lesson 24: area and distances
TRANSCRIPT
Section 5.1Areas and Distances
V63.0121.002.2010Su, Calculus I
New York University
June 16, 2010
Announcements
I Quiz Thursday on 4.1–4.4
. . . . . .
. . . . . .
Announcements
I Quiz Thursday on 4.1–4.4
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 2 / 31
. . . . . .
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 asdistance = (rate)(time) andletting the time intervalsover which oneapproximates tend to zero.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 3 / 31
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 4 / 31
. . . . . .
Easy Areas: Rectangle
DefinitionThe area of a rectangle with dimensions ℓ and w is the product A = ℓw.
..ℓ
.w
It may seem strange that this is a definition and not a theorem but wehave to start somewhere.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 5 / 31
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
..b
.b
.h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
..b
.b
.h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b .b
.h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b
.b
.h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b
.b
.h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
. . . . . .
Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summingthe areas of the triangles:
.
.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 8 / 31
. . . . . .
Hard Areas: Curved Regions
.
.
???
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 9 / 31
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
. . . . . .
Archimedes and the Parabola
.
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A =
1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
. . . . . .
Archimedes and the Parabola
.
.1
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1
+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
. . . . . .
Archimedes and the Parabola
.
.1.18 .18
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18
+ 4 · 164
+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
. . . . . .
Archimedes and the Parabola
.
.1.18 .18
.164 .164
.164 .164
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
. . . . . .
Archimedes and the Parabola
.
.1.18 .18
.164 .164
.164 .164
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
. . . . . .
Summing the series
[label=archimedes-parabola-sum] We would then need to know thevalue of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
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− rTherefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43
as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
. . . . . .
Summing the series
[label=archimedes-parabola-sum] We would then need to know thevalue of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
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+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43
as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
. . . . . .
Summing the series
[label=archimedes-parabola-sum] We would then need to know thevalue of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
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− rTherefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4
→ 13/4
=43
as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
. . . . . .
Summing the series
[label=archimedes-parabola-sum] We would then need to know thevalue of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
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− rTherefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43
as n → ∞.V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
. . . . . .
Cavalieri
I Italian,1598–1647
I Revisited theareaproblem witha differentperspective
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 13 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.13
.
.23
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.13
.
.23
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.14
.
.24
.
.34
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.14
.
.24
.
.34
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.15
.
.25
.
.35
.
.45
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.15
.
.25
.
.35
.
.45
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
Cavalieri's method
.
.y = x2
..0
..1
.
.12
.
.
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n.
Therectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+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)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+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)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+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)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+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)
6So
Ln =n(n− 1)(2n− 1)
6n3
→ 13
as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+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)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
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
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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
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
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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
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
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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
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
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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
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
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 → ∞.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
. . . . . .
Cavalieri's method with different heights
.
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
4as n → ∞.
So even though the rectangles overlap, we still get the same answer.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 17 / 31
. . . . . .
Cavalieri's method with different heights
.
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
4as n → ∞.
So even though the rectangles overlap, we still get the same answer.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 17 / 31
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 18 / 31
. . . . . .
Cavalieri's method in general.
.
Let f be a positive function defined on the interval [a,b]. We want to find thearea between 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− an
. For each i between 1 and n, let xi be the nth step between a andb. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
. . . . . .
Cavalieri's method in general.
.
Let f be a positive function defined on the interval [a,b]. We want to find thearea between 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− an
. For each i between 1 and n, let xi be the nth step between a andb. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
. . . . . .
Cavalieri's method in general.
.
Let f be a positive function defined on the interval [a,b]. We want to find thearea between 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− an
. For each i between 1 and n, let xi be the nth step between a andb. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
. . . . . .
Cavalieri's method in general.
.
Let f be a positive function defined on the interval [a,b]. We want to find thearea between 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− an
. For each i between 1 and n, let xi be the nth step between a andb. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
. . . . . .
Cavalieri's method in general.
.
Let f be a positive function defined on the interval [a,b]. We want to find thearea between 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− an
. For each i between 1 and n, let xi be the nth step between a andb. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
. . . . . .
Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆xRn = 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
. . . . . .
Forming Riemann sums
We have many choices of how to approximate the area:
Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆xRn = 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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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
..x7.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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
..x7
..x8.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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
..x7
..x8
..x9.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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
..x7
..x8
..x9
..x10.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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
..x7
..x8
..x9
..x10
..x11.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.
.x16.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.
.x16.
.x17.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.
.x16.
.x17.
.x18.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.
.x16.
.x17.
.x18.
.x19.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Theorem of the Day
TheoremIf 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..x7..x8..x9.
.x10.
.x11.
.x12.
.x13.
.x14.
.x15.
.x16.
.x17.
.x18.
.x19.
.x20.a .b
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 23 / 31
. . . . . .
Distances
Just like area = length× width, we have
distance = rate× time.
So here is another use for Riemann sums.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 24 / 31
. . . . . .
Application: Dead Reckoning
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 25 / 31
. . . . . .
Computing position by Dead Reckoning
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 2:00Speed (knots) 4 8 12 6 4Direction E E E E WTime 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E
Estimate the ship’s position at 4:00pm.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 26 / 31
. . . . . .
Solution
SolutionWe 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 nmihr
)(12hr)
= 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 15.5nmi east of its original position.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 27 / 31
. . . . . .
Analysis
I This method of measuring position by recording velocity wasnecessary until global-positioning satellite technology becamewidespread
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 28 / 31
. . . . . .
Other uses of Riemann sums
Anything with a product!I Area, volumeI Anything with a density: Population, massI Anything with a “speed:” distance, throughput, powerI Consumer surplusI Expected value of a random variable
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 29 / 31
. . . . . .
Surplus by picture
..quantity (q)
.price (p)
.
.demand f(q)
.supply
.equilibrium
..q∗
..p∗
.consumer surplus
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 30 / 31
. . . . . .
Summary
I We can compute the area of a curved region with a limit ofRiemann sums
I We can compute the distance traveled from the velocity with alimit of Riemann sums
I Many other important uses of this process.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 31 / 31