1. Section 5.1
Areas and Distances
V63.0121.002.2010Su, Calculus I
New York University
June 16, 2010
Announcements
Quiz Thursday on 4.1–4.4
. . . . . .
2. Announcements
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
3. Objectives
Compute the area of a
region by approximating it
with rectangles and letting
the size of the rectangles
tend to zero.
Compute the total distance
traveled by a particle by
approximating it as
distance = (rate)(time) and
letting the time intervals
over which one
approximates tend to zero.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 3 / 31
4. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 4 / 31
5. Easy Areas: Rectangle
Definition
The area of a rectangle with dimensions ℓ and w is the product A = ℓw.
w
.
.
.
ℓ
It may seem strange that this is a definition and not a theorem but we
have to start somewhere.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 5 / 31
6. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
7. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
.
b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
8. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
9. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
.
b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
10. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
.
.
b
.
So
Fact
The 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
11. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
.
b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
12. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
h
.
.
b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
13. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
h
.
.
b
.
So
Fact
The area of a triangle of base width b and height h is
1
A= bh
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
14. Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summing
the areas of the triangles:
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 8 / 31
15. Hard Areas: Curved Regions
.
???
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 9 / 31
16. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
17. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
18. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC
– 212 BC (after Euclid)
Geometer
Weapons engineer
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
19. Archimedes and the Parabola
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
20. Archimedes and the Parabola
1
.
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
21. Archimedes and the Parabola
1
.
.1
8 .1
8
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1
A=1+2·
8
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
22. Archimedes and the Parabola
1 1
.64 .64
1
.
.1
8 .1
8
1 1
.64 .64
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1 1
A=1+2· +4· + ···
8 64
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
23. Archimedes and the Parabola
1 1
.64 .64
1
.
.1
8 .1
8
1 1
.64 .64
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1 1
A=1+2· +4· + ···
8 64
1 1 1
=1+ + + ··· + n + ···
4 16 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
24. Summing the series
[label=archimedes-parabola-sum] We would then need to know the
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
25. Summing the series
[label=archimedes-parabola-sum] We would then need to know the
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
But for any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
26. Summing the series
[label=archimedes-parabola-sum] We would then need to know the
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
But for any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1
1+ + + ··· + n =
4 16 4 1 − 1/4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
27. Summing the series
[label=archimedes-parabola-sum] We would then need to know the
value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
But for any number r and any positive integer n,
(1 − r)(1 + r + · · · + rn ) = 1 − rn+1
So
1 − rn+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1 1 4
1+ + + ··· + n = →3 =
4 16 4 1− 1/4 /4 3
as n → ∞. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
28. Cavalieri
Italian,
1598–1647
Revisited the
area
problem with
a different
perspective
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 13 / 31
29. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
. .
0
. 1
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
30. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
. . .
0
. 1 1
.
.
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
31. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
L3 =
. . . .
0
. 1 2 1
.
. .
3 3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
32. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
. . . .
0
. 1 2 1
.
. .
3 3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
33. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
L4 =
. . . . .
0
. 1 2 3 1
.
. . .
4 4 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
34. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . 64 64 64 64
0
. 1 2 3 1
.
. . .
4 4 4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
35. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . . 64 64 64 64
0
. 1 2 3 4 1
. L5 =
. . . .
5 5 5 5
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
36. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . . . . . 64 64 64 64
1 4 9 16 30
0
. 1 2 3 4 1
. L5 = + + + =
. . . . 125 125 125 125 125
5 5 5 5
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
37. Cavalieri's method
Divide up the interval into
2
y
. =x pieces and measure the area of
the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
. . 64 64 64 64
1 4 9 16 30
0
. 1
. L5 = + + + =
. 125 125 125 125 125
Ln =?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
38. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
39. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
40. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
41. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1)
Ln =
6n3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
42. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width . The
n
rectangle over the ith interval and under the parabola has area
( )
1 i − 1 2 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
The Arabs knew that
n(n − 1)(2n − 1)
1 + 22 + 32 + · · · + (n − 1)2 =
6
So
n(n − 1)(2n − 1) 1
Ln = 3
→
6n 3
as n → ∞. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
43. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
44. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
45. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
46. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
47. Cavalieri's method for different functions
Try the same trick with f(x) = x3 . We have
( ) ( ) ( )
1 1 1 2 1 n−1
Ln = · f + ·f + ··· + · f
n n n n n n
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3
1 + 2 + 3 + · · · + (n − 1)3
=
n4
The formula out of the hat is
[ ]2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)
So
n2 (n − 1)2 1
Ln = →
4n4 4
as n → ∞. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
48. Cavalieri's method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as n → ∞.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 17 / 31
49. Cavalieri's method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
3 3 3
1 + 2 + 3 + ··· + n 3
=
n4
1 [1 ]2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
.
as 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
50. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 18 / 31
51. Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
52. Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
53. Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
54. Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
55. Cavalieri's method in general
.
Let f be a positive function defined on the interval [a, b]. We want to find the
area between x = a, x = b, y = 0, and y = f(x).
For each positive integer n, divide up the interval into n pieces. Then
b−a
∆x = . For each i between 1 and n, let xi be the nth step between a and
n
b. So
x0 = a
b−a
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
. ······
b−a
xi = a + i ·
n
. . . . . . . . ······
a
.
