Expected consultant bill given distribution of time taken 
The cdf of the number of hours it takes a consultant to complete a project is given by $F(x)= \dfrac{x^2}{16}$ for o to 4. The consultant bills $300 per hour, rounded up to the nearest half hour, for the project. What is the expected amount of the total bill?
(a)900
(b)800
(c)872
(d)950
(e)1100

My work:
$f(x)= \dfrac{dF(x)}{dx}$
$f(x)=\dfrac{x}{8}$
so
integral of $x^2/8$ from 0 to 4 = $x^3/24$ from 0 to 4 = 64/24 = 2.6667
round 2.6667 to nearest half hour is 2.5
so
2.5*300 = 800
But that's wrong; the answer to the question is 872.
Can I please have help understanding why my method is incorrect, so I can try another method while understanding why my last attempt was incorrect.
 A: Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150\left(\frac{0.5^2}{16}-\frac{0^2}{16}\right)+$300\left(\frac{1^2}{16}-\frac{0.5^2}{16}\right)+\cdots+$1200\left(\frac{4^2}{16}-\frac{3.5^2}{16}\right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150\left(\frac{3.5^2}{16}+\cdots+\frac{0.5^2}{16}+\frac{0^2}{16}\right)=$871.875\approx$872$$
A: The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$\text{Since }Y=300X,\quad E[Y]=300E[X].$$
However, we actually have $Y=300\,\mathtt{rnd}(X)$, where $\mathtt{rnd}$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$\text{Since }Y=300\,\mathtt{rnd}(X),\quad E[Y]=300\,\mathtt{rnd}(E[X])\quad\color{red}{\text{False!}}$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
A: Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, \ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
A: The probability for the duration to be between $\frac{k-1}2$ and $\frac k2$ is $\frac{2k-1}{64}$ and the charge would be $150k$. Thus, the expected charge would be
$$
\begin{align}
\frac{75}{32}\sum_{k=1}^8\left(2k^2-k\right)
&=\frac{75}{32}\sum_{k=1}^8\left[4\binom{k}{2}+\binom{k}{1}\right]\\
&=\frac{75}{32}\left[4\binom{k+1}{3}+\binom{k+1}{2}\right]_{k=0}^{k=8}\\
&=\frac{75}{32}\left[4\binom{9}{3}+\binom{9}{2}\right]\\
&=\frac{75}{32}\cdot372\\[9pt]
&=871\tfrac78
\end{align}
$$
