5.2: Sum Rectangles
This activity is intended to supplement Calculus, Chapter 4, Lesson 4.
ID: 12099
Time Required: 15 minutes
Activity Overview
In this activity, students will graphically and numerically explore Riemann sums and develop an understanding of summation notation for adding these rectangles.
Topic: Riemann sums
 Graphically exploring Riemann sums
 Summation notation
Teacher Preparation and Notes
 The file area.89p is a program file. It contains several approximation methods using Riemann sums, the Trapezoidal Rule, and Simpson’s Rule. This program should be sent to student graphing calculators prior to beginning this activity
 To download the calculator file, go to http://www.education.ti.com/calculators/downloads/US/Activities/Detail?id=12099 and select main.area.89p
Associated Materials
 Student Worksheet: Sum Rectangles http://www.ck12.org/flexr/chapter/9729, scroll down to the second activity.
 main.area.89p
Part 1 – Graphical Riemann Sums
Students will use the program area to complete this part of the activity. To run the program enter area() on the Home screen. To use the program, select to view the approximation either graphically or numerically. Next, enter the equation that you are examining and press \begin{align*}\div\end{align*}
Many questions can be asked and observations made. This section can be extended by asking about other functions. Students should graph other functions that have different concavity or slope to complete Exercise 3.
Students get a visual of what Riemann sums are. If only a few rectangles are graphed students may be able to better see and understand why the midpoint is an over approximation in some situations and an under approximation at other times.
Student Solutions
 The midpoint approximation is between the left and right endpoint approximations. The left is too big and the right too small.
 The Riemann sums converge (to the definite integral) as \begin{align*}n \to \infty \end{align*}
n→∞ . With thinner widths, the rectangles approach the true area.  (a) over, (b) under, (c) over, (d) under. For example, when the function is decreasing and concave down, the function curves more steeply for the second half of the rectangle than the first. Therefore, since there is more rectangle above than below the graph, it over estimates the area.
Part 2 – Summation notation
In Part 2, students examine the summation notation. By reading and answering questions on the student worksheet, students will better understand the sigma notation.
To make entering the summation easier, encourage students to define the function before entering the sum. To enter a summation on the HOME screen, students should select \begin{align*}F3\end{align*}
\begin{align*}\sum\end{align*}
Student Solutions

\begin{align*}\triangle x = h = 1\end{align*}
△x=h=1 
\begin{align*}y(a) + y(a + 1) + y(a + 2) + y(a + 3) + y(a + 4)\end{align*}
y(a)+y(a+1)+y(a+2)+y(a+3)+y(a+4)  The right Riemann sums from \begin{align*}a\end{align*}
a to \begin{align*}b\end{align*}b with 5 subintervals, looks like \begin{align*}y(a + 1) + y(a + 2) + y(a + 3) + y(a + 4) + y(a + 5)\end{align*}y(a+1)+y(a+2)+y(a+3)+y(a+4)+y(a+5) . Note \begin{align*}a + 5 = b\end{align*}a+5=b since \begin{align*}\triangle x = 1\end{align*}△x=1 
\begin{align*}\sum_{i=1}^n (h \cdot y(a + (i  1) h))  h = \frac{6 1}{n}\end{align*}
∑i=1n(h⋅y(a+(i−1)h))h=6−1n
When \begin{align*}n = 10\end{align*}
When \begin{align*}n = 20\end{align*}
When \begin{align*}n = 50\end{align*}
When \begin{align*}n = 100\end{align*}
Extension – Area Programs
Students will use the area program to compare area approximation methods.
Student Solutions

\begin{align*}n & = 10 && \qquad n = 50 && \qquad n = 100 \\
\quad \text{mid} & = 71.5625 && \quad \text{mid} = 71.6625 && \quad \text{mid} = 71.6656 \\
\quad \text{trap} & = 71.875 && \quad \text{trap} = 71.675 && \quad \text{trap} = 71.6688 \end{align*}
nmidtrap=10=71.5625=71.875n=50mid=71.6625trap=71.675n=100mid=71.6656trap=71.6688  For \begin{align*}y = x^2\end{align*}
y=x2 , an increasing concave up function in this domain, the midpoint is approaching \begin{align*}71\frac{2}{3}\end{align*}7123 from the left and the trapezoid method is approaching the value of the definite integral from the right. Midpoint gives a better approximation.
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