Koshko 1 Jack Koshko Mr. Acre AP Calculus 27 March, 2017 Riemann Sums German mathematician Bernhard Riemann was a poor, shy, and religious man. When his family had enough money to send him to college, he went to the University of Göttingen in 1846, where he first met, and attended the lectures of, Carl Friedrich Gauss (Mastin). He gradually worked his way up the University's hierarchy to become a professor and, eventually, head of the mathematics department at Göttingen (Mastin). From there he would go on to discover many new techniques in the math world, but he became famous for Riemann Sums.
Part 1: Formal Definitions A Riemann Sum is “a sum of the form
∑ f ( x ) dx
in which each term of the sum
represents the area of a rectangle of altitude f(x) and base dx” (Foerster). It is used to find “an approximate value for a definite integral.” This means that when given a function (f(x)) and an interval (n), in order to find the approximation, simply draw “n” evenly spaced rectangles under the curve and find a point on the graph inside each interval. Then plug in the “n” points into the equation for f(x), solve each one, add them together, and multiply by the width of each interval (dx). This will give the Riemann sum approximation.
Koshko 2 The Trapezoid Rule states that “the definite integral of f(x) from
x=a ¿ x=b
1 1 approximately equal to T n=∆ x 2 f ( a )+ f ( x 1 ) +f ( x 2 ) + f ( x 3 ) + …+ f ( x n−1 ) + 2 f ( b )
(
the number of increments (trapezoids), ∆ x=
values of
x 1 , x 2 , x 3 , … are spaced ∆ x
( b−a) n
)
is
where n is
is the width of each increment, and the
units apart” (Foerster). Basically, what this means is
that when the area under a curve is broken up into trapezoidal pieces, all the values of f(x) are added together (first and last values are only half) and then multiplied by the width of the increments to find the estimated total area for that curve. Simpson’s Rule states that “if the interval [a, b] is divided into an even number, n, of subintervals of equal width ∆ x , then the integral of f(x) from
equal to
x=a ¿ x=b
is approximately
1 ( ∆ x)( y 0 + 4 y 1+ 2 y 2 +4 y 3+ 2 y 4+ …+2 y n−2+ 4 y n−1 + y n ) ” (Foerster). This means 3
that to use Simpson’s rule, multiply each y-value by the appropriate number, add them together, then multiply the answer by one third of the spacing between the x-values. Note that there must be an even number of increments, so there must be an odd number of data points in order to use Simpson’s rule. These three processes have both similarities and differences. All three estimate the area under a given curve. Also, all three involve adding up y-values and then multiplying by the
Koshko 3 change in x to find the approximation. However, they all use different techniques. Riemann sums find the area approximation by using rectangles and adding up the area of each rectangle of altitude f(x) and base dx. The trapezoid rule finds the area by using trapezoids, where all the values of f(x) are added together (first and last values are only half) and then multiplied by the width of the increments. Simpson’s rule finds the area using quadratic line segments (parabolas), which involves multiplying each y-value a certain number in the pattern, adding them together, and multiplying that by one third of the change in x. Simpson’s rule is the most accurate because it uses line segments of quadratic functions (parabolas), resulting in a better fit for the curve.
Part 2: Five Different Riemann Sums Graphs 4 3 Given the equation f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 and using two intervals, there are five different Riemann sums that can be used to
approximate the area under the curve. These five ways are left, right, lower, upper, and midpoint. The graphs are drawn out and broken into two intervals shown in the figures below.
Koshko 4 Figure 1. Left Riemann Sum Graph Figure 1 shows the left Riemann sum graph using two intervals. Simply look at the left most point on the graph in each of the two intervals and draw the rectangle from there so that is stretches from point a to point b and covers the whole interval.
Figure 2. Right Riemann Sum Graph Figure 2 shows the right Riemann sum graph using two intervals. Simply look at the right most point on the graph in each of the two intervals and draw the rectangle from there so that is stretches from point a to point b and covers the whole interval.
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Figure 3. Lower Riemann Sum Graph Figure 3 shows the lower Riemann sum graph using two intervals. Simply look at the lowest point on the graph in each of the two intervals and draw the rectangle from there so that is stretches from point a to point b and covers the whole interval.
Figure 4. Upper Riemann Sum Graph Figure 4 shows the upper Riemann sum graph using two intervals. Simply look at the highest point on the graph in each of the two intervals and draw the rectangle from there so that is stretches from point a to point b and covers the whole interval.
