We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 22: Areas and Distances
1. Section 5.1
Areas and Distances
V63.0121.006/016, Calculus I
New York University
April 13, 2010
Announcements
Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May 10, 12:00noon
2. Announcements
Quiz April 16 on §§4.1–4.4
Final Exam: Monday, May
10, 12:00noon
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 3 / 30
4. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 4 / 30
5. 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 we
have to start somewhere.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 5 / 30
6. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
7. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
8. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
9. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
h
b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
11. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
12. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
h
b
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
14. Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summing the
areas of the triangles:
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 8 / 30
15. Hard Areas: Curved Regions
???
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 9 / 30
16. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC –
212 BC (after Euclid)
Geometer
Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
17. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC –
212 BC (after Euclid)
Geometer
Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
18. Meet the mathematician: Archimedes
Greek (Syracuse), 287 BC –
212 BC (after Euclid)
Geometer
Weapons engineer
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
19. Archimedes found areas of a sequence of triangles inscribed in a parabola.
A=
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
20. 1
Archimedes found areas of a sequence of triangles inscribed in a parabola.
A=1
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
21. 1
1 1
8 8
Archimedes found areas of a sequence of triangles inscribed in a parabola.
1
A=1+2·
8
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
22. 1 1
64 64
1
1 1
8 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
23. 1 1
64 64
1
1 1
8 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
24. We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
25. 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 + · · · + r n ) = 1 − r n+1
So
1 − r n+1
1 + r + · · · + rn =
1−r
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
26. 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 + · · · + r n ) = 1 − r n+1
So
1 − r n+1
1 + r + · · · + rn =
1−r
Therefore
1 1 1 1 − (1/4)n+1
1+ + + ··· + n =
4 16 4 1 − 1/4
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
27. 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 + · · · + r n ) = 1 − r n+1
So
1 − r n+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 → ∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
28. Cavalieri
Italian,
1598–1647
Revisited the
area problem
with a
different
perspective
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 13 / 30
29. Cavalieri’s method
Divide up the interval into pieces
y = x2 and measure the area of the
inscribed rectangles:
0 1
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
30. Cavalieri’s method
Divide up the interval into pieces
y = x2 and measure the area of the
inscribed rectangles:
1
L2 =
8
0 1 1
2
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
31. Cavalieri’s method
Divide up the interval into pieces
y = x2 and measure the area of the
inscribed rectangles:
1
L2 =
8
L3 =
0 1 2 1
3 3
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
32. Cavalieri’s method
Divide up the interval into pieces
y = x2 and measure the area of the
inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
0 1 2 1
3 3
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
33. Cavalieri’s method
Divide up the interval into pieces
y = x2 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
34. Cavalieri’s method
Divide up the interval into pieces
y = x2 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
35. Cavalieri’s method
Divide up the interval into pieces
y = x2 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
36. Cavalieri’s method
Divide up the interval into pieces
y = x2 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
37. Cavalieri’s method
Divide up the interval into pieces
y = x2 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 =?
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
38. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
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
2
1 i −1 (i − 1)2
· = .
n n n3
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
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
2
1 i −1 (i − 1)2
· = .
n n n3
So
1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
Ln = + 3 + ··· + =
n3 n n3 n3
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
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
2
1 i −1 (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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
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
2
1 i −1 (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 = →
6n3 3
as n → ∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
43. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + ·f
n n n n n n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
44. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + · f
n n n n n n
1 1 1 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
45. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + · f
n n n n n n
1 1 1 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (n − 1)3
=
n4
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
46. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + · f
n n n n n n
1 1 1 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (n − 1)3
=
n4
The formula out of the hat is
2
1 + 23 + 33 + · · · + (n − 1)3 = 1
2 n(n − 1)
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
47. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
1 1 1 2 1 n−1
Ln = ·f + ·f + ··· + · f
n n n n n n
1 1 1 2 3 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
1+2 3 + 33 + · · · + (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 → ∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
48. Cavalieri’s method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
13 + 23 + 33 + · · · + n3
=
n4
1 1 2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
= →
4n4 4
as n → ∞.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
49. Cavalieri’s method with different heights
1 13 1 23 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
13 + 23 + 33 + · · · + n3
=
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
50. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 18 / 30
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn xn = a + n · =b
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn xn = a + n · =b
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn xn = a + n · =b
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn xn = a + n · =b
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
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).
b−a
For each positive integer n, divide up the interval into n pieces. Then ∆x = .
n
For each i between 1 and n, let xi be the nth step between a and 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 b b−a
x0 x1 x2 . . . xi xn−1xn xn = a + n · =b
n
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
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 b
x1
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 x1 b
x2
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 x1 x2 b
x3
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 x1 x2 x3 b
x4
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 b5
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 b6
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
a x x x x x x x
matter what choice of ci we 1 2 3 4 5 6 b7
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
matter what choice of ci we 1 2 3 4 5 6 7 b8
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
matter what choice of ci we 1 2 3 4 5 6 7 8b9
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
a x 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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 x x xb
matter what choice of ci we 1 2 3 4 5 6 7 8 9 1011
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 101112
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
ax x x x x x x x xx x x x x xb
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14
11 13 15
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
axxxxxxxxxxxxxxxxb
matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14 16
11 13 15
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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 xxxxxxxxxxxxxxxxb
x1 2 3 4 5 6 7 8 910 12 14 16
matter what choice of ci we 11 13 15 17
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
x1 2 3 4 5 6 7 8x10 12 14 16 18
matter what choice of ci we 9 11 13 15 17
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
x12345678910 12 14 16 18
x 11 13 15 17 19
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
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
a xxxxxxxx xxxxxxxxxxb
x123456789101214161820
x 1113151719
matter what choice of ci we
made.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
78. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4)
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
79. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4)
Want the slope of a curve
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
80. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
81. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the slope of lines
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
82. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope of lines
polygons
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
83. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope of lines
polygons
Approximate curve with a
line
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
84. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope of lines
polygons
Approximate curve with a
Approximate region with
line
polygons
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
85. Analogies
The Area Problem (Ch. 5)
The Tangent Problem
(Ch. 2–4) Want the area of a curved
region
Want the slope of a curve
Only know the area of
Only know the slope 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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
86. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 23 / 30
87. 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
89. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 26 / 30
90. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 27 / 30
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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 28 / 30
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
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 29 / 30
93. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 30 / 30