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Integral Question

  1. Dec 31, 2011 #1
    When integrating f from a to b, b > a, I understand this is the area under the curve
    y = f(x).

    How ever say that I am computing the area of a graph where f is both positive and negative in the interval, like in the picture I posted....

    Refering to the picture say I was calculating the integral from -3 to 4. From -3 to -2 the area is pos, say equal to 4. from -2 to -1 it is negative say equal to -3. -1 to 3 equal to -50 and from 3 to 4 whose area is say 10.

    Then is the area in this interval: 4 - 3 - 50 + 10 = -39 (total area)
    or is it 4 -(-3) -(-50) + 10 = + 67 (total area)
     

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  3. Dec 31, 2011 #2
    The area under the x-axis counts as "negative area". It subtracts from the positive area. In order to find the actual area of the graph you have to integrate |f|, the absolute value of the function.
     
  4. Dec 31, 2011 #3
    If ∫f(x)dx from a to b = a(b) - a(a)

    Then from my graph if I integrated from -3 to 4 which would be a(4) - a(-3), I would get an answer of -39 correct?
     
  5. Dec 31, 2011 #4
    And if I'm asked to evaluate an integral does that mean evaluate individually the absolute value of the pos and neg areas then sum them together?
     
  6. Jan 1, 2012 #5
    You're getting confused between two concepts. If you break down the area into parts as you did, you simply add them all together to get the total area. (You did it right the first time.)

    On the other hand, if you have an antiderivative [itex]F[/itex], the Fundamental Theorem of Calculus says that you can evaluate [itex]F(b)-F(a)[/itex] instead of calculating the area directly as you did. This is the only time you subtract.

    In other words, if you have that [itex]F[/itex] is an antiderivative for [itex]f[/itex], then [itex]\int\limits_a^bf\,dx=F(b)-F(a)[/itex].

    But if you evaluate the integral directly, as you did, it's a sum. [itex]\int\limits_{-3}^{4}f\,\,dx=\int\limits_{-3}^{-2}f\,\,dx+\int\limits_{-2}^{-1}f\,\,dx+\int\limits_{-1}^{3}f\,\,dx+\int\limits_{3}^{4}f\,\,dx[/itex].

    Hope this helps!
     
  7. Jan 1, 2012 #6
    This might be looking too far into the future, but thinking of an integral as just being the area underneath a curve is very dangerous. It can be used to compute this sort of thing, but as far as I am concerned an integral is nothing more or less than the continuous analog of summation.

    When you're using it compute "areas", you work under the assumption that areas cannot be negative. To facilitate this difficulty, you must introduce absolute values accordingly. If the single-valued function you wish to integrate is negative along any interval, you must take its absolute value in taking an area. Otherwise you're not computing an area, you're merely computing the integral of the function without regard to what you're being asked.

    How do you know the limits of integration for the negative parts? Just compute where the function intersects the x-axis. Be very careful not to just compute the integral and take it's absolute value, that will not do the same thing.

    A single integral is, in some low brow respect, the "net area" underneath a curve (that is allowing for the concept of "negative area"), if you merely take the absolute value of this your answer will make absolutely no sense whatsoever in the respect of area as a decent human being views it.
     
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