Why Is dx Necessary?Date: 01/31/2002 at 17:03:28 From: Lee Townsend Subject: Integral notation... dx Dr. Math, In integral notation in which you have one variable after an integral sign you must always end with the change in your variable (i.e.: INT[] x^2 dx). Why is the 'dx' necessary? True, it comes in handy when using the substitution method, but how did this notation develop? Is it just there because the original formal definition of an integral (using Reimann sums) multiplied the height of the rectangles f(xi) and then length of the rectangles dx? Please explain... Lee Date: 02/01/2002 at 02:06:13 From: Doctor Jeremiah Subject: Re: Integral notation... dx Hi Lee, It _is_ all because of how an integral is defined. Let's think about the definition of an integral. An integral tells you the area between the x axis and a graph of an equation. Say this is your graph: | /---------------\ | / \ |-------/ \ | . \ | . \------------- y=f(x) | . . | . . | . . +---|-----------------------------|------------- a b The integral of y=f(x) between x=a and x=b tells you the area that is enclosed. But if you didn't know about integrals, how would you calculate the area? Well, you could multiply the change in x (a-b) by the average height of y=f(x), or you could break it down into smaller pieces. Let's start with 3: | /---------------\ | / . . \ |-------/ . . \ | . . . \ | . . . \------------- y=f(x) | . . . . | . . . . | . . . . +---|--------|---------|----------|------------- a |<- dx ->| b The width of these smaller areas we will call delta-x, because the width is the change in x and the Greek letter delta is used a lot for "change in," but delta-x is too long so we shoerten it to dx. dx is (a-b)/3, so now the area is the sum of the smaller areas. The smaller areas have a width of (a-b)/3 (the width of a section) times the average value of y=f(x) for that section. We could write that as a sum: # of areas +---- \ / small_area(i) +---- i=1 3 areas +---- \ / height(i) width(i) +---- i=1 And since all the sections are the same width: 3 +---- \ / average(f(x)) (a-b)/3 +---- i=1 3 +---- \ / average(f(x)) dx +---- i=1 But what happens if we keep making more and more, smaller and smaller areas to sum up? Eventually we get many, many really narrow areas, each having a width that is a tiny change in x (called "dx" remember) times the average height. But since there are so many tiny areas we can't do this, so instead we come up with a new symbol: infinity +---- / \ | / ====> | +---- / i=1 For a really, really narrow section the average height is y=f(x), so the area of the section is f(x)dx. To get the total area we sum up all the little narrow areas. So we are actually summing up areas of size f(x)dx: / | | really_small_areas | / / | | height width | / Or, since the average value of a really thin section is f(x): / | | f(x) dx | / So it does really mean something. If you want more details let me know. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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