Given the following density function for continuous random variable x
f(x)=2/3($x^4$+5$x^3$+2$x^2$+3x-2) for x between [2,3]
Determine the c.df and use the c.df to determine p(2.25 < x < 2.5)
So first I realize I can't take the integral between 2 and 3 since the area under the curve would be 1 for a continuous random variable. Therefore I'm going to integrate between my lower limit 2 and my upper limit x. Would that be a good first step? Then from there I would just take my c.d.f and plug in the lower bounds and upper bounds. Not asking anyone to actually work through the problem just want to understand c.d.f better and how to tackle this sort of problem. Advice would be very helpful.
I notice that
$$\int_2^3\frac23(x^4+5x^3+2x^2+3x-2) \ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = \int_a^x f(x) \, dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$