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RIEMANN SUMS
Aaron Perry
Mr. Acre
AP Calculus
March 17, 2014
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Introduction:
Often, in Calculus, one is asked to approximate the area under a curve. There are various
methods of ways to do this, such as the Trapezoid Rule, Simpson’s Rule, and Riemann Sums.
However, there are certain cases in which each one should be used and why that is so. The
various uses of these particular approximation techniques for the integral will be discussed,
described, and demonstrated in order to enhance any readers knowledge about these Calculus
topics.
Part 1: Riemann Sums, Trapezoid Rule, and Simpson’s Rule
A Riemann sum is generally defined as an approximation of the area under a curve
between two points. Instead of calculating the exact value of the area under the curve, a Riemann
sum uses multiple rectangles in order to calculate the area. Figure 1 below shows an example of
a Riemann Sum.
Figure 1. Riemann Sum Graph Example
The graph shows the general idea of a Riemann sum. An approximation would be created
by finding the area of the rectangles and adding them together. The only difference between
rectangles is their height because in general the change in x between each rectangle is the same.
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The height of each rectangle would depend on what type of Riemann sum is selected. Figure 2
depicts the five types of Riemann sums.
Figure 2. Different Types of Riemann Sums
The figure depicts the five different types of Riemann sums. The red circles depict the point
that would be selected for each rectangle for each Riemann sum situation. Generally a Riemann
sum can be used as a time saver when trying to approximate the integral for a function whose
integral is hard to calculate via the Fundamental Theorem of Calculus.
The Trapezoid rule is similar as the Riemann sum as they are both approximations of the
definite integral. However, the only difference for the Trapezoid rule is that instead of finding the
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area of multiple rectangles, one is now finding the area of multiple trapezoids. Figure 3 below
depicts the equation for the Trapezoid rule with N number of trapezoids from point A to point B.
𝑇𝑛 = ∆𝑥(.5𝑓(𝐴) + 𝑓(𝑥1 ) + 𝑓(𝑥2 ) + 𝑓(𝑥3 ) + ⋯ + 𝑓(𝑥𝑛−1 ) + .5𝑓(𝐵))
Figure 3. Trapezoid Rule General Equation
Tn represents the area of under the curve being found with n number of trapezoids. F(A)
and F(B) represent the starting and stopping points respectively. ∆𝑥 represents the change in x for
each trapezoid. Finally, F(x1), F(x2), etc. represents the y-value for greatest x-value for the
trapezoid.
Simpson’s rule is also similar to the Trapezoid rule and Riemann sums because of the fact
that it is also an approximation of the definite integral. Simpson’s rule works by taking the
summation of various areas of parabolas under and above the original function. This method is the
most accurate as it takes accounts for the curves of the original function whereas Trapezoid rule
and Riemann sums attempt cannot fully do so due to the shapes (rectangles and trapezoids) for
each method not being curved. Also in order to use Simpson’s rule, there must be an even number
of increments which means there must be an odd number of data points. Figure 4 below depicts
the general equation used for Simpson’s rule.
𝑆𝑛 =
1
∆𝑥(𝑦0 + 4𝑦1 + 2𝑦2 + 4𝑦3 + ⋯ + 4𝑦𝑛−1 + 𝑦𝑛 )
3
Figure 4. Simspon’s Rule General Equation
Sn represents the area of under the curve being found with n number of equal intervals. ∆𝑥
represents the change in x for each interval. Finally, Y0, Y1, etc. represent the y=values for the end
points of the intervals.
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Part 2: Riemann Sum Example Problem
Given 𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5 on the interval from [1,5], illustrate
the five Riemann sums with 2 intervals. In each case, draw the appropriate rectangles. It must be
clear which value is being sued for the height of each rectangle.
Figure 5 through Figure 9 represent the left, right, midpoint, lower, and upper Riemann
sum graphs respectively, with the heights for each rectangle, for the given equation.
Figures 5-9. The Left, Right, Midpoint, Lower, and Upper Riemann Sum Graphs
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In order to calculate the area of each individual Riemann sum graph, the areas of the two
rectangles must be added together. Figure 10 shows the equations for the rectangle summation for
each Riemann sum graph pictured above.
Figure 10. Riemann Sum Graph Rectangle Summations
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Part 3: Trapezoid Rule Example Problem
For the equation 𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5, illustrate the Trapezoid
rule with 4 intervals. It must be clear which value is being used for the “bases” of each trapezoid.
Write the sum in the appropriate form using f(x). Figure 11 below represents the graph of the
equation when using the Trapezoid rule with 4 intervals.
Figure 11. Trapezoid Rule Graph
Figure 11 shows the labeled Trapezoid rule graph for the equation f(x). A represents the
starting point and B represents the ending point of the entire interval for the problem. X1, X2, and
X3 represent the other three bases for the trapezoids. Figure 12 below shows the summation of
each of the trapezoids using the Trapezoid rule equation.
