Lesson 24: Areas, Distances, the Integral (Section 041 slides)
Lesson 23: Areas and Distances (Section 10 version)
-
Upload
matthew-leingang -
Category
Education
-
view
786 -
download
0
description
Transcript of Lesson 23: Areas and Distances (Section 10 version)
![Page 1: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/1.jpg)
. . . . . .
Section5.1AreasandDistances
V63.0121, CalculusI
April13, 2009
Announcements
I Movingto624today(noOH)I Quiz5thisweekon§§4.1–4.4I BackHW returnedinrecitationthisweek
![Page 2: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/2.jpg)
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
![Page 3: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/3.jpg)
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
![Page 4: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/4.jpg)
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
![Page 5: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/5.jpg)
. . . . . .
Meetthemathematician: Archimedes
I 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
![Page 6: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/6.jpg)
. . . . . .
.
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A =
1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
![Page 7: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/7.jpg)
. . . . . .
.
.1
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1
+ 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
![Page 8: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/8.jpg)
. . . . . .
.
.1.18 .18
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
![Page 9: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/9.jpg)
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
![Page 10: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/10.jpg)
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
![Page 11: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/11.jpg)
. . . . . .
Wewouldthenneedtoknowthevalueoftheseries
1 +14
+116
+ · · · + 14n
+ · · ·
Butforanynumber r andanypositiveinteger n,
(1− r)(1 + r + · · · + rn) = 1− rn+1
So
1 + r + · · · + rn =1− rn+1
1− r
Therefore
1 +14
+116
+ · · · + 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=
43
as n → ∞.
![Page 12: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/12.jpg)
. . . . . .
Wewouldthenneedtoknowthevalueoftheseries
1 +14
+116
+ · · · + 14n
+ · · ·
Butforanynumber r andanypositiveinteger n,
(1− r)(1 + r + · · · + rn) = 1− rn+1
So
1 + r + · · · + rn =1− rn+1
1− r
Therefore
1 +14
+116
+ · · · + 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=
43
as n → ∞.
![Page 13: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/13.jpg)
. . . . . .
Wewouldthenneedtoknowthevalueoftheseries
1 +14
+116
+ · · · + 14n
+ · · ·
Butforanynumber r andanypositiveinteger n,
(1− r)(1 + r + · · · + rn) = 1− rn+1
So
1 + r + · · · + rn =1− rn+1
1− r
Therefore
1 +14
+116
+ · · · + 14n
=1− (1/4)n+1
1− 1/4
→ 13/4
=43
as n → ∞.
![Page 14: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/14.jpg)
. . . . . .
Wewouldthenneedtoknowthevalueoftheseries
1 +14
+116
+ · · · + 14n
+ · · ·
Butforanynumber r andanypositiveinteger n,
(1− r)(1 + r + · · · + rn) = 1− rn+1
So
1 + r + · · · + rn =1− rn+1
1− r
Therefore
1 +14
+116
+ · · · + 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=
43
as n → ∞.
![Page 15: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/15.jpg)
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
![Page 16: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/16.jpg)
. . . . . .
Cavalieri
I Italian,1598–1647
I Revisitedtheareaproblemwithadifferentperspective
![Page 17: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/17.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 18: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/18.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 19: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/19.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 20: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/20.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 21: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/21.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 22: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/22.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 23: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/23.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
![Page 24: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/24.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
1625
=30125
Ln =?
![Page 25: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/25.jpg)
. . . . . .
Cavalieri’smethod
.
.y = x2
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
1625
=30125
Ln =?
![Page 26: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/26.jpg)
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
![Page 27: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/27.jpg)
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
![Page 28: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/28.jpg)
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
![Page 29: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/29.jpg)
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3
→ 13
as n → ∞.
![Page 30: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/30.jpg)
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
![Page 31: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/31.jpg)
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
Ln =1n· f
(1n
)+
1n· f
(2n
)+ · · · + 1
n· f
(n− 1n
)
=1n· 1n3
+1n· 2
3
n3+ · · · + 1
n· (n− 1)3
n3
=1 + 23 + 33 + · · · + (n− 1)3
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
![Page 32: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/32.jpg)
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
Ln =1n· f
(1n
)+
1n· f
(2n
)+ · · · + 1
n· f
(n− 1n
)=
1n· 1n3
+1n· 2
3
n3+ · · · + 1
n· (n− 1)3
n3
=1 + 23 + 33 + · · · + (n− 1)3
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
![Page 33: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/33.jpg)
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
Ln =1n· f
(1n
)+
1n· f
(2n
)+ · · · + 1
n· f
(n− 1n
)=
1n· 1n3
+1n· 2
3
n3+ · · · + 1
n· (n− 1)3
n3
=1 + 23 + 33 + · · · + (n− 1)3
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
![Page 34: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/34.jpg)
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
Ln =1n· f
(1n
)+
1n· f
(2n
)+ · · · + 1
n· f
(n− 1n
)=
1n· 1n3
+1n· 2
3
n3+ · · · + 1
n· (n− 1)3
n3
=1 + 23 + 33 + · · · + (n− 1)3
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2
So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
![Page 35: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/35.jpg)
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
Ln =1n· f
(1n
)+
1n· f
(2n
)+ · · · + 1
n· f
(n− 1n
)=
1n· 1n3
+1n· 2
3
n3+ · · · + 1
n· (n− 1)3
n3
=1 + 23 + 33 + · · · + (n− 1)3
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
![Page 36: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/36.jpg)
. . . . . .
