### Home > CALC > Chapter 3 > Lesson 3.3.2 > Problem3-106

3-106.

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?

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

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

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

$a = 3$ is easy to spot in a Riemann sum. Now find the end-value, $b$.

General form of a left-endpoint Riemann sum:

$\sum_{i=0}^{n-1}\Delta xf(a+\Delta xi)$

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

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