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Desiree Boyd
Tallman
AP Calculus
23 March 2015
Riemann Sums Essay
An important topic discussed in calculus is the Riemann sum. A Riemann sum is related
to the integral, which calculates the area under a function. A Riemann sum is the process of
calculating and adding the area of rectangles that are placed under a function with a given
number of rectangles and width of each one. Riemann sums can have a left, right, upper, lower,
or midpoint height. These heights are where the curve of the function and top of the rectangles
meet.
Figure 1. Left and Right Riemann Sums
In Figure 1, the right and left Riemann sums from 1 to 5 are displayed for the function
𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5. The left Riemann sum shown on the left has four
rectangles with a width of one. The area of the rectangle when x = 1 would be f(1)(1) with the
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width being one as well. The area of first two rectangles in the left Riemann sum would be
f(1)(1) + f(2)(1). The area of the first two rectangles in the right Riemann sum would be a similar
f(2)(1) + f(3)(1).
Figure 2. Upper and Lower Riemann Sum
Figure 3. Midpoint Riemann Sum
Figures 2 and 3 show the upper, lower, and midpoint Riemann sums for the function used
in Figure 1: 𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5. Upper Riemann sums overestimate
the area with the maximum height whereas lower Riemann sums underestimate with the
minimum height ("Riemann Sums - HMC Calculus Tutorial."). In the upper Riemann sum, when
x is 1, 2, and 3, the rectangle is at its highest on the left side of the rectangle. When x is 5, the
height of the rectangle is at its highest on the right side of the rectangle. The sum of the first two
rectangles with a width of one for the upper Riemann sum is f(1)(1) + f(2)(1). The midpoint
Riemann sum is a little different from the other Riemann sums. Instead of the height being 1 or
2, the height is f(1.5), the middle of the rectangle. The sum of the first two rectangles of the
midpoint Riemann sum with a width of one is f(1.5)(1) + f(2.5)(1).
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The Riemann sum uses rectangles to find the area under a curve. There are other methods
to calculate area with different shapes for a more precise calculation. For example, the trapezoid
rule uses trapezoids in the place of rectangles.
Figure 4. Trapezoid Rule for f(x)
Figure 4 shows the function 𝑓(𝑥) = (𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5 with four
trapezoids with a width of one from 1 to 5. The area of a trapezoid is the following:
𝐴=
1
[𝑏 +𝑏 ]ℎ
2 1 2
The b1 and b2 represent the two bases of a trapezoid which are added together and multiplied by
one half and the height of the trapezoid. In Figure 4, the height of the trapezoid is equivalent to
the width of the rectangles in the Riemann sum process. The two bases of a trapezoid are the f(x)
values at the beginning and end of the height. With this knowledge of how to calculate a single
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trapezoid, there is a way to combine all the trapezoids that are desired under a function in a
designated interval. If the addition of multiple trapezoids is simplified, the equation is as follows:
𝐴=
1
[𝑓(𝑎) + 2𝑓(𝑥) … + 𝑓(𝑏)](ℎ)
2
The a represents the first term and the b represents the last. Each term is multiplied by 2 except
for the first and the last term. The common height of each trapezoid is represented by h.
Another alternative for a Riemann sum is Simpson’s rule. Simpson’s rule involves taking
the sum of parabolas. It is the most accurate of the three methods discussed with the Riemann
sum being the least accurate.
Figure 5. Comparing the Trapezoid and Simpson Rule (Dawkins)
Figure 5 shows the overlay of trapezoids and parabolas to clearly visualize the difference
between the trapezoid rule and Simpson’s rule. Simpson’s rule requires an even number of
values. This is because one parabola covers the length of two trapezoids as shown in the figure.
The equation for Simpson’s rule is derived from the quadratic equation y = ax2 + bx + c and is
similar to the trapezoid rule:
𝐴=
1
ℎ[𝑦0 + 4𝑦1 + 2𝑦2 + 4𝑦3 + 2𝑦4 + 4𝑦5 + ⋯ 4𝑦𝑛−1 + 𝑦𝑛 ]
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The h, like in the trapezoid rule and Riemann sum, is the common width of the parabolas. Instead
of multiplying the values by a half like when using the trapezoid rule, the values are multiplied
by a third. The values are multiplied by an alternation of 4 and 2 except the first and last term.
The Mean Value Theorem for integrals is the average f(x) value on [a, b]. An integral
calculates the area under a curve. The Mean Value Theorem’s equation is as follows and states
that there will exist a number c in [a, b] such that:
𝑏
1
∫ 𝑓(𝑡)𝑑𝑡 = 𝑓(𝑐)
(𝑏 − 𝑎)
𝑎
To evaluate the average value, the integral must be taken of the function by taking the antiderivative. Once the anti-derivative is found, it must be evaluated at b and a. The final step is
subtracting a from b to find the average f(x) value on [a, b].
𝑏
1
=
∫ 𝑓(𝑥)𝑑𝑥
𝑏−𝑎
𝑎
2
1
=
∫(𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5 𝑑𝑥
2−1
1
=
1 (𝑥 − 3)5 2(𝑥 − 3)4 4(𝑥 − 3)2
2
∗
+
−
+ 5𝑥 |
1
5
4
2
1
(2 − 3)5 2(2 − 3)4 4(2 − 3)2
(1 − 3)5 2(1 − 3)4 4(1 − 3)2
=[
+
−
+ 5 ∗ 2] − [
+
−
+ 5]
5
4
2
5
4
2
= 9.7 units
Figure 6. Calculating the First Rectangle Using the MVT for Integrals
Figure 6 shows the calculations for finding the average height of a rectangle of the
function f(x) used in the beginning of this paper. See Figure 1 if necessary.
