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A Few Quick Questions: Integral Notation and Integrals

  1. Dec 19, 2006 #1
    1. The problem statement, all variables and given/known data

    The integral (2x) dx between x= -2 and 1

    and indefinate integral (dx)/((x+3)^2)

    2. Relevant equations

    None really.

    3. The attempt at a solution

    I am working through practice problems for my final later this week. I have come across a few integral problems where the answer comes out as a negative number. I have no problems doing these questions I am just having an issue understanding how an area can be negative.

    And for the second one does this simply mean that it is integral (1/((x+3)^2) dx??

  2. jcsd
  3. Dec 19, 2006 #2


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    Staff Emeritus
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    We define a negative area to mean the area underneath the graph when the function is negative, and so is below the x axis. As an example, consider the integral of the function sin(x) between pi and 2*pi:
    [tex] \int_\pi^{2\pi}sin(x)dx=[-cos(x)]_\pi^{2\pi}=-2 [/tex]

    Looking at the graph of the sine function, we see that the given integral is in the region where the function is negative, and so this gives rise to the "negative" area under the curve.

    Last edited: Dec 19, 2006
  4. Dec 19, 2006 #3

    So when you are asked to find the area under a curve between two points and the function happens to be both above and below the x-axis on that interval. You are really being asked to find the difference between the area above and below the graph to the x axis?

    I guess I shouldn't be thinking in terms of area then.
  5. Dec 19, 2006 #4


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    Thinking of integration in terms of area can actually be quite helpful at times, consider:
    [tex] \int_0^{2\pi}sin(x)dx[/tex]

    Because the area above the graph is the same as the area below the graph, you immediately know the integral evaluates to 0, without doing any integration.
  6. Dec 19, 2006 #5


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    Yes, then if the difference is negative, it shows the area below the x axis is greater, and vice versa. If you consider the above example with sin(x), but this time integrate between 0 and 2*pi, then the value of the integral is zero. This is because the area below and above the axis are equal, and so cancel out.
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