different heights. You multiply the heights, [itex]1^2[/itex] and [itex]2^2[/itex] by this infinitely small width [itex]dx[/itex], 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 [itex]x[/itex] between 1 and 2 has a rectangle of height [itex]x^2[/itex], we notate this infinite sum of infinitely thin rectangles by writing
    [tex]\int_1^2 x^2 dx[/tex]

    So as a summary of the above text, just to wrap it all up:
    Each point [itex]x[/itex] between a and b has a rectangle of height [itex]f(x)[/itex], so you multiply that with the infinitely small width, [itex]dx[/itex], to get [itex]f(x)dx[/itex] and add them using integration going between a and b, [itex]\int_a^bf(x)dx[/itex].
    So just imagine these blue lines, coming from the [itex]x[/itex]-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 [itex]x[/itex].
     
  6. Jul 7, 2011 #5

    HallsofIvy

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    The crucial point is what daldce said and it bears repeating.

    When you integrate [itex]\int dx[/itex], you are really integrating [itex]\int f(x)dx[/itex] with [itex]f(x)= 1[/itex]. The function you are integrating is [itex]f(x)= 1[/itex] for all x, not "dx".
     
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