PDF for a continuous random X, where $a$ is a constant such that $1< a \leqslant 5$ I have a pretty decent length question - so if you don't feel like answering each part in its entirety... the first few steps might be all I need. :)
The information given
Consider the probability density function for a continuous random variable, X.
$$f(x) =
\begin{cases}
\frac{a+3}{4}x,  & \text{0 $\le$ $x$ $\le$ 1} \\[2ex]
\frac{a-5}{4(a-1)^2}(x-a), & \text{1 $<$ $x$ $\le$ a}
\end{cases}$$
Where $a$ is a constant such that  $1< a \le 5$
Now the questions
(a) What is P$(X < 1)$ when $a = 2$
Given that this is continuous I assume we must do some sort of integration. Do we just substitute a=2 and integrate between X and 1? I don't get it...
(b) What is P$(1 < X < 2)$ when $a=2$
Again, do we evaluate the integral between $1$ and $2$, and sub $a$ for $2$?
(c) What is P$(X > 1)$ when $a = 5$
(d) Show that $f(x)$ is a valid PDF, for all values of $a$
Now, I know that to satisfy the criteria for a valid PDF there must...
1) $f(x) \ge 0 $ for all $x$,
and
2) $\int_{-\infty}^\infty f(x)\, dx =1$.
But that's about all I know...
Conclusion
I know my small contributions will not go far in getting the answer, so for that I apologise. To be honest...I'm just stuck and need some pointers / answers.
Thanks so much guys!
 A: a) The problem should have specified that  the density function is $0$ when $x\lt 0$ and when $x\gt a$. Your integration suggestion is the right one. We have
$$\Pr(X\lt 1)=\Pr(0\le X\lt 1)=\int_0^1 f(x)\,dx.$$
For $a=2$, the density function is $\frac{5}{4}x$ on the interval $[0,1]$. It follows that
$$\Pr(0\le X\lt 1)=\int_0^1 \frac{5}{4}x\,dx=\frac{5}{8}.$$
b) This is very similar to a). We want
$$\int_1^2 -\frac{3}{4}(x-2)\,dx.$$
However, there is a somewhat simpler way. For $\Pr(0\le X\le 1)+\Pr(X\gt 1)=1$, and now we can use the result of the computation a).
c) The relevant part of the density function is from $x=1$ to $x=5$. We get lucky, since for $a=5$ the density function is $0$ on the interval $[1,5]$.
d) You wrote down the things we need to verify. The given density function is non-negative. This is clear for $0\le x\le 1$. For $1\lt x\le a$, there is the term $a-5$, which is $\le 0$. But the term $x-a$ is also $\le 0$, so the product is $\ge 0$.
Finally, you need to show that 
$$\int_0^1 \frac{a+3}{4}x\,dx+\int_1^a \frac{a-5}{4(a-1)^2}(x-a)\,dx=1.$$
We now calculate. The algebra is somewhat messy.
