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Setting up the correct limit for an area (double) integral

  1. Apr 14, 2012 #1
    I'm confused regarding the limits required for the following question:

    Find the area in the plane between the graphs of y = x2 and y = x3 for x non-negative.

    The limit for y would be from x^2 to x^3 but what about x? It's lower bound would be zero but as for the upper bound I have no idea. The answer of the above question comes out to be 1/12. Help?
  2. jcsd
  3. Apr 14, 2012 #2

    You need to answe to

    1) Where do the graphs of [itex]x^2\,,\,\,x^3\,\,,\,\,x\geq 0[/itex] meet?

    2) What function's graph is over the other's in the interval between both points of intersection you found in (1)?

    3) Now do the integral between both x-coordinates of the intersection points of the function above minus the function below, dx, and that's the area you're looking for.

  4. Apr 15, 2012 #3
    Thanks for the reply Don.
    Now I just have another question. When we set up the integral, obviously we take the xlimits as x^2 to x^3 as generally x^3 is bigger then x^2. However in this case where we are to integrate from 0 to 1 this generalization fails us as x^2 is bigger than x^3. Now the question is that if we still integrate from x^2 to x^3 the area comes out to be -1/12 whch is plain wrong. We can do either one of the following two options:

    1) Just take the absolute value of -1/12 and present 1/12 as our answer.

    2) Or we set up our integral again from x^3 to x^2 in light that x^2 is bigger than x^3 and the answer is 1/12 which is correct.

    Which path do we take or are both methods correct?
  5. Apr 15, 2012 #4

    I'm not completely sure what you mean by "...take the x-limits as x^2, x^3", or "integrate from x^2 to x^3" ...??

    The limits are exactly what I told you in my first post (i.e., they are numbers!) of the difference [itex](x^2-x^3)\,dx[/itex] since, as you correctly

    stated, in the unit interval the parabola's graph is above the cubic's.

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