Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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 $x^*_i=x_{i-1}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.