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Figure 5. Midpoint Riemann Sum Graph Figure 5 shows the midpoint Riemann sum graph using two intervals. Simply look at the middle point on the graph in each of the two intervals and draw the rectangle from there so that is stretches from point a to point b and covers the whole interval. Part 3: Five Different Riemann Sum Calculations 4 3 Given the equation f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 and using two intervals, there are five different Riemann sums that can be used to
approximate the area under the curve. These five ways are left, right, lower, upper, and midpoint. Once the graph is drawn out and broken into its two intervals, finding the different Riemann Sums are easy to calculate. See figures below. L2=2 [ f ( 1 ) +f (3) ] L2=2 [ 13+5 ]
Koshko 7 2
L2=36 units
Figure 6. Left Riemann Sum for two intervals Figure 6 shows how to find a left Riemann sum. Simply look at the left most point in each interval that touches the graph. These two points are separately plugged into the equation for f(x), solved, added together, and finally multiplied by the width of each increment, which in this case is two. The answer for the left Riemann sum approximation is 36 units2. R2=2 [ f ( 3 ) + f (5) ] R2=2 [ 5+29 ]
R2=68 units
2
Figure 7. Right Riemann Sum for two intervals Figure 7 shows how to find a right Riemann sum. Simply look at the right most point in each interval that touches the graph. These two points are separately plugged into the equation for f(x), solved, added together, and finally multiplied by the width of each increment, which in this case is two. The answer for the right Riemann sum approximation is 68 units2. L2=2 [ f ( 3 ) + f (3.68) ] L2=2 [ 5+ 3.12 ]
L2=16.24 units2 Figure 8. Lower Riemann Sum for two intervals
Koshko 8 Figure 8 shows how to find a lower Riemann sum. Simply look at the lowest point in each interval that touches the graph. These two points are separately plugged into the equation for f(x), solved, added together, and finally multiplied by the width of each increment, which in this case is two. The answer for the lower Riemann sum approximation is 16.24 units2. U 2=2 [ f ( 1 ) + f (5) ] U 2=2 [ 13+29 ] 2
U 2=84 units
Figure 9. Upper Riemann Sum for two intervals Figure 9 shows how to find an upper Riemann sum. Simply look at the highest point in each interval that touches the graph. These two points are separately plugged into the equation for f(x), solved, added together, and finally multiplied by the width of each increment, which in this case is two. The answer for the upper Riemann sum approximation is 84 units2.
M 2=2 [ f ( 2 ) +f (4) ] M 2=2 [ 8+ 4 ] 2
M 2=24 units
Koshko 9 Figure 10. Midpoint Riemann Sum for two intervals Figure 10 shows how to find a midpoint Riemann sum. Simply look at the middle point in each interval that touches the graph. These two points are separately plugged into the equation for f(x), solved, added together, and finally multiplied by the width of each increment, which in this case is two. The answer for the midpoint Riemann sum approximation is 24 units2.
Part 4: Trapezoid and Simpson’s Rule 4 3 Given the equation f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 and using four intervals, the trapezoid rule and Simpson’s rule can be used to
approximate the area under the curve. The calculations and graphs for each are shown in the figures below. T n=∆ x
( 12 f ( a )+ f ( x )+f ( x ) + f ( x )+ …+ f ( x 1
2
3
T 4=1
( 12 f (1 )+ f ( 2)+ f ( 3) + f ( 4) + 12 f (5 ))
T 4=1
( 12 ( 13) +8+ 5+4 + 12 ( 29))
T 4=38 units2 Figure 11. Trapezoid Rule Calculation
1
n−1
)+ 2 f ( b)
)
Koshko 10 Figure 11 shows how to use the trapezoid rule to calculate the area of the equation f ( x )=( x−3 )4 +2 ( x −3 )3−4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 using four intervals.
2 It gives an area of 38 units .
Figure 12. Trapezoid Rule Graph Figure 12 shows the area under the curve using the trapezoid rule for the equation 4
3
f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 using four intervals.
1 I = ( ∆ x)( y 0 + 4 y 1+ 2 y 2 +4 y 3+ 2 y 4+ …+2 y n−2+ 4 y n−1 + y n ) 3 1 I = (1)( f (1)+ 4 f (2)+2 f (3)+ 4 f (4 )+ f (5)) 3 1 I = (1)(13+ 4(8)+2(5)+ 4( 4)+(29)) 3 I =33.33 units2 Figure 13. Simpson’s Rule Calculation
Koshko 11 Figure 13 shows how to use Simpson’s rule to calculate the area of the equation f ( x )=( x−3 )4 +2 ( x −3 )3−4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 using four intervals.
2 It gives an area of 33.33units .
Figure 14. Simpson’s Rule Graph Figure 14 shows the area under the curve using Simpson’s rule for the equation f ( x )=( x−3 )4 +2 ( x −3 )3−4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 using four intervals.
The approximation for finding the area under the curve using the trapezoid rule compared to Simpson’s rule came out to be fairly similar to each other. The trapezoid rule gave a value of 38 units2, while Simpson’s rule gave a value of 33.33 units2. When the definite integral was calculated for f(x) from one to five, it yielded a value of 32.8 units2 shown below in Figure 15. Both the trapezoid rule and Simpson’s rule were over approximations compared to the definite
Koshko 12 integral, but Simpson’s rule was much closer. The definite integral is the most accurate and Simpson’s rule is a close second when finding the area under a curve. b
¿∫ f ( x ) dx a 5
¿∫ ( x−3 )4 +2 ( x−3 )3 −4 ( x−3 )+ 5 dx 1
¿ 32.8units 2 Figure 15. Definite integral Figure 15 shows how to use a definite integral to calculate the area of the equation f ( x )=( x−3 )4 +2 ( x −3 )3−4 ( x−3 ) +5 on the interval from
x=1 ¿ x =5 . It gives an area of
3 2.28units 2 .
