Thus begins the second "half" of calculus—in which we attempt to find areas of curved regions. Like with derivatives, we use a limiting process starting from things we know (areas of rectangles) and finer and finer approximations.

1. Section 5.1–5.2
Areas and Distances
The Definite Integral
V63.0121.034, Calculus I
November 30, 2009
Announcements
Quiz 5 this week in recitation on 4.1–4.4, 4.7
Final Exam, December 18, 2:00–3:50pm
. . . . . .

2. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Properties of the integral
. . . . . .

3. 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.
. . . . . .

4. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a
rectangle.
.
b
.
. . . . . .

5. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a
rectangle.
h
.
.
b
.
. . . . . .

6. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a
rectangle.
h
.
.
. . . . . .

7. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a
rectangle.
h
.
.
b
.
. . . . . .

8. Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a
rectangle.
h
.
.
b
.
So
A = bh
. . . . . .

9. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a
parallelogram.
.
b
.
. . . . . .

10. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a
parallelogram.
h
.
.
b
.
. . . . . .

11. Easy Areas: Triangle
By copying and pasting, a triangle can be made into a
parallelogram.
h
.
.
b
.
So
1
A= bh
2
. . . . . .

12. Easy Areas: Polygons
Any polygon can be triangulated, so its area can be found by
summing the areas of the triangles:
.
. . . . . .

13. Hard Areas: Curved Regions
.
???
. . . . . .

14. Meet the mathematician: Archimedes
Greek (Syracuse), 287
BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

15. Meet the mathematician: Archimedes
Greek (Syracuse), 287
BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

16. Meet the mathematician: Archimedes
Greek (Syracuse), 287
BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

17. .
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=
. . . . . .

18. 1
.
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=1
. . . . . .

19. 1
.
.1
8 .1
8
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1
A=1+2·
8
. . . . . .

20. 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
. . . . . .

21. 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
. . . . . .

22. We would then need to know the value of the series
1 1 1
1+ + + ··· + n + ···
4 16 4
. . . . . .

23. 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
. . . . . .

24. 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
. . . . . .

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 + · · · + 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 = → =
4 16 4 1− 1/4 3/4 3
as n → ∞.
. . . . . .

26. Cavalieri
Italian,
1598–1647
Revisited
the area
problem
with a
different
perspective
. . . . . .

27. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y
of the inscribed rectangles:
. .
0
. 1
.
. . . . . .

28. Cavalieri’s method
Divide up the interval into
. = x2
y pieces and measure the area
of the inscribed rectangles:
1
L2 =
8
. . .
0
. 1 1
.
.
2
. . . . . .

29. Cavalieri’s method
Divide up the interval into
. = x2
y pieces and measure the area
of the inscribed rectangles:
1
L2 =
8
L3 =
. . . .
0
. 1 2 1
.
. .
3 3
. . . . . .

30. Cavalieri’s method
Divide up the interval into
. = x2
y 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
. . . . . .

31. Cavalieri’s method
Divide up the interval into
. = x2
y 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
. . . . . .

32. Cavalieri’s method
Divide up the interval into
. = x2
y 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
. . . . . .

33. Cavalieri’s method
Divide up the interval into
. = x2
y 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
. . . . . .

34. Cavalieri’s method
Divide up the interval into
. = x2
y 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
. . . . . .

35. Cavalieri’s method
Divide up the interval into
. = x2
y 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 =?
. . . . . .

36. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
. . . . . .

37. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
The rectangle over the ith interval and under the parabola has
area ( )
1 i − 1 2 (i − 1)2
· = .
n n n3
. . . . . .

38. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
The 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
. . . . . .

39. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
The 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
. . . . . .

40. What is Ln ?
1
Divide the interval [0, 1] into n pieces. Then each has width .
n
The 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 = →
6n3 3
as n → ∞. . . . . . .

41. 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
. . . . . .

42. 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
. . . . . .

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
1 1 1 23 1 (n − 1)3
= · 3 + · 3 + ··· + ·
n n n n n n3
3 3 3
1 + 2 + 3 + · · · + (n − 1 )