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Question about May 1992, Problem 12

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  • Question about May 1992, Problem 12

    Hey all. I'm having a bit of a problem. In Dr. Ostazewski's manual, it's number 13 in practice exam 5. I'm confused about the area of integration. Since y >= x, I thought the area should have been x: (0 to w), y: (x to 1), and for the second part x: (w to sqrt w), y: (x to w/x)? In simpler terms, I thought the shaded region should be to the upper left of the line y = x instead of to the lower right. I don't get it. I would be more than grateful if someone could help me out. Thanks very much all.

  • #2
    Question about May 1992, Problem 12

    Dear Robertr24:

    I do not know which edition of my manual you have: first or second edition, but
    I think this is probably a typo that I have corrected in one of these posted errata:

    First edition:
    http://www.math.ilstu.edu/krzysio/P-Manual-Errata.pdf

    Second edition:
    http://www.math.ilstu.edu/krzysio/P-...dEd-Errata.pdf

    I hope this helps. If it does not, please reply, or send me an e-mail.

    Yours most sincerely,
    Krzys' Ostaszewski


    Originally posted by robertr24
    Hey all. I'm having a bit of a problem. In Dr. Ostazewski's manual, it's number 13 in practice exam 5. I'm confused about the area of integration. Since y >= x, I thought the area should have been x: (0 to w), y: (x to 1), and for the second part x: (w to sqrt w), y: (x to w/x)? In simpler terms, I thought the shaded region should be to the upper left of the line y = x instead of to the lower right. I don't get it. I would be more than grateful if someone could help me out. Thanks very much all.
    Want to know how to pass actuarial exams? Go to: smartURL.it/pass

    Comment


    • #3
      Thanks. That clears things up a bit. However, I'm still confused about the actual integration. In the manual you have x:={0 to sqrt w}, y:={0 to x} for the first part, x:={sqrt w to 1}, y:={0 to x/w} for the second part. Doesn't this correspond to the area below the line y = x?

      Comment


      • #4
        November 1992, Problem No. 12

        The intersection of the line y = x and the hyperbola xy = w is at the point
        ( w^0.5, w^.05). First part is the integral over the shaded region between
        the origin and the line y = w^.05, the second one is above that line
        and still under the hyperbola. I still had typos in this solution, so I updated
        the errata, and you can download corrected versions here:
        http://www.math.ilstu.edu/krzysio/P-...dEd-Errata.pdf
        http://www.math.ilstu.edu/krzysio/P-Manual-Errata.pdf

        Yours,
        Krzys'


        In the first part, we have

        Originally posted by robertr24
        Thanks. That clears things up a bit. However, I'm still confused about the actual integration. In the manual you have x:={0 to sqrt w}, y:={0 to x} for the first part, x:={sqrt w to 1}, y:={0 to x/w} for the second part. Doesn't this correspond to the area below the line y = x?
        Want to know how to pass actuarial exams? Go to: smartURL.it/pass

        Comment


        • #5
          Thanks very much. I knew something was wrong there. It makes much more sense with the dx's and dy's interchanged. Sorry to drive you crazy. I will definitely look through ALL the errata before I post questions from now on.

          Comment


          • #6
            Drive crazy? Not at all.

            Dear Robert24:

            Unfortunately, errors and typos still show up in my manuals, despite my best efforts.
            So, I view your inquiry as a very nice gesture, helping me correct those errors. I appreciate it very much.

            Yours very truly,
            Krzys' Ostaszewski

            Originally posted by robertr24
            Thanks very much. I knew something was wrong there. It makes much more sense with the dx's and dy's interchanged. Sorry to drive you crazy. I will definitely look through ALL the errata before I post questions from now on.
            Want to know how to pass actuarial exams? Go to: smartURL.it/pass

            Comment

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