Lesson 22: Areas and Distances
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Transcript of Lesson 22: Areas and Distances
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Hard Areas: Curved Regions
???
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 9 / 30
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
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
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
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
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
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
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
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
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
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
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
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 23 / 30
Distances
Just like area = length× width, we have
distance = rate× time.
So here is another use for Riemann sums.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 24 / 30
Application: Dead Reckoning
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 25 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 26 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 27 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 28 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 29 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 30 / 30