Sum of all probabilities is not 1 in homework Excerpt from "A first course in Probability, 8ed"

A student is taking a one-hour-time-limit makeup examination. Suppose the probability that the student will finish the exam in less than $x$ hours is $x/2$, for all $0 ... x ... 1$. Then, given that the student is still working after .75 hour, what is the conditional probability that the full hour is used?

What I am concerning here is $\int_0^1\frac{x}{2}dx=\frac{1}{4}$, which is not $1$. Therefore, this is not a probability distribution function.
Am I correct? Thank you
 A: The information provided by the statement "Suppose the probability that the student will finish the exam in less than $x$ hours is $x/2, ...$ refers to the cumulative distribution function of the amount of time required to finish the exam, not the probability distribution function.  With a one-hour time limit, $x$ cannot be greater than $1$, at which time we can see that only half the students have finished the exam at all.  This is OK!  Students sometimes don't!  
We can model this by observing that if a student doesn't finish the exam in the full hour, they have still used the full hour.  Consequently, the cumulative distribution function has a jump at $x=1$:
$$P(x) = {x \over 2} + {1 \over 2}1(x = 1)$$
where $1(a)$ is the indicator function that that takes on the value $1$ if its argument $a$ evaluates to "true" and $0$ if it evaluates to "false".
The cumulative distribution of time-to-completion given that the student has worked for $3/4$ of an hour will be:
$$P(x|x \geq 0.75) = {P(x) - P(0.75) \over 1-P(0.75)} = {P(x) - P(0.75) \over 1-0.375} = \left({8 \over 5}\right)(P(x) - 0.375)$$
Taking the derivative w.r.t. $x$, and remembering the discrete mass at $x=1$, gives us the probability function:
$$\begin{eqnarray}
p(x|x\geq 0.75) &=& {4 \over 5},\; x < 1 \\
&=& \left({8 \over 5}\cdot{1\over 2}=0.8\right),\; x=1
\end{eqnarray}$$
and the answer is evidently $0.8$.
A: Edit2: The function x/2 is not a probability density function, but a part of a cumulative distribution function. It's integral thus does not need to sum up to 1. (And a note, for density function the integral which should add up to 1 should be from $-\infty$ to $+\infty$, not from 0 to 1).