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Figure 7. Mean Value Theorem Visual
Figure 7 shows what the Mean Value Theorem is describing. The value c is the average
height of the function on the interval [a, b].
Figure 8. MVT on f(x) First Rectangle
Figure 8 applies the Mean Value Theorem to the function f(x) that was calculated in
Figure 6. The interval [a, b] in the case of this problem is when x is equal to 1 and 2. As shown
in Figure 6, the value of c is 9.7 units.
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1
𝑏
= 𝑏−𝑎 ∫𝑎 𝑓(𝑥)𝑑𝑥
3
1
=
∫(𝑥 − 3)4 + 2(𝑥 − 3)3 − 4(𝑥 − 3) + 5 𝑑𝑥
3−2
2
= 6.7 𝑢𝑛𝑖𝑡𝑠
Figure 9. Calculating the MVT for the Second Rectangle
Figure 9 shows the calculations for finding the average height of the rectangle in the
interval [3, 4]. The average height is equal to 6.7 units and seems accurate when looking at the
graph in Figure 1.
Figure 10. MVT of f(x) Second Rectangle
Figure 10 applies the Mean Value Theorem to the function f(x) that was calculated in
Figure 9. The interval [a, b] in the case of this problem is when x is equal to 3 and 4. As shown
in Figure 9, the value of c is 6.7 units.
The following is an example problem that will include the method discussed previously
in this paper: a 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
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seconds. For 0 < t < 12, the graph is concave down. When t = 7, the radius of the balloon is 32
feet.
Table 1
Selected Values of the Rate of Change, r’(t)
0
1
Time (s)
5.7
4.0
R’(t) (ft/s)
4
2.0
7
1.4
11
0.5
12
0.4
Table 1 provides the given values of the rate of change, r’(t), of the radius of the balloon
over the time interval 0 < t < 12. If the radius of the balloon when t = 7.2 is to be estimated using
the tangent line approximation at t = 7, it’s important to remember that the radius of the balloon
is 32 at t = 7. The point (7, 32) will be the point used in the tangent line approximation. The
slope of the tangent line will the r’(t), or 1.4 as given in the table. With these, the tangent line can
be composed as this:
𝑟(𝑡) = 1.4𝑡 + 22.2
With the tangent line, if 7.2 is plugged into the equation, the approximation for the radius would
be 32.28 feet. When applying the tangent line to its graph, the graph would be concave down.
This is because the radius was filled up with hot air and then depletes as time goes on. The
table’s values show the decreasing values of the radius. When a tangent line is applied to a
concave down graph, the approximation is an overestimate.
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It is known that the volume of a sphere is 𝑉 = 3 𝜋𝑟 3. To find the rate of change of the
volume, the equation requires implicit differentiation. Once the equation is differentiated, the
rate of change of the volume of the balloon with respect to time when t = 7 can be found.
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4
𝑉 = 3 𝜋𝑟 3
𝑑𝑉 4 2 𝑑𝑟
= 𝜋𝑟
𝑑𝑡 3
𝑑𝑡
𝑑𝑉
= 4𝜋(32)2 (1.4)
𝑑𝑡
𝑑𝑉
𝑓𝑡 3
= 5734.4 𝜋
𝑑𝑡
𝑠
Figure 11. Implicitly Differentiating the Volume Equation
Figure 11 show the process of implicitly differentiating and plugging values into the
volume equation. When implicitly differentiated, the rate of change of the volume can be found
when t = 7 by plugging in 32 for the radius and 1.4 for the rate of change of the radius. This
ultimately produces a rate of change of the volume of 5734.4π
𝑓𝑡 3
𝑠
.
Figure 12.
Figure 12 shows the graph of the given values in Table 1. To evaluate this graph, a right
Riemann sum can be used. All of the y-values are given in Table 1.
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𝑅5 = 𝑓(1)∆𝑥 + 𝑓(4)∆𝑥 + 𝑓(7)∆𝑥 + 𝑓(11)∆𝑥 + 𝑓(12)∆𝑥
𝑅5 = 4(1) + 2(3) + 1.4(3) + 0.5(4) + 0.4(1)
𝑅5 = 4.0 + 6.0 + 4.2 + 2.0 + 0.4 = 16.6 𝑓𝑡
Figure 13. Right Riemann Sum Evaluation
Figure 13 shows the values of the right Riemann sum calculated for the graph in Figure
12. Because the widths of the rectangles are of different lengths, the symbol ∆x is used to
represent each width. The sum of the rectangles, or just Riemann sum, is equal to 16.6 ft. This
approximation is an underestimate. This is because each of the right rectangles are below the
function (red line). Therefore, the radii of the rectangles are below the true radius values.
This paper discussed the definition of a Riemann sum, the trapezoid rule, Simpson’s rule,
and the Mean Value Theorem for Integrals. The Riemann sum, the trapezoid rule, and Simpson’s
rule are estimates for finding the area under a curve in a specific interval. They can either
overestimate or underestimate depending on the function’s shape. The Mean Value Theorem will
produce the value c which is the average f(x) value on [a, b]. All of the topics discussed can be
applied to real life situations as demonstrated previously.
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Works Cited
"Riemann Sums - HMC Calculus Tutorial." Riemann Sums - HMC Calculus Tutorial. Harvey
Mudd College, 2014. Web. 22 Mar. 2015.
<https://www.math.hmc.edu/calculus/tutorials/riemann_sums/>.
Dawkins, Paul. "Integration Techniques." Pauls Online Notes : Calculus II - Approximating
Definite Integrals. N.p., 2003. Web. 21 Mar. 2015.
<http://tutorial.math.lamar.edu/Classes/CalcII/ApproximatingDefIntegrals.aspx>.