. 0 . 1 . 2 . . . . i. n−1..n
xx x b b−a
x x x xn = a + n · =b
n
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
56. 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
( ) ( ) ( )
x0 + x1 x1 + x2 xn−1 + xn
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
57. 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
( ) ( ) ( )
x0 + x1 x1 + x2 xn−1 + xn
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2
In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the
Riemann sum
Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x
∑
n
= f(ci )∆x
i=1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
58. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . .
a
. ..1
xb
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
59. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . .
a
. x
.1 ..2
xb
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
60. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . .
a
. x
.1 x
.2 ..3
xb
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
61. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . .
a
. x
.1 x
.2 x
.3 ..4
xb
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
62. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . .
a x x x x x
. . . . . ..
matter what choice of ci we 1 2 3 4 b 5
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
63. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . .
a x x x x x x
. . . . . . ..
matter what choice of ci we 1 2 3 4 5 b 6
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
64. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . . .
ax x x x x x x
. . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 b 7
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
65. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . . . .
ax x x x x x x x
. . . . . . . . ..
b
matter what choice of ci we 1 2 3 4 5 6 7 8
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
66. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . . . . .
ax x x x x x x x x
. . . . . . . . . ..
b
matter what choice of ci we 1 2 3 4 5 6 7 8 9
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
67. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . . . . . .
ax x x x x x x x x xb
. . . . . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 9 10
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
68. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . . . . . . . . . . . .
ax x x x x x x x xx xb
. . . . . . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 9 1011
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
69. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ............
ax x x x x x x x xx x xb
. . . . . . . . . . . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 9 10 12
11
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
70. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . .............
ax x x x x x x x xx x x xb
.. . . . . . . . .. . . ..
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12
11 13
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
71. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ..............
ax x x x x x x x xx x x x xb
.. . . . . . . . .. . . . . .
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14
11 13
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
72. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ...............
a xxxxxxxxxxxxxxb
.. . . . . . . . .. . . . . . .
x1 2 3 4 5 6 7 8 910 12 14
matter what choice of ci we 11 13 15
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
73. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ................
a xxxxxxxx xxxxxxb
. . . . . . . . . .. . . . . . . .
x1 2 3 4 5 6 7 8x10 12 14 16
matter what choice of ci we 9 11 13 15
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
74. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . .................
a xxxxxxxx xxxxxxxb
.. . . . . . . . .. . . . . . . . .
x1 2 3 4 5 6 7 8x10 12 14 16
matter what choice of ci we 9 11 13 15 17
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
75. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ..................
a xxxxxxxx xxxxxxxxb
.. . . . . . . . .. . . . . . . . . .
x12345678910 12 14 16 18
x 11 13 15 17
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
76. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ...................
a xxxxxxxx xxxxxxxxxb
.. . . . . . . . .. . . . . . . . . . .
x1234567891012141618
x 1113151719
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
77. Theorem of the Day
Theorem
If f is a continuous function or
has finitely many jump
discontinuities on [a, b], then
∑
n
lim Sn = lim f(ci )∆x
n→∞ n→∞
i=1
exists and is the same value no . ....................
axxxxxxxx xxxxxxxxxxb
.. . . . . . . . .. . . . . . . . . . . .
x123456789 1113151719
x101214161820
matter what choice of ci we
made.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
78. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
79. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4)
Want the slope of a curve
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
80. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
81. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of
lines
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
82. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of Only know the area of
lines polygons
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
83. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of Only know the area of
lines polygons
Approximate curve with a
line
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
84. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of Only know the area of
lines polygons
Approximate curve with a Approximate region with
line polygons
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
85. Analogies
The Tangent Problem The Area Problem (Ch. 5)
(Ch. 2–4) Want the area of a curved
Want the slope of a curve region
Only know the slope of Only know the area of
lines polygons
Approximate curve with a Approximate region with
line polygons
Take limit over better and Take limit over better and
better approximations better approximations
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
86. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 23 / 31
87. 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
88. Application: Dead Reckoning
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 25 / 31
89. Computing position by Dead Reckoning
Example
A sailing ship is cruising back and forth along a channel (in a straight
line). At noon the ship’s position and velocity are recorded, but shortly
thereafter a storm blows in and position is impossible to measure. The
velocity 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 26 / 31
90. Solution
Solution
We estimate that the speed of 4 knots (nautical miles per hour) is
maintained from 12:00 until 12:30. So over this time interval the ship
travels ( )( )
4 nmi 1
hr = 2 nmi
hr 2
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 27 / 31
91. Analysis
This method of measuring position by recording velocity was
necessary until global-positioning satellite technology became
widespread
If we had velocity estimates at finer intervals, we’d get better
estimates.
If we had velocity at every instant, a limit would tell us our exact
position 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
92. Other uses of Riemann sums
Anything with a product!
Area, volume
Anything with a density: Population, mass
Anything with a “speed:” distance, throughput, power
Consumer surplus
Expected value of a random variable
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 29 / 31
93. Surplus by picture
c
. onsumer surplus
p
. rice (p)
s
. upply
.∗ .
p . . quilibrium
e
d
. emand f(q)
. .
.∗
q q
. uantity (q)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 30 / 31
94. Summary
We can compute the area of a curved region with a limit of
Riemann sums
We can compute the distance traveled from the velocity with a
limit of Riemann sums
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
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