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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, $x_i = 2 + 2i$.

The image is from the solutions manual, not the textbook.

Stewart Calculus Early Trans. 7th Edition, Chapter 5, Section 2, Problem 1

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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:

  1. left Riemann sum: $x^*_i=x_{i-1}.$
  2. right Riemann sum: $x^*_i=x_{i}.$
  3. middle Riemann sum: $x^*_i=\dfrac{x_i+x_{i-1}}{2}.$

Your particular case is (1), hence $x^*_i=x_{i-1}$.

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