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    • Thread Starter

    OK I've done P3 integration but this S2 integration is doing my head in!!! For example in the S2 Blue book for Edexcel on Exercise 2B Question 9 where you have to convert a PDF to CDF. The integration in part is fine except there is a constant that I always miss....What am I doing wrong and could someone be kind enough to do the question above Exercise 2B 9)a)b) Thanks guys

    (a) For the range -3 <= x <= 3 we have F(x) = INT{-3 to x} f(t) dt = INT{-3 to x} (9-t^2)/36 dt, you can work out this integral. Then, F(x) = 0 for x < -3 and F(x) = 1 for x > 3.

    (b) For x <= 0, F(x) = 0. In the range 0 < x < 3 we have F(x) = INT{0 to x} t/9 dt. For the range 3 <= x <= 6 we work out INT{3 to x} (6-t)/9 dt. Then, we note that up to the point x = 3 there has already been some area which adds to 1/3. So the entry for the range 3 <= x <= 6 is 1/3 plus the last integral you worked out. And then it is F(x) = 1 for x < 6.
    • Thread Starter

    Sorry but I could not understand that. Is there any way you can explain it a bit more clearly?? Thanks
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Updated: June 18, 2004
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