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Integrating for area of intersection

  1. Jan 12, 2016 #1
    How does it work that you can subtract y2 from y1 and integrate the product within defined limits for the area of their intersection (within those limits)?
    Maybe that's not the right terminology - you arrive at the area for the region bounded by both functions.

    Is it just the same in practice as subtracting the integral of y2 from the integral of y1?
     
  2. jcsd
  3. Jan 12, 2016 #2

    462chevelle

    User Avatar
    Gold Member

    I'm not 100% what you're asking, but it sounds to me like you are trying to find the area bound by the intersection of two curves. One thing you may have to consider is if they only intersect at two points within those limits. If they only intersect at two points within your limits, you can simply just do the integral of y2 minus the integral of y1. With y2 being the upper function on the graph.
     
  4. Jan 12, 2016 #3
    How's that?

    Untitled.png
     
  5. Jan 12, 2016 #4

    Mark44

    Staff: Mentor

    Let's call the line f(x) and the curve g(x). Assuming that the line and curve intersect at a and b, the area of the region in blue is ##\int_a^b (f(x) - g(x)) dx##. This is exactly the same as ##\int_a^b f(x) dx - \int_a^b g(x)dx##, due to the linearity properties of the definite integral.

    The product that you refer to is the area of the typical area element, ##\Delta A = (f(x) - g(x)\Delta x##. When you integrate, you are essentially summing all of these incremental area elements to get the total area.
     
  6. Jan 13, 2016 #5
    Due to the linearity properties of the definite integral?
     
  7. Jan 13, 2016 #6

    Mark44

    Staff: Mentor

    ##\int_a^b (f(x) + g(x))~dx = \int_a^b f(x)~dx + \int_a^b g(x)~dx##
    and
    ##\int_a^b k \cdot f(x)~dx = k\int_a^b f(x)~dx## where k is any constant
    The same properties hokl for indefinite integrals, as well.
     
  8. Jan 13, 2016 #7
    LOL. I take it I'm punching well above my weight again.
     
  9. Jan 13, 2016 #8

    462chevelle

    User Avatar
    Gold Member

    I think you aren't as confused as you think. He us just showing properties of the integral in a general sense. That way of writing might be intimidating if you don't have any formal knowledge on the subject.
     
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