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    Let X be a continuous random variate with pdf f_X (x) = 6x(1-x) for  0\le x \le 1 and  0 otherwise.

    I need to find the cdf F_{X}(x) of X.

    I know that, in general, F_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

    But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.
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    (Original post by MR1999)
    Let X be a continuous random variate with pdf f_X (x) = 6x(1-x) for  0\le x \le 1 and  0 otherwise.

    I need to find the cdf F_{X}(x) of X.

    I know that, in general, F_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

    But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.
    Your general formula is correct, and in this case the lower bound will be 0, since f(x) is 0 below x=0.

    I can't make sense of "I tried integrating between [0, x] and [x, 1]" - sounds like you're integrating between two intervals???

    Your limits should just be 0 and x, and you end up a function of x.

    I'm going to have to keep my responses brief, as I keep getting "Service not available", and even when it is, the screen and cursor are jumping around all over as it's so f**king slow!
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    (Original post by MR1999)
    Let X be a continuous random variate with pdf f_X (x) = 6x(1-x) for  0\le x \le 1 and  0 otherwise.

    I need to find the cdf F_{X}(x) of X.

    I know that, in general, F_{X}(x)= \displaystyle \int_{-\infty}^{x} f_{X}(t)dt

    But in this case, I'm not sure where to integrate between. I tried integrating between [0, x] and [x, 1] but that just gives 1. So any help would be appreciated.
    Just evaluate \displaystyle  \int_0^x 6t(1-t).dt, which is valid for  x \in [0,1]

    EDIT: Obviously integrating over [0,x] then adding on the integral over [x,1] will give you 1. This is the same as integrating your density function over \mathbb{R} in this case, and we know this must be 1 since you're summing all the probabilities!
 
 
 
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