1. ## Integral calculus

Hi I'm Heather and I'm doing my Masters in Statistics.
I have a calculus prob.
I have this question about pdf:
Suppose X be a continuous random variable with the pdf:
f(x) = {k (2 +x^2 -x) -1 <= x <= 2
0 otherwise
I need to find the value of k, which is a positive constant and then obtain the cumulative distribution function (F (x)) of X.

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2. ## Re: Integral calculus

$\displaystyle \int_{-\infty}^{\infty}~f(x) ~dx = 1$

that let's you evaluate $k$ and thus get $f(x)$, then

$F(x) = \displaystyle \int_{-\infty}^x~f(u)~du$

3. ## Re: Integral calculus

Originally Posted by romsek
$\displaystyle \int_{-\infty}^{\infty}~f(x) ~dx = 1$

that let's you evaluate $k$ and thus get $f(x)$, then

$F(x) = \displaystyle \int_{-\infty}^x~f(u)~du$
So I have tried to work this out and I believe k =3/8. Is that right?
Do I then have to work out 3/8 (2+x^2-x)? Or is that the pdf?
Thanks.

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4. ## Re: Integral calculus

is $f(x) = k(x^2 - x + 2),~x \in [-1,2]$ ?

If so I get $k=\dfrac {2}{15}$

so

$f(x) = \dfrac{2}{15}(x^2 - x + 2),~x \in [-1,2]$

integrating we get

$F(x) = \displaystyle \int_{-\infty}^x~\dfrac{2}{15}(x^2 - x + 2)~dx = \begin{cases} 0 &x<-1 \\ \dfrac{1}{45} \left(2 x^3-3 x^2+12 x+17\right) &x\in [-1,2] \\ 1 &2 < x \end{cases}$

5. ## Re: Integral calculus

Thanks fur that romsek, I'm going to try and figure this all out not.

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6. ## Re: Integral calculus

Sorry my post should have said - Thanks for that romsek, I'm going to try and figure this all out now.

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