lesson 22: areas and distances
DESCRIPTION
We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.TRANSCRIPT
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Section 5.1Areas and Distances
V63.0121.006/016, Calculus I
New York University
April 13, 2010
Announcements
I Quiz April 16 on §§4.1–4.4
I Final Exam: Monday, May 10, 12:00noon
![Page 2: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/2.jpg)
Announcements
I Quiz April 16 on §§4.1–4.4
I Final Exam: Monday, May10, 12:00noon
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30
![Page 3: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/3.jpg)
Objectives
I Compute the area of aregion by approximating itwith rectangles and lettingthe size of the rectanglestend to zero.
I Compute the total distancetraveled by a particle byapproximating it as distance= (rate)(time) and lettingthe time intervals over whichone approximates tend tozero.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 3 / 30
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Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 4 / 30
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Easy Areas: Rectangle
Definition
The area of a rectangle with dimensions ` and w is the product A = `w .
`
w
It may seem strange that this is a definition and not a theorem but wehave to start somewhere.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 5 / 30
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Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b
b
h
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
![Page 7: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/7.jpg)
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b
b
h
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
![Page 8: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/8.jpg)
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b b
h
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
![Page 9: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/9.jpg)
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b
b
h
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
![Page 10: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/10.jpg)
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b
b
h
So
Fact
The area of a parallelogram of base width b and height h is
A = bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
![Page 11: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/11.jpg)
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
b
h
So
Fact
The area of a triangle of base width b and height h is
A =1
2bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
![Page 12: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/12.jpg)
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
b
h
So
Fact
The area of a triangle of base width b and height h is
A =1
2bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
![Page 13: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/13.jpg)
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
b
h
So
Fact
The area of a triangle of base width b and height h is
A =1
2bh
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
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Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summing theareas of the triangles:
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 8 / 30
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Hard Areas: Curved Regions
???
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 9 / 30
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Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC –212 BC (after Euclid)
I Geometer
I Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
![Page 17: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/17.jpg)
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC –212 BC (after Euclid)
I Geometer
I Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
![Page 18: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/18.jpg)
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC –212 BC (after Euclid)
I Geometer
I Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
![Page 19: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/19.jpg)
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A =
1 + 2 · 1
8+ 4 · 1
64+ · · ·
= 1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
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1
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A = 1
+ 2 · 1
8+ 4 · 1
64+ · · ·
= 1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
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118
18
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A = 1 + 2 · 1
8
+ 4 · 1
64+ · · ·
= 1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
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118
18
164
164
164
164
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A = 1 + 2 · 1
8+ 4 · 1
64+ · · ·
= 1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
![Page 23: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/23.jpg)
118
18
164
164
164
164
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A = 1 + 2 · 1
8+ 4 · 1
64+ · · ·
= 1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
![Page 24: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/24.jpg)
We would then need to know the value of the series
1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
But for any number r and any positive integer n,
(1− r)(1 + r + · · ·+ rn) = 1− rn+1
So
1 + r + · · ·+ rn =1− rn+1
1− r
Therefore
1 +1
4+
1
16+ · · ·+ 1
4n=
1− (1/4)n+1
1− 1/4→ 1
3/4=
4
3
as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
![Page 25: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/25.jpg)
We would then need to know the value of the series
