Definite Integral, Sample Points Problem: Evaluate the Riemann sum for $f(x) = 3 - 1/2x$, $2 \le x \le 14$, with six subintervals, taking the sample points to be left endpoints.
Question: How is it that $x_i = x_{i-1}$?
My logic: $\Delta x = 2$, therefore, $x_i = 2 + 2i$.

1. $\begin{align}
\displaystyle f(x)=3-\frac{x}{2},2\le x\le 14 \\
\Delta x=\frac{b-a}{n}=\frac{14-2}{6}=2
\end{align}$
Since we are using endpoints, $x_i^{*}=x_{i-1}$
$\begin{align}
\displaystyle L_6=& \sum_{i=1}^6 f(x_{i-1}) \Delta x \\
&=(\Delta x) [f(x_0)+f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)] \\
&=2[f(2)+f(4)+f(6)+f(8)+f(10)+f(12)] \\
&=2[2+1+0+(-1)+(-2)+(-3)] \\
&=2(-3)=-6
\end{align}$
The Riemann sum represents the sum oh the area of the two rectangles above the x-axis minus the sum of the areas of the three rectangles below the x-axis; that is, the net areaof the rectangles with respect to the x-axis.
Font: Stewart Calculus Early Trans. 7th Edition, Chapter 5, Section 2, Problem 1
 A: Your logic
Your logic is correct, but you didn't realize that the solution does not say $x_i=x_{i-1}$. It says $x_i^{\color{blue}{*}}=x_{i-1}$. In the definition of Riemann sums, we have $x_{i-1} \le x_i^{*} \le x_i$. When we choose a left Riemann sum (i.e. take sample points to be left endpoints), we are actively making the choice $x_i^{*}=x_{i-1}$ since $x_i^{*}$ is otherwise arbitrary.
For more
See this.

Explicit Details
Here's the definition of a Riemann sum as defined by Wiki:
Let $f: D \to \mathbb{R}$ be a function defined on a subset, $D$, of the real line, $\mathbb{R}.$ (Note that $f:D \to \mathbb{R}$ just means $f$ is a map from $D$ to $\mathbb{R}$.) Let $I=[a,b]$ be a closed interval contained in $D$ (meaning $I \subset D$), and let $P=\{[x_0,x_1),[x_1,x_2),\dots,[x_{n-1},x_{n}]\}$ be a partition of $I$, where $a=x_0<x_1<x_2<\cdots<x_n=b$. (This just means you're breaking $P$ into cells $[x_i, x_{i+1})$ with the special case $[x_{n-1},x_n\color{blue}{]}$ since $x_n=b$ and $b \in I$.)
The Riemann sum of $f$ over $I$ with partition $P$ is defined as $$S=\sum_{i=1}^{n}f(x_i^{*})(x_i-x_{i-1}), \quad x_{i-1}\le x_i^*\le x_i.$$
As emphasized, the choice of $x^*_i$ can be anything within the interval $[x_{i-1},x_i]$. Depending on our choices, we can have a vast variety of sums. However, there are three main sums that relate to your particular case:

*

*left Riemann sum: $x^*_i=x_{i-1}.$

*right Riemann sum: $x^*_i=x_{i}.$

*middle Riemann sum: $x^*_i=\dfrac{x_i+x_{i-1}}{2}.$
Your particular case is (1), hence $x^*_i=x_{i-1}$.
