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Homework Help: Integration notation

  1. Sep 5, 2007 #1
    1. The problem statement, all variables and given/known data

    or http://img.photobucket.com/albums/v245/lilblonderoxy/ibntergrate.jpg

    3. The attempt at a solution
    For the first question, i looked at the graphs and which one was above the other one. in the first intersection, g(x) is above f(x) so i assumed the first area solution would be [tex]\int[/tex]g(x) - f(x) dx + [tex]\int[/tex] f(x) - g(x)
    So i guessed option B. However i do not think that this reasoning is correct, as the boundaries for the integration of the second part is not e-c but c-e :S

    the second question i also selected B, on the the basis that the reverse area is being worked out would require a negative sign in front of the g(x). Is this correct?
  2. jcsd
  3. Sep 5, 2007 #2


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    Okay, [itex]\int (g(x)-f(x))dx+ \int (f(x)+ g(x))dx[/itex] is good for the two parts. Now, what are the two parts? g(x)>f(x) between what two values of x? f(x)> g(x) between what two values of x? You also need to be careful about the direction. Remember that [itex]\int_a^b f(x)dx= -\int_b^a f(x)dx[/itex]. "B" is NOT correct for the first but is correct for the second question.
  4. Sep 5, 2007 #3


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    The problem is badly drawn. >"< For the first problem, all choices are false for me.

    It seems like a is placed on the Oy axis, while the rest, b, c, d, and e lie on Ox axis.

    If the picture is correct, then all the a's in the options, should be better replaced by 0.
  5. Sep 5, 2007 #4
    Then the first anser would be E, as it within the boundaries, but negative because the lowest point is being taken as the highest when integrating? Is that correct, if you can understand my reasoning:S
  6. Sep 5, 2007 #5
    I could be wrong but speaking when you have f(x) and g(x) the order is generally the function on top minus the function on the bottom. The reasoning behind this is that the top function has the area of everything below it and then bottom function has the area under its curve. So to only get the area in between the curves you have to cancel out the fact that the top function has the area of everything, so you subtract the bottom function's area.

    Or the integral is f(x)-g(x) if f(x) is the function on top and g(x) is the function on the bottom.

    So your assumption for the first question is wrong.
    Last edited: Sep 5, 2007
  7. Sep 5, 2007 #6
    I agree if the functions were being integrated with respect to y it would make more sense for a to c.

    Then if you look the two graphs go beyond e which would leave to believe that there should be an f at the final intersection and if placed like the a then it to could be integrate with respect to y.

    Like stated above the problem cant be right by the picture.
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