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The Meaning of 'dx' in an Integral

Date: 02/22/2002 at 02:06:11
From: Nosson
Subject: What does the dx in an integral really mean?

No matter how many times its explained to me, and even though I've 
taken several advanced math courses (diff eq, linear algebra, etc), 
nobody has ever given me a satisfactory explanation for the meaning 
of the notation in which an integral has dx appended to the end if x 
is the variable which we are integrating with respect to. In physics, 
for example, dx seems to mean a very small amount of x, and then we 
use it in an integral to integrate whatever physical quantity is 
being discussed. I just don't understand. 

Or, when a differential is defined, all of a sudden the dx has a 
meaning, but then when an integral is being evaluated, the teacher 
says, "Oh, the dx is just a formality." 

So, sometimes it's a formality, sometimes a vital concept, sometimes a 
physical quantity, sometimes a derivative: What is it?

Please help me.

Date: 02/22/2002 at 05:00:53
From: Doctor Jeremiah
Subject: Re: What does the dx in an integral really mean?

Hi Nosson,

Think about it this way:

An integral gives you the area between the horizontal axis
and the curve.  Most of the time this is the x axis.


                           |                    |
                         --|--              ----|---- f(x)
                       /   |   \          /     |
                      /    |     --------       |
            |        /     |                    |
       -----|-------       |                    |
            |              |                    |
            |              |                    |
  ----------|--------------+--------------------|----- x
            a                                   b

And the area enclosed is:

 Area = | f(x) dx

But say you didn't want to use an integral to measure the
area between the x axis and the curve.  Instead you just
caclulate the average value of the graph between a and b
and draw a striaght flat line y = avg(x)   (the average
value of x in that range).

Now you have a graph like this:


                           |                    |
                         - | -              - - | - - f(x)
            |          /   |   \          /     |
       -----|-----------------------------------|---- avg(x)
            |        /     |                    |
       - - -|- - - -       |                    |
            |              |                    |
            |              |                    |
  ----------|--------------+--------------------|----- x
            a                                   b

And the area enclosed is a rectangle:

 Area = avg(x) w          where w is the width iof the section

The height is avg(x) and the width is w = b-a or in English,
"the width of a slice of the x axis going from a to b."

But say you need a more accurate area.  You could break the
graph up into smaller sections and make rectangles out of them.
Say you make 4 equal sections:


                           |                    |
                      |----|---|        |-------|---- f(x)
                      |    |   |        |       |
                      |    |   |--------|       |
            |         |    |   |        |       |
       -----|---------|    |   |        |       |
            |         |    |   |        |       |
            |         |    |   |        |       |
  ----------|---------|----+---|--------|-------|----- x
            a                                   b

And the area is:

 Area = section 1  + section 2  + section 3  + section 4
      = avg(x,1) w + avg(x,2) w + avg(x,3) w + avg(x,4) w

where w is the width of each section.  The sections are all
the same size, so in this case w=(b-a)/4 or in English,
"the width of a thin slice of the x axis or 1/4 of the
width from a to b."

And if we write this with a sumation we get:

 Area =  /   avg(x,n) w

But it's still not accurate enough.  Let's use
an infinite number of sections.  Now our area
becomes a summation of an infinite number of sections.
Since it's an infinite sum, we will use the
integral sign instead of the summation sign:

 Area =  | avg(x) w

where avg(x) for an infinitely thin section will be
equal to f(x) in that section, and w will be "the width of
an infinitely thin section of the x axis."

So instead of avg(x) we can write f(x), because they are
the same if the average is taken over an infinitely small width.

And we can rename the w variable to anything we want.
The width of a section is the difference between the
right side and the left side.  The difference between
two points is often called the delta of those values.
So the difference of two x values (like a and b) would
be called delta-x.  But that is too long to use in an
equation, so when we have an infinitely small delta,
it is shortened to dx.

If we replace avg(x) and w with these equivalent things:

 Area =  | f(x) dx

So what the equation says is:

Area equals the sum of an infinite number of rectangles
that are f(x) high and dx wide (where dx is an infinitely
small distance).

So you need the dx because otherwise you aren't summing
up rectangles and your answer wouldn't be total area.

dx literally means "an infinitely small width of x".

It even means this in derivatives.  A derivative of
a function is the slope of the graph at that point.
Slope is usually measured as the y difference of two
points divided by the x difference of those points:

  Slope = (y2 - y1) / (x2 - x1)

But the closer these points get the smaller these
differences get.  Let's start calling them deltas,
because the difference between two points is often
called the delta of those values.

  Slope = delta-y / delta-x

The deltas get smaller and smaller as these two x,y
points get closer and closer.  When they are an
infinitely small distance apart, then the delta-y
and delta-x is shortened to dy and dx:

  Slope = dy / dx

The slope is still Slope = (y2 - y1) / (x2 - x1)
but these points are infinitely close together, so
we use dy and dx to tell ourselves that they are
infinitely close or "differential distances."

Does that help?  Let me know if you still have questions.

- Doctor Jeremiah, The Math Forum

Date: 02/22/2002 at 05:40:21
From: Nosson
Subject: What does the dx in an integral really mean?

Thanks for your answer, Jeremiah. That makes a lot of sense. I 
appreciate the time that you took to write out the answer, as 
well as your promptness.

Associated Topics:
College Calculus
High School Calculus

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