Examine the Riemann sum below for the area under $f\left(x\right)$.

$\displaystyle \sum _ { i = 0 } ^ { 11 } \frac { 6 - 3 } { 12 } f ( 3 + \frac { 6 - 3 } { 12 } \cdot i )$

1. How many rectangles were used?

   Hint 1(a):

   Notice that the index starts at $0$ and ends at $11$.

   Hint 2(a):

   Also recall that $\Delta x= \frac{b-a}{n}$, where $n$ represents number of rectangles.

2. If the area being approximated is $A(f, a ≤ x ≤ b)$, what are $a$ and $b$?

   Hint 1(b):

   $a = 3$ is easy to spot in a Riemann sum. Now find the end-value, $b$.
   General form of a left-endpoint Riemann sum:
   $\displaystyle\sum_{i=0}^{n-1}\Delta xf(a+\Delta xi)$

   Hint 2(b):

   You already computed how many rectangles there are part (a).

   Hint 3(b):

   Notice that $\Delta x= \frac{6-3}{12}=\frac{1}{4}$, so each rectangle has a width of $\frac{1}{4}$.