Cavalieri’smethodwithdifferentheights
.
Rn =1n· 1
3
n3+
1n· 2
3
n3+ · · · + 1
n· n
3
n3
=13 + 23 + 33 + · · · + n3
n4
=1n4
[12n(n + 1)
]2=
n2(n + 1)2
4n4→ 1
4
as n → ∞.
Soeventhoughtherectanglesoverlap, westillgetthesameanswer.
![Page 37: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/37.jpg)
. . . . . .
Cavalieri’smethodwithdifferentheights
.
Rn =1n· 1
3
n3+
1n· 2
3
n3+ · · · + 1
n· n
3
n3
=13 + 23 + 33 + · · · + n3
n4
=1n4
[12n(n + 1)
]2=
n2(n + 1)2
4n4→ 1
4
as n → ∞.Soeventhoughtherectanglesoverlap, westillgetthesameanswer.
![Page 38: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/38.jpg)
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
![Page 39: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/39.jpg)
. . . . . .
Cavalieri’smethodingeneralLet f beapositivefunctiondefinedontheinterval [a,b]. Wewanttofindtheareabetween x = a, x = b, y = 0, and y = f(x).Foreachpositiveinteger n, divideuptheintervalinto n pieces.
Then ∆x =b− an
. Foreach i between 1 and n, let xi bethe nth
stepbetween a and b. So
..a .b. . . . . . ..x0 .x1 .x2 .xi.xn−1.xn
x0 = a
x1 = x0 + ∆x = a +b− an
x2 = x1 + ∆x = a + 2 · b− an
· · · · · ·
xi = a + i · b− an
· · · · · ·
xn = a + n · b− an
= b
![Page 40: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/40.jpg)
. . . . . .
FormingRiemannsums
Wehavemanychoicesofhowtoapproximatethearea:
Ln = f(x0)∆x + f(x1)∆x + · · · + f(xn−1)∆x
Rn = f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x
Mn = f(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum
Sn = f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x
=n∑
i=1
f(ci)∆x
![Page 41: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/41.jpg)
. . . . . .
FormingRiemannsums
Wehavemanychoicesofhowtoapproximatethearea:
Ln = f(x0)∆x + f(x1)∆x + · · · + f(xn−1)∆x
Rn = f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x
Mn = f(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum
Sn = f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x
=n∑
i=1
f(ci)∆x
![Page 42: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/42.jpg)
. . . . . .
TheoremoftheDay
TheoremIf f isacontinuousfunctionon [a,b] orhasfinitelymanyjumpdiscontinuities, then
limn→∞
Sn = limn→∞
{f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x}
existsandisthesamevaluenomatterwhatchoiceof ci wemade.
![Page 43: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/43.jpg)
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
![Page 44: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/44.jpg)
. . . . . .
Distances
Justlike area = length×width, wehave
distance = rate× time.
SohereisanotheruseforRiemannsums.
![Page 45: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/45.jpg)
. . . . . .
ExampleA sailingshipiscruisingbackandforthalongachannel(inastraightline). Atnoontheship’spositionandvelocityarerecorded, butshortlythereafterastormblowsinandpositionisimpossibletomeasure. Thevelocitycontinuestoberecordedatthirty-minuteintervals.
Time 12:00 12:30 1:00 1:30 2:00Speed(knots) 4 8 12 6 4Direction E E E E W
Time 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E
Estimatetheship’spositionat4:00pm.
![Page 46: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/46.jpg)
. . . . . .
SolutionWeestimatethatthespeedof4knots(nauticalmilesperhour)ismaintainedfrom12:00until12:30. Sooverthistimeintervaltheshiptravels (
4 nmihr
) (12hr
)= 2 nmi
Wecancontinueforeachadditionalhalfhourandget
distance = 4× 1/2 + 8× 1/2 + 12× 1/2
+ 6× 1/2− 4× 1/2− 3× 1/2 + 3× 1/2 + 5× 1/2
= 15.5
Sotheshipis 15.5 nmi eastofitsoriginalposition.
![Page 47: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/47.jpg)
. . . . . .
Analysis
I Thismethodofmeasuringpositionbyrecordingvelocityisknownas deadreckoning.
I Ifwehadvelocityestimatesatfinerintervals, we’dgetbetterestimates.
I Ifwehadvelocityateveryinstant, alimitwouldtellusourexactpositionrelativetothelasttimewemeasuredit.
![Page 48: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/48.jpg)
. . . . . .
Outline
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Distances
Otherapplications
![Page 49: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/49.jpg)
. . . . . .
OtherusesofRiemannsums
Anythingwithaproduct!I Area, volumeI Anythingwithadensity: Population, massI Anythingwitha“speed:” distance, throughput, powerI ConsumersurplusI Expectedvalueofarandomvariable
![Page 50: Lesson 23: Areas and Distances (Section 10 version)](https://reader033.fdocuments.net/reader033/viewer/2022042816/5597dd331a28abbf1e8b46b7/html5/thumbnails/50.jpg)
. . . . . .
Summary
I Riemannsumscanbeusedtoestimateareas, distance, andotherquantities
I ThelimitofRiemannsumscangettheexactvalueI Comingup: givingthislimitanameandworkingwithit.