Let X=the time to failure and probability distribution function f(x)=3/(x+3)^2 where x>0. i) Determine the cumulative distribution function.
ii) Calculate the probability that the time to failure is between is between 2 and 5 years.
iii) Obtain the expected time to failure
iv) Obtain the time before which 90% of the components will fail.


*

*I know how to find the cumulative distribution function, all i need to do is integrate the given pdf, but what i'm really confused by here is the fact that there are no defined limits. And my lecture told me to determine it, so do i leave it as a indefinite integral, or is there something that i'm missing from the question?

*The second part is easier as the limits are defined here, so all i need to do is find the integral between 2 and 5 to find P(2

*This part i just do the same thing as the previous part but this time find the integral with the pdf multiplied by x, with the same limits.
-This part is the one that confuses me the most. I dont know how i would do this. Ive gotten as far as knowing that i need to find P(X<0.9)
 A: As $x > 0$ for the pdf, call it $p(x)$, the cdf looks like it will be just
$$c(x) = \int_0^x p(t) \ dt = \int_0^x \frac{3}{(t+3)^2} dt = 3\left( \frac{1}{3} - \frac{1}{x+3}\right) = \frac{x}{x+3}$$ As we would expect $\lim_{x\to\infty} c(x) = 1$.
For the second part integrate the pdf between 2 and 5, assuming the variable was years to begin with.
For the third part, calculate as usual $\displaystyle E[x] = \int_0^\infty xp(x) \ dx$.
For the fourth part, use the cdf.
A: 
•I know how to find the cumulative distribution function, all i need to do is integrate the given pdf, but what i'm really confused by here is the fact that there are no defined limits. And my lecture told me to determine it, so do i leave it as a indefinite integral, or is there something that i'm missing from the question

You know that the support for $f(x)$ is $x\geq 0$, so you want to integrate:
$$F(x) = \Pr(0\leq X\leq x) = \int_0^x f(t)\operatorname d t$$ 

•The second part is easier as the limits are defined here, so all i need to do is find the integral between 2 and 5 to find P(2

Yes.  You want $F(5)-F(2)$   (assuming that $X$ is measured in years.)

•This part i just do the same thing as the previous part but this time find the integral with the pdf multiplied by x, with the same limits.

Mostly.  Expectation is integrated over the entire support; which again is all $x\geq 0$.  Hence:
$$\mathsf E(X) = \int_0^\infty x\;f(x)\operatorname d x$$

This part is the one that confuses me the most. I dont know how i would do this. Ive gotten as far as knowing that i need to find P(X<0.9)

No, what you want to find is the time the probability of failure passes $0.90$.
That is, find the $x$ such that $F(x) = \Pr(0\leq X\leq x) = 0.9$