1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
But for any number r and any positive integer n,
(1− r)(1 + r + · · ·+ rn) = 1− rn+1
So
1 + r + · · ·+ rn =1− rn+1
1− r
Therefore
1 +1
4+
1
16+ · · ·+ 1
4n=
1− (1/4)n+1
1− 1/4→ 1
3/4=
4
3
as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
![Page 26: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/26.jpg)
We would then need to know the value of the series
1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
But for any number r and any positive integer n,
(1− r)(1 + r + · · ·+ rn) = 1− rn+1
So
1 + r + · · ·+ rn =1− rn+1
1− r
Therefore
1 +1
4+
1
16+ · · ·+ 1
4n=
1− (1/4)n+1
1− 1/4
→ 13/4
=4
3
as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
![Page 27: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/27.jpg)
We would then need to know the value of the series
1 +1
4+
1
16+ · · ·+ 1
4n+ · · ·
But for any number r and any positive integer n,
(1− r)(1 + r + · · ·+ rn) = 1− rn+1
So
1 + r + · · ·+ rn =1− rn+1
1− r
Therefore
1 +1
4+
1
16+ · · ·+ 1
4n=
1− (1/4)n+1
1− 1/4→ 1
3/4=
4
3
as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
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Cavalieri
I Italian,1598–1647
I Revisited thearea problemwith adifferentperspective
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 13 / 30
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Cavalieri’s method
y = x2
0 1
1
2
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =
1
27+
4
27=
5
27
L4 =
1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 30: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/30.jpg)
Cavalieri’s method
y = x2
0 11
2
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =
1
27+
4
27=
5
27
L4 =
1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 31: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/31.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
3
2
3
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =
1
27+
4
27=
5
27
L4 =
1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 32: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/32.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
3
2
3
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =
1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 33: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/33.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
4
2
4
3
4
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =
1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 34: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/34.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
4
2
4
3
4
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 35: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/35.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
5
2
5
3
5
4
5
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =1
64+
4
64+
9
64=
14
64
L5 =
1
125+
4
125+
9
125+
16
125=
30
125Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 36: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/36.jpg)
Cavalieri’s method
y = x2
0 1
1
2
1
5
2
5
3
5
4
5
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =1
64+
4
64+
9
64=
14
64
L5 =1
125+
4
125+
9
125+
16
125=
30
125
Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 37: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/37.jpg)
Cavalieri’s method
y = x2
0 1
1
2
Divide up the interval into piecesand measure the area of theinscribed rectangles:
L2 =1
8
L3 =1
27+
4
27=
5
27
L4 =1
64+
4
64+
9
64=
14
64
L5 =1
125+
4
125+
9
125+
16
125=
30
125Ln =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
![Page 38: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/38.jpg)
What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width1
n.
The
rectangle over the ith interval and under the parabola has area
1
n·(
i − 1
n
)2
=(i − 1)2
n3.
So
Ln =1
n3+
22
n3+ · · ·+ (n − 1)2
n3=
1 + 22 + 32 + · · ·+ (n − 1)2
n3
The Arabs knew that
1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)
6
So
Ln =n(n − 1)(2n − 1)
6n3→ 1
3as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
![Page 39: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/39.jpg)
What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width1
n. The
rectangle over the ith interval and under the parabola has area
1
n·(
i − 1
n
)2
=(i − 1)2
n3.
So
Ln =1
n3+
22
n3+ · · ·+ (n − 1)2
n3=
1 + 22 + 32 + · · ·+ (n − 1)2
n3
The Arabs knew that
1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)
6
So
Ln =n(n − 1)(2n − 1)
6n3→ 1
3as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
![Page 40: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/40.jpg)
What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width1
n. The
rectangle over the ith interval and under the parabola has area
1
n·(
i − 1
n
)2
=(i − 1)2
n3.
So
Ln =1
n3+
22
n3+ · · ·+ (n − 1)2
n3=
1 + 22 + 32 + · · ·+ (n − 1)2
n3
The Arabs knew that
1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)
6
So
Ln =n(n − 1)(2n − 1)
6n3→ 1
3as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
![Page 41: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/41.jpg)
What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width1
n. The
rectangle over the ith interval and under the parabola has area
1
n·(
i − 1
n
)2
=(i − 1)2
n3.
So
Ln =1
n3+
22
n3+ · · ·+ (n − 1)2
n3=
1 + 22 + 32 + · · ·+ (n − 1)2
n3
The Arabs knew that
1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)
6
So
Ln =n(n − 1)(2n − 1)
6n3
→ 1
3as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
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What is Ln?
Divide the interval [0, 1] into n pieces. Then each has width1
n. The
rectangle over the ith interval and under the parabola has area
1
n·(
i − 1
n
)2
=(i − 1)2
n3.
