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Integral question

  1. Jul 2, 2013 #1
    We can find an area under a curve with integral, but how if the curve is under the x axis? Do e still use integral?

    In general where can we use integral?
     
  2. jcsd
  3. Jul 2, 2013 #2

    Yes, and you will get a negative number. This may seem confusing (how can we have negative area?), but here is a good thread about it: https://www.physicsforums.com/showthread.php?t=292737

    Check out HallsofIvy's excellent response. I'll quote him if that's ok

    If a function is continuous on an interval, it will be integrable on that interval. (That's a mouthful!)

    Here's a neat wolfram demonstration as to why that's true: http://demonstrations.wolfram.com/ContinuousFunctionsAreIntegrable/

    -Dave K
     
  4. Jul 2, 2013 #3
    If the curve is under the x-axis the integral has a negative value.

    Basically this gives you the net area under the curve.

    If you want to find the total area both under and above the x-axis, you'll need to integrate the modulus of the function.

    i.e [tex] \displaystyle \int^{2\pi}_{0} sin(x) = 0[/tex]

    But..

    [tex] \displaystyle \int^{2\pi}_{0} |sin(x)| = 4 [/tex]
     
  5. Jul 6, 2013 #4
    I don't mean to be rude, but you need dx in the integral for that to be true. You could confuse some people, otherwise.
     
  6. Jul 6, 2013 #5
    Oops I was being sloppy again....
     
  7. Jul 6, 2013 #6
    In addition to area it can be used to find volume and surface area of solids, arc lengths, mass, average value of a function, the centroid, just to name a few simple applications.
     
  8. Jul 6, 2013 #7

    HallsofIvy

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    Notice that the integral gives signed area- negative if the graph goes below the x- axis. Strictly speaking, area is always positive so if the problem actually asks for the area between a curve and the x-axis, where the curve goes both above and below the x-axis, is [itex]\int |f(x)| dx[/itex]
     
  9. Jul 6, 2013 #8

    WannabeNewton

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    This is only a sufficient condition; it is not a necessary condition. The condition of Riemann integrable is much more general. The indicator function associated with the Cantor set has uncountably many discontinuities but it is Riemann integrable.
     
    Last edited: Jul 6, 2013
  10. Jul 6, 2013 #9

    micromass

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    A bit nitpicking, but you of course want your interval to be closed and bounded! I'm sure you mean this.
     
  11. Jul 6, 2013 #10

    jbunniii

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    Indeed, we can go a bit further and state a necessary and sufficient condition for Riemann integrability of a bounded function ##f## defined on a closed, bounded interval: ##f## is Riemann integrable if and only if the set of points where it is discontinuous has Lebesgue measure zero. The indicator function of the Cantor set qualifies because its set of discontinuities is precisely the Cantor set, which has measure zero.
     
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