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How are jump discontinuities important when determining the area of a curve?

  1. Sep 9, 2010 #1

    I'm a complete n00b at calculus (first day of the first year) so this question may appear a little...stupid. But I want to hammer into my head every little thing I hear in the lecture hall, so I might as well give it a go.

    Today the professor explained how jump discontinuities must be taken into consideration when determining the area under the curve using the limit of #of rectangles definition. I did not really understand, or at least as well as I would have liked. Basically I want to know why a function can be integrated only on a continuous interval, and not on an interval with a discontinuity in it.

    Also, a function with a jump discontinuity is usually a piecewise defined function. Actually I think this isn't always so, as is the case with (x/|x|). Is it true that a piecewise defined function can't be integrated in one step, and each part has to be analyzed separately?
    Last edited: Sep 9, 2010
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  3. Sep 9, 2010 #2


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    If g is continuous on [a,b], f has only finitely many jump discontinuities on [a,b], and f(x)=g(x) for all x where f is continuous, then the integral of g from a to b is equal to the integral of f from a to b. Or ...

    [tex]\int_a^bg = \int_a^bf[/tex]
  4. Sep 11, 2010 #3
    I'm sorry, I don't really follow. Are you saying that if I find a continuous function equivalent to the continuous parts of the jump function, and integrate it on [a,b], it would equal the integral of the jump function on that same interval? Could you perhaps provide a pictorial representation?
  5. Sep 11, 2010 #4


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    Sorry! I got my terms mixed up! I meant removable discontinuity in my post. However, there is a similar result for jump discontinuities. I'll work on posting it in the next few hours when I find the time. Again, sorry for the confusion.
  6. Sep 12, 2010 #5


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    The best thing to do with jump discontinuities is to break the problem into "pieces" with the boundary at the points of discontinuity. Then you have several "regular" areas and the entire area is the sum of those.

    First, I would say that your example of |x|/x is not an example of a jump discontinuity since the value is not defined at x= 0. If you were, for example to define f(x)= x/|x| for x not 0, f(x)= a for x= 0, then there is, indeed a jump discontinuity at x= 0. But now the function can be written as a piecwise continuous function- f(x)= 1 for x> 0, f(0)= a, f(x)= -1 for x< 0.

    Also, I would not say that can't be integrated directly, only that it is much easier, using the usual techniques of integration, to divide it into "continuous parts". And I suspect that was what your teacher was trying to say. Integration is done pretty much by "remembering" anti-derivatives- and we typically learn continuous functions because those are easiest.
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