Middle Boxes

The Riemann sum determined by the midpoints is given by

$\displaystyle {\frac{{b-a}}{{n}}}$$\displaystyle \sum_{{i=0}}^{{n-1}}$f$\displaystyle \left(\vphantom{ a+\frac{b-a}{2n}+i\frac{b-a}{n}}\right.$a + $\displaystyle {\frac{{b-a}}{{2n}}}$ + i$\displaystyle {\frac{{b-a}}{{n}}}$$\displaystyle \left.\vphantom{ a+\frac{b-a}{2n}+i\frac{b-a}{n}}\right)$

which is the sum of the areas of rectangles whose heights are determined by midpoints of subintervals.

$\blacktriangleright$ To make a middle-boxes plot

1.
Leave the insertion point inside the expression x sin x.

2.
From the Calculus submenu, choose Plot Approx. Integral.

3.
Click the plot to select the frame, or double-click the plot to select the view.

4.
Click the properties tool itbpF0.3009in0.3009in0.0701inproperty.wmf.

5.
Reset the interval and number of boxes as desired.

6.
Choose OK.

$\blacktriangleright$ Calculus + Plot Approx. Integral

x sin x

dtbpF3in2.0003in0pt

Applied to the expression x sin x, with four rectangles and limits 0 and 3, the approximating Riemann sum is

$\displaystyle {\frac{{3}}{{4}}}$$\displaystyle \sum_{{i=0}}^{{3}}$$\displaystyle \left(\vphantom{ \frac{3}{8}+i\frac{3}{4}}\right.$$\displaystyle {\frac{{3}}{{8}}}$ + i$\displaystyle {\frac{{3}}{{4}}}$$\displaystyle \left.\vphantom{ \frac{3}{8}+i\frac{3}{4}}\right)$sin$\displaystyle \left(\vphantom{ \frac{3}{8}+i\frac{3}{4}}\right.$$\displaystyle {\frac{{3}}{{8}}}$ + i$\displaystyle {\frac{{3}}{{4}}}$$\displaystyle \left.\vphantom{ \frac{3}{8}+i\frac{3}{4}}\right)$ = 3.17839115

Direct evaluation using Evaluate Numerically produces

$\displaystyle \int_{{0}}^{{3}}$x sin xdx = 3.111097498