So
Ln =1
n3+
22
n3+ · · ·+ (n − 1)2
n3=
1 + 22 + 32 + · · ·+ (n − 1)2
n3
The Arabs knew that
1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)
6
So
Ln =n(n − 1)(2n − 1)
6n3→ 1
3as n→∞.V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
![Page 43: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/43.jpg)
Cavalieri’s method for different functions
Try the same trick with f (x) = x3. We have
Ln =1
n· f(
1
n
)+
1
n· f(
2
n
)+ · · ·+ 1
n· f(
n − 1
n
)
=1
n· 1
n3+
1
n· 23
n3+ · · ·+ 1
n· (n − 1)3
n3
=1 + 23 + 33 + · · ·+ (n − 1)3
n4
The formula out of the hat is
1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)
]2So
Ln =n2(n − 1)2
4n4→ 1
4as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
![Page 44: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/44.jpg)
Cavalieri’s method for different functions
Try the same trick with f (x) = x3. We have
Ln =1
n· f(
1
n
)+
1
n· f(
2
n
)+ · · ·+ 1
n· f(
n − 1
n
)=
1
n· 1
n3+
1
n· 23
n3+ · · ·+ 1
n· (n − 1)3
n3
=1 + 23 + 33 + · · ·+ (n − 1)3
n4
The formula out of the hat is
1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)
]2So
Ln =n2(n − 1)2
4n4→ 1
4as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
![Page 45: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/45.jpg)
Cavalieri’s method for different functions
Try the same trick with f (x) = x3. We have
Ln =1
n· f(
1
n
)+
1
n· f(
2
n
)+ · · ·+ 1
n· f(
n − 1
n
)=
1
n· 1
n3+
1
n· 23
n3+ · · ·+ 1
n· (n − 1)3
n3
=1 + 23 + 33 + · · ·+ (n − 1)3
n4
The formula out of the hat is
1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)
]2So
Ln =n2(n − 1)2
4n4→ 1
4as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
![Page 46: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/46.jpg)
Cavalieri’s method for different functions
Try the same trick with f (x) = x3. We have
Ln =1
n· f(
1
n
)+
1
n· f(
2
n
)+ · · ·+ 1
n· f(
n − 1
n
)=
1
n· 1
n3+
1
n· 23
n3+ · · ·+ 1
n· (n − 1)3
n3
=1 + 23 + 33 + · · ·+ (n − 1)3
n4
The formula out of the hat is
1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)
]2
So
Ln =n2(n − 1)2
4n4→ 1
4as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
![Page 47: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/47.jpg)
Cavalieri’s method for different functions
Try the same trick with f (x) = x3. We have
Ln =1
n· f(
1
n
)+
1
n· f(
2
n
)+ · · ·+ 1
n· f(
n − 1
n
)=
1
n· 1
n3+
1
n· 23
n3+ · · ·+ 1
n· (n − 1)3
n3
=1 + 23 + 33 + · · ·+ (n − 1)3
n4
The formula out of the hat is
1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)
]2So
Ln =n2(n − 1)2
4n4→ 1
4as n→∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
![Page 48: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/48.jpg)
Cavalieri’s method with different heights
Rn =1
n· 13
n3+
1
n· 23
n3+ · · ·+ 1
n· n3
n3
=13 + 23 + 33 + · · ·+ n3
n4
=1
n4
[12n(n + 1)
]2=
n2(n + 1)2
4n4→ 1
4
as n→∞.
So even though the rectangles overlap, we still get the same answer.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
![Page 49: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/49.jpg)
Cavalieri’s method with different heights
Rn =1
n· 13
n3+
1
n· 23
n3+ · · ·+ 1
n· n3
n3
=13 + 23 + 33 + · · ·+ n3
n4
=1
n4
[12n(n + 1)
]2=
n2(n + 1)2
4n4→ 1
4
as n→∞.So even though the rectangles overlap, we still get the same answer.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
![Page 50: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/50.jpg)
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 18 / 30
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Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a
n.
For each i between 1 and n, let xi be the nth step between a and b. So
a bx0 x1 x2 . . . xixn−1xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n· · · · · ·
xi = a + i · b − a
n· · · · · ·
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
![Page 52: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/52.jpg)
Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a
n.
