<s>[INST] How to find the $f\left(x_i\right)$ for midpoint Riemann sum 
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\mbox{Time} & \mbox{9am} & \mbox{10:30 am} & \mbox{Noon} & \mbox{1:30 pm} & \mbox{3 pm} & \mbox{4:30 pm} & \mbox{6 pm} & \mbox{7:30 pm} & \mbox{9 pm} \\
\hline
t & 0 & 1.5 & 3 & 4.5 & 6 & 7.5 & 9 & 10.5 & 12 \\
\hline
P\left(t\right) & 200 & 728 & 1193 & 1329 & 1583 & 1291 & 804 & 256 & 0 \\
\hline
\end{array} 
$ 
Use a midpoint Riemann sum with four intervals of equal size to estimate the total number of people seeking care during the 12-hour period. 

I know the formula for the midpoint Riemann sum is $ \int_a^b f\left(x\right) \; \mathrm{d}x \approx \frac{b-a}{n} \left[ f\left(x_1\right) + f\left(x_2\right) + \ldots + f\left(x_n\right) \right]$
I know that $b = 12$ and $a = 0$ and $n = 4$, but what is my $f\left(x_1\right), \; f\left(x_2\right), \; f\left(x_3\right), \; f\left(x_4\right)$
 [/INST] For $n$ points from $a$ to $b$, 
$x_i = a + \frac{b-a}{n}(i-\frac1{2})$.
Note that $x_1 = a + \frac1{2}\frac{b-a}{n}$
is just to the right of $a$,
and $x_n = a+(n-\frac1{2})\frac{b-a}{n}
= b - \frac1{2}\frac{b-a}{n}$
is just to the left of $b$.
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