Part 5: Mean Value Theorem for Integrals The Mean Value Theorem for Integrals states that “if y=f(x) is continuous on the closed
b
∫ f (x) dx
interval [a, b], then there is at least one point x=c in [a, b] for which f ( c )= y = a av
b
equivalently, f ( c ) ( b−a )=∫ f ( x ) dx ” (Foerster). a
b−a
or
Koshko 13 4
3
Given the equation f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 on the interval from x=1 ¿ x =5 and using two different intervals, the Mean Value Theorem for Integrals can be
used to approximate the area under the curve. The two intervals would be from and
x=1 ¿ x =3
x=3 ¿ x=5 . To do this, use the equation for f(c) above, plug in f(x), and use the interval
3
from
x=1 ¿ x =3 , which looks like f ( c )=
∫ ( x−3 ) 4 +2 ( x−3 )3−4 ( x−3 ) +5 dx 1
. This will
3−1
2 yield an area of 8.2 units . Then repeat this process again, except use the interval
5
x=3 ¿ x=5 , which looks like f ( c )=
2
yield an area of 8.2 units
∫ ( x−3 ) 4 +2 ( x−3 )3−4 ( x−3 ) +5 dx 3
. This will also
5−3
. Since both intervals have a width equal to 2 (3-1 and 5-3) and a
height equal to 8.2, both rectangles have the same area of 16.4 units2. This means that to find the overall area, both the rectangles are added together to get a total area of 32.8 units2.
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Figure 16. Mean Value Theorem for Integrals Graph Figure 16 shows the use of the Mean Value Theorem for Integrals for the equation 4
3
f ( x )=( x−3 ) +2 ( x −3 ) −4 ( x−3 ) +5 . There are two intervals, represented by the two
rectangles, that go from
x=1 ¿ x =3 and
x=3 ¿ x=5 . Since the areas of these two intervals
are the same, the rectangles are also the same.
Part 6: Solving Problems Problem 1: Table 1 t(seconds) r’(t) (feet/second)
0 5.7
1 4.0
4 2.0
7 1.4
11 0.5
12 0.4
' Table 1 shows the selected values of the rate of change, r ( t ) , of the radius of the
balloon over the time interval 0 ≤t ≤12 .
Koshko 15 1a: r ( 7.2 )=r ( 7 ) +r ' (7 ) ∆ t r ( 7.2 )=32+1.4(0.2)
r ( 7.2 )=32.28 feet
Figure 17. Radius Estimation Figure 17 shows the estimate of the radius of the balloon when t=7 .2 using a tangent line approximation at t ¿ 7 . This estimate of 32.28 feet is greater than the true value because ' r ( t ) is decreasing over the interval 7
1b: 4 3 v= π r 3 dv 4 2 dr = πr whent=7 dt 3 dt dv =4 π (32)2 × 1.4 dt dv ft 3 =5734.3 π dt sec Figure 18. Rate of Change of the Balloons Volume
Koshko 16 Figure 18 shows the rate of change of the volume of the balloon with respect to time when t=7 . 1c: 12
∫ r ' ( t ) dt=( 1−0 )( 4 ) +( 4−1 )( 2 ) +(7−4)(1.4)+( 11−7 ) ( 0.5 ) +(12−11)(0.4) 1
¿ 16.6 feet Figure 19. right Riemann Sum Figure 19 shows a right Riemann sum with five subintervals indicated by the data in the
12
12
' table to approximate ∫ r ( t ) dt . In terms of the radius of the balloon, 0
∫ r ' ( t ) dt 0
represents
the change in radius from t=0 to t=12 seconds. 1d: 12
The approximation of 16.6 feet found in part c is less than
∫ r ' ( t ) dt 0
because r is concave
down and r’ is decreasing on the interval from 0
In conclusion, a Riemann sum is used to find the area under a curve. They are very similar to the trapezoid rule and Simpson’s rule, but use rectangles to estimate the area under the
Koshko 17 curve rather than trapezoids or quadratic line segments (parabolas). Simpson’s Rule is the most accurate because it uses quadratic line segments (parabolas), which better fit the curve.
Koshko 18 Works Cited Foerster, Paul A. Calculus: Concepts and Applications. 2nd ed. Dubuque, Iowa: Kendall hunt, 2010. Print. Mastin, Luke. "19th Century Mathematics - Riemann." The Story of Mathematics. N.p., 2010. Web. 25 Mar. 2017. .
