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Area under the curve problem, but there's a twist

  1. Sep 1, 2004 #1
    I have a unique area under a curve problem. Say we have a function y = x^2. Ok, so, the area under that curve is just the integral of x^2 from 0 to some value of x. That's easy. Lets call this area A.

    Now normally if we wanted to find the area under a height, we would simply mulitply height by the value of x and subtract the integrated form to find the are above the curve to that height. Lets call this area B.

    I cant use this method. I cant simply say h*x minus the integral of whatever to find area B. Is there another way to find area B while knowing the function but not involving area A in any way?
  2. jcsd
  3. Sep 1, 2004 #2


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    try making this confusing question a little more precise. an example could help.
  4. Sep 1, 2004 #3
    I included a diagram. I need to find area A and area B separately. I cant find area B by subtracting A out of it and conversely I cant find area A by subtracting B out of it.

    Attached Files:

  5. Sep 1, 2004 #4


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    Why can't you?
  6. Sep 1, 2004 #5
    I just cant.
  7. Sep 1, 2004 #6
    Why not? Try it for some value of x and h. What's the problem? On the other hand, maybe you could try expressing the f(x) in terms of y and integrating wrt y instead to find B directly.

    PS. What did you use to create the picture? It looks rather neat.
  8. Sep 1, 2004 #7
    I cant do it the normal way because my problem that this assists boils down to 1=1, a true yet uninspiring finding. I will try it with your suggestions, Ethereal.

    That chart? Eh, that's nothing. I made it in Adobe Illustrator. You can draw ANYTHING in that program.
  9. Sep 1, 2004 #8


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    If you found one of them via the integral, couldn't you find the other by subtracting?
  10. Sep 1, 2004 #9
    I said I cant use the subtraction method.
  11. Sep 2, 2004 #10


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    Do you mean you don't know how? Or aren't you allowed to use it for solving the problem?

    If you're looking for another way to find area B, here's a hint:
    Think of the line that gives the height as your x-axis and stand on your head.
  12. Sep 2, 2004 #11
    I am not allowed.

    Thanks Gilleo, will do.
  13. Sep 2, 2004 #12
    wouldn't the inverse function of x^2 have the same area if you integrated along the x-axis and used the height as the upper bound limit?
  14. Sep 3, 2004 #13
    Here's how

    Find [itex] f^{-1}(x) [/itex]. In this case, you are using [itex] f(x) = x^2 [/tex]. The inverse of this function is [itex] f(x) = \sqrt{x} [/itex]. You can now integrate, but make sure to adjust your limits of integration. If you wanted from 0 to 3 of the original function, that would translate to 0 to 9 of the inverse. Just plug in the original limits to the inverse (y) to obtain your new limits.

    Hope this helps!
  15. Sep 3, 2004 #14


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    I don't know if inversion will help. Inversion mirrors the function in (through?) the line y=x.
    My idea was to mirror the function in the x-axis, then add the height h to the function and then integrate. (or alternatively subtract the height from the function and then mirror it).

    But the calculation would be exactly the same if you were calculating the area of the rectangle and subtract area A. So it seems silly not to use this method and do use the above method.

    Subtract height: f(x)-h
    Mirror: h-f(x)
    Integrate: [itex]h\Delta x - \int f(x)dx[/itex]

    It's the same :)
    Last edited: Sep 3, 2004
  16. Sep 3, 2004 #15
    Well basically it is the same exact thing as what you planed on doing. The area between the function and the y-axis is the same as its inverse function and the x-axis.
  17. Sep 8, 2004 #16
    ok i'm not sure if anyone mentioned htis but....

    lets say the equation was y = x^2.... you wanted to find area B

    could you not find the integral from 0 -> x by changing the equation ... by flipping it upside down (reflect upon x-axis) and shift it up h units?

    thus you get
    y = -x^2 + h

    and find integral from 0-->x now?

    correct me if i'm wrong
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