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𝑇𝑛 = ∆𝑥(.5𝑓(𝐴) + 𝑓(𝑥1 ) + 𝑓(𝑥2 ) + 𝑓(𝑥3 ) + .5𝑓(𝐵))
𝑇𝑛 = 1(.5𝑓(1) + 𝑓(2) + 𝑓(3) + 𝑓(4) + .5𝑓(5))
𝑇𝑛 = 1(6.5 + 8 + 5 + 4 + 14.5) = 38𝑢2
Figure 12. Trapezoid Rule Equation and Solution
The approximate integral from the Trapezoid rule was 38u2. However, the definite integral
for f(x) on the interval [1,5] is actually 32.8u2. The reason that the Trapezoid rule overestimates
the graph is that when the graph for the original function is concave up (the curve looks like a
smile) then the trapezoid will overestimate the area for that interval. When the graph is concave
down (the curve looks like a frown) then the trapezoid will underestimate the area for that interval.
The reason being that the entire graph is overestimated is because of the significant increase in the
y-value from x = 4 to x = 5. This significant increase causes the trapezoid to significantly
overestimate the area for that particular interval, thus leaning the approximation towards being an
overestimate overall.
Part 4: Mean Value Theorem
The Mean Value Theorem for integrals is essentially a modified form of the Mean Value
Theorem for derivatives combined with the first fundamental theorem of calculus. It essentially
states that there is a point c where the value of the function is the same as the average value of
the function within the given interval [a,b]. Figure 13 below states the Mean Value Theorem
Equation for integrals.
𝑓(𝑐) =
𝑏
𝑏
1
∫ 𝑓(𝑥)𝑑𝑥 = 𝑓𝑎𝑣𝑔 𝑡ℎ𝑒𝑛 ∫ 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑐)(𝑏 − 𝑎)
𝑏−𝑎 𝑎
𝑎
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Figure 13. Mean Value Theorem Equation for Integrals
The three values a, b, and c all represent a point on the given graph f. The a represents the
starting point, b represents the ending point, and c represents a point on f. Figure 14 below shows
the Mean Value Theorem for integrals can be used for the equation 𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 −
3)3 − 4(𝑥 − 3) + 5.
Figure 14. Mean Value Theorem for Integrals
Figure 14 shows how it is possible to separate the graph into two rectangles and calculate
the definite integral from those two rectangles. After plugging in the values into the MVT equation,
the average value turned out to be 8.2. By plotting 8.2 on the graph f(x) and comparing the
intersection, the values of 1.9 and 4.37 were the x-values that corresponded with it. Figure 15
shows how now the definite integral can be calculated by using the point 1.9 to create the two
rectangles for this particular scenario (using 4.37 will result in the same answer but different
numbers to multiply and add).
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𝑓(1.9)(. 9) + 𝑓(1.9)(3.1)
. 9 ∗ 8.2 + 3.1 ∗ 8.2 = 7.38 + 25.42 = 32.8𝑢2 (𝐴𝑝𝑝𝑟𝑜𝑥. ) = 32.8𝑢2 (𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙)
Figure 15. Mean Value Theorem for Integrals Calculations
Part 5: Given Problems
A single problem, listed below, will now be worked out solved using the methods
explained in the past four parts.
The volume of a spherical hot air balloon expands as the air inside the balloon is heated.
The radius of the balloon, in feet, is modeled by a twice-differentiable function r of time t, where
t is measured in minutes. For 0 < t < 12 the graph is concave down. The table below gives
selected values of the rate of change, r’ (t), of the radius of the balloon over the time interval 0 ≤
t ≤ 12. The radius of the balloon is 32 feet when t = 7. (The volume of a sphere of radius r is
given by V = 4/3 π r3.)
Table 1
Given r’(t) Values
t (minutes)
0
r’ (t)
(ft/min)
5.7
1
4
7
11
12
4.0
2.0
1.4
0.5
0.4
A) Estimate the radius of the balloon when t – 7.2 using the tangent line approximation at t –
7. Is your estimate greater than or less than the true value? Give a reason for your answer.
B) Find the rate of change of the volume of the balloon with respect in time when t – 7.
Indicate the units of measure.
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C) Use a right Riemann sum with 5 subintervals indicated by the data in the table to
approximate the integral from 0 to 12 of r’(t). Using the correct units, explain the
meaning of this integral in terms of the radius of the balloon.
D) Is your Approximation in part c greater than the definite integral for r’(t)? Give a reason.
Figure 15-18 shows the work and calculations for parts A, B, C and D respectively.
Figure 15. Work and Calculations for Part A
Figure 16. Work and Calculations for Part B
Figure 17. Work and Calculations for Part C
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Figure 19. Work and Calculations for Part D
Conclusion:
Although some of this information in regards to finding approximations of the definite
integral may seem pointless. There are various situations in real life which these topics can be
applied. Hopefully the reader will be able to understand, visualize, and demonstrate these
concepts when the situation calls for it.