For each i between 1 and n, let xi be the nth step between a and b. So
a bx0 x1 x2 . . . xixn−1xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n· · · · · ·
xi = a + i · b − a
n· · · · · ·
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
![Page 53: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/53.jpg)
Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a
n.
For each i between 1 and n, let xi be the nth step between a and b. So
a bx0 x1 x2 . . . xixn−1xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n· · · · · ·
xi = a + i · b − a
n· · · · · ·
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
![Page 54: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/54.jpg)
Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a
n.
For each i between 1 and n, let xi be the nth step between a and b. So
a bx0 x1 x2 . . . xixn−1xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n· · · · · ·
xi = a + i · b − a
n· · · · · ·
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
![Page 55: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/55.jpg)
Cavalieri’s method in general
Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).
For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a
n.
For each i between 1 and n, let xi be the nth step between a and b. So
a bx0 x1 x2 . . . xixn−1xn
x0 = a
x1 = x0 + ∆x = a +b − a
n
x2 = x1 + ∆x = a + 2 · b − a
n· · · · · ·
xi = a + i · b − a
n· · · · · ·
xn = a + n · b − a
n= b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
![Page 56: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/56.jpg)
Forming Riemann sums
We have many choices of how to approximate the area:
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
In general, choose ci to be a point in the ith interval [xi−1, xi ]. Form theRiemann sum
Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x
=n∑
i=1
f (ci )∆x
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
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Forming Riemann sums
We have many choices of how to approximate the area:
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
In general, choose ci to be a point in the ith interval [xi−1, xi ]. Form theRiemann sum
Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x
=n∑
i=1
f (ci )∆x
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
![Page 60: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/60.jpg)
Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2 x3a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2 x3 x4a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2 x3 x4 x5a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
![Page 63: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/63.jpg)
Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2 x3 x4 x5 x6a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1 x2 x3 x4 x5 x6 x7a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
![Page 67: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/67.jpg)
Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16a b
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17a b
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Theorem of the Day
Theorem
If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then
limn→∞
Sn = limn→∞
n∑i=1
f (ci )∆x
exists and is the same value nomatter what choice of ci wemade.
x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20a b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
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![Page 79: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/79.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 80: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/80.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 81: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/81.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 82: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/82.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 83: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/83.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 84: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/84.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 85: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/85.jpg)
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope of lines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
![Page 86: Lesson 22: Areas and Distances](https://reader030.vdocuments.us/reader030/viewer/2022020110/5482f00db47959140d8b48d1/html5/thumbnails/86.jpg)
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
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Distances
Just like area = length× width, we have
distance = rate× time.
So here is another use for Riemann sums.
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Application: Dead Reckoning
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Example
A sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s position and velocity are recorded, but shortlythereafter a storm blows in and position is impossible to measure. Thevelocity continues to be recorded at thirty-minute intervals.
Time 12:00 12:30 1:00 1:30 2:00
Speed (knots) 4 8 12 6 4
Direction E E E E W
Time 2:30 3:00 3:30 4:00
Speed 3 3 5 9
Direction W E E E
Estimate the ship’s position at 4:00pm.
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Solution
We estimate that the speed of 4 knots (nautical miles per hour) ismaintained from 12:00 until 12:30. So over this time interval the shiptravels (
4 nmi
hr
)(1
2hr
)= 2nmi
We can continue for each additional half hour and get
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
So the ship is 15.5 nmi east of its original position.
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Analysis
I This method of measuring position by recording velocity was necessaryuntil global-positioning satellite technology became widespread
I If we had velocity estimates at finer intervals, we’d get betterestimates.
I If we had velocity at every instant, a limit would tell us our exactposition relative to the last time we measured it.
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Other uses of Riemann sums
Anything with a product!
I Area, volume
I Anything with a density: Population, mass
I Anything with a “speed:” distance, throughput, power
I Consumer surplus
I Expected value of a random variable
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Summary
I We can compute the area of a curved region with a limit of Riemannsums
I We can compute the distance traveled from the velocity with a limitof Riemann sums
I Many other important uses of this process.
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