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 area
of the rectangles with respect to the x-axis
.
Font: Stewart Calculus Early Trans. 7th Edition, Chapter 5, Section 2, Problem 1