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Keara Miley
AP Calculus
Mrs. Tallman
17 March 2013
Riemann Sums
The formal definition of a Riemann sum states that it is a sum of the
form  f ( x)dx . Each term of the sum represents the area of a rectangle with
altitude f(x) and base dx. A Riemann sum gives an approximate value for a
definite integral. Therefore, Riemann sums are used to find the area under the
curve. The limit of a Riemann sum as dx approaches zero is the basis for the
formal definition of a definite integral (Foerster.). Figure 1 is illustration of
Riemann sums.
Figure 1. Riemann Sums (Keisler)
The trapezoid rule is another technique used for finding definite integrals.
The trapezoid rule is a numerical way of approximating a definite integral by
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slicing the region under the graph into trapezoids and adding the areas of the
trapezoids. Figure 2 is an illustration of the trapezoid rule.
Figure 2. Trapezoid Rule (Dawkins)
As shown in Figure 2, the area under the curve is broken up into trapezoids of
equal width. The area of each trapezoid is calculated and then added together to
find the total area under the curve on the interval given.
Simpson’s rule is very similar to the trapezoid rule but the graph is
replaced by segments of quadratic functions rather than by segments of linear
functions (Foerster.). Simpson’s rule is again a numerical way of approximating
a definite integral by replacing the graph of the integrand with segments of
parabolas, then adding the areas of the regions under each segment. Figure 3 is
an example of Simpson’s rule. Simpson’s is the most accurate of these three
processes in finding the area under the curve.
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Figure 3. Simpson’s Rule (Richardson)
In summary, all three of these methods are used to calculate the area undert the
curve. Riemann sums calculate the area under a curve using rectangles. The
trapezoid rule calculates the area under the curve using trapezoids and
Simpson’s rule calculates the area under the curve using parabolas.
There are different types of Riemann sums: left, right, midpoint, upper,
and lower. Left Riemann sums have rectangles at which the height is
determined by the left most point of each interval, see Figure 4. Right Riemann
sums have rectangles at which the height is determined by the right most point of
each interval, see Figure 5. Midpoint Riemann sums have rectangles at which
the height is determined by the midpoint of each interval, see Figure 6. Upper
Riemann sums have rectangles at which the height is determined by the highest
point of each interval, see Figure 7. Lower Riemann sums have rectangles at
which the height is determined by the lowest point of each interval, see Figure 8.
A Riemann sum is expressed by
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f (c1 ) * x1  f (c2 ) * x2  ...  f (cn ) * xn
Where c is the x-value of the point at which the height is determined and Δx is
the distance between each point.
Figure 4. Left Riemann sum with two intervals
The sum of these rectangles is
f (1) * 2  f (3) * 2
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Figure 5. Right Riemann sum with two intervals
The sum of these rectangles is
f (3) * 2  f (5) * 2
Figure 6. Midpoint Riemann sum with two intervals
The sum of these rectangles is
f (2) * 2  f (4) * 2
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Figure 7. Upper Riemann sum with two intervals
The sum of these rectangles is
f (1) * 2  f (5) * 2
Figure 8. Lower Riemann sum with two intervals
The sum of these rectangles is
f (3) * 2  f (3.68) * 2
The trapezoid rule can be solved using the equation
1
Area  [ f (a)  2 f ( x1 )  2 f ( x 2 )  2 f ( x3 )  ...  f (b)]x
2
Where a is the lower limit of the integral, b is the upper limit of the integral, and
Δx is the width of each trapezoid. An example of the trapezoid rule is shown in
Figure 9 below.
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Figure 9. Trapezoid rule with four intervals
1
Area  [ f (a)  2 f ( x1 )  2 f ( x 2 )  2 f ( x3 )  ...  f (b)]x
2
1
Area  [ f (1)  2 f (2)  2 f (3)  2 f (4)  f (5)] *1
2
All of these approximations are estimating the integral from 1 to 5 of f(x).
As the intervals get closer to zero, the answer of each Riemann sum or trapezoid
rule gets closer to the actual definite integral.
The Mean Value Theorem for Integrals says that if a function is continuous
in the interval [a,b] then there exists c on [a,b] such that f (c) 
1 b
f ( x)dx . In
b  a a
other words, there is at least one point where the function equals the average
value of the function. So, in a Riemann sum, the rectangles will have the same
height and width. This can then be used to find the height of the 2 rectangles
used in a midpoint Riemann sum of f ( x)  ( x  3) 4  2( x  3) 3  4( x  3)  5 . This is
shown below:
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f (c ) 
f (c ) 
1 b
f ( x)dx
b  a a
1 5
( x  3) 4  2( x  3) 3  4( x  3)  5dx
5  1 1
1
f (c)  (32.8)
4
f ( c )  8.2
This means that the height of the 2 rectangles is 8.2 units. This is illustrated on
the graph below.
Figure 11. MVT for Integrals Example
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 twicedifferentiable 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
4
given by V   * r 3 .
3
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Table 1
Values for the rate of change of the radius
t (minutes)
0
1
4
7
r’(t) (ft/min)
5.7
4.0
2.0
1.4
11
0.5
12
0.4
First, estimate the radius of the balloon when t=7.2 using the tangent line
approximation at t=7. The equation for the tangent line is
y  1.4( x  7)  32
The values of 7 and 32 were given in the problem and the slope was found using
the table at t=7. Now, plug in 7.2 into the equation.
y  1.4( x  7)  32
y  1.4(7.2  7)  32
y  32.28
So, the estimated radius at t=7.2 using the tangent line at x=7 is 32.38 feet. This
is more than the true answer because the graph is concave down so the tangent
line would fall above the actual graph.
Now, find the rate of change of the volume of the balloon with respect to
time when t=7. First, take the derivative of the volume equation.
V 
4
 *r3
3
V '  4 * r 2 * r '
For r, plug in 32 since it was given in the problem and plug in 1.4 for r’, which
was found in the table at t=7.
V '  4 * (32) 2 *1.4
V ' 5734.4
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So, the rate of change of the volume of the balloon with respect to time at x=7 is
ft 3
5734.4 
.
min
Using the values given in the table, use a right Riemann sum to
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approximate  r ' (t )dt .
0
12

0
r ' (t )dt  f (1) *1  f (4) * 3  f (7) * 3  f (11) * 4  f (12) *1
 4(1)  2(3)  1.4(3)  0.5(4)  0.4(1)
 4  6  4.2  2  0.4
 16.6
So, the Riemann sum estimated that the integral of r’(t) from 0 to 12 is
29.8 feet. From 0 to 12 minutes, the radius of the balloon increased by 29.8 feet.
This estimated value is less than the real value because the original function is
concave down and a right Riemann sum of r’(t) will underestimate the area.
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Works Cited
Dawkins, Paul. "Calculus II - Approximating Definite Integrals." Paul's Online
Notes. Paul Dawkins, n.d. Web. 17 Mar. 2013.
Foerster, Paul A. Calculus: Concepts and Applications. Berkeley, CA: Key
Curriculum, 1998. Print.
Keisler, H. J. "Elementary Calculus: Riemann Sum." Elementary Calculus:
Riemann Sum. N.p., 26 Sept. 2010. Web. 17 Mar. 2013.
Richardson, James. "Integration." Computing for Engineers @ JCU /. N.p., n.d.
Web. 17 Mar. 2013.