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Interpretation of integral as area?

  1. Jul 28, 2004 #1
    I know the definite integral of a function can be thought of as the area between the function and y=0 and between the lower and upper bounds of integration, so long as the function is positive in that region.

    However, I also know that:

    [tex]\int_{a}^{b} f(x) ~dx = -\int_{b}^{a} f(x) ~dx[/tex]

    And I find it hard to reconcile these two things. It would be nice to think of the integral as a process where you're adding up a whole bunch of little slices between the function and y=0 from the lower bound to the upper bound, but under this rationalization, you'd be adding up exactly the same thing whether you add them from left to right or right to left. The "area under the curve" would remain exactly the same even if you mirrored the whole thing horizontally.

    How do you think of this? Do you have a way of looking at it that neatly accounts for the fact that inverting the bounds causes the sign to change?
     
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  3. Jul 28, 2004 #2

    Hurkyl

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    The trick is to use the idea of an oriented region. If you take the boundary of a region, there are two ways you can go around it, clockwise, and counterclockwise. If we are going around the boundary counterclockwise, we say the region has a positive orientation. Otherwise, it has a negative orientation.

    The direction of integration allows us to determine the orientation of the boundary. For example, if we take the unit square given by:

    0 <= x <= 1 and 0 <= y <= 1.

    We can find the area via

    [tex]
    \int_0^1 1 dx
    [/tex]

    In this integral, we're going from 0 to 1, so we're going to the right along the x-axis. If we continue around the square, we go around it counterclockwise, so it is positively oriented, and we get the positive area, +1.

    If, instead, we do

    [tex]
    \int_1^0 1 dx
    [/tex]

    We're going left along the x-axis, thus go around the square clockwise, getting a negative area, -1.
     
  4. Jul 28, 2004 #3

    plover

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    Start with:
    [tex]
    \int_{a}^{b} f(x) ~dx\ +
    \ \int_{b}^{c} f(x) ~dx\ =
    \ \int_{a}^{c} f(x) ~dx
    [/tex]​
    If you then set c = a you have
    [tex]
    \int_{a}^{b} f(x) ~dx\ +
    \ \int_{b}^{a} f(x) ~dx\ =
    \ \int_{a}^{a} f(x) ~dx\ =\ 0
    [/tex]​
    which implies your original equation.
     
    Last edited: Jul 28, 2004
  5. Jul 28, 2004 #4
    Well, when you flip the order of the limits then you are basically taking all the little slices you are adding up and switching them from having a positive width to a negative width. Sort of analogous to how when you integrate over a negative function, you are "flipping" the slices over so that they have negative height instead of positive height.
     
  6. Jul 28, 2004 #5

    Gokul43201

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    Treat each area slice as a vector; the cross product of the functional height and the width of the slice. When you reverse the direction of the width, you switch the direction of the area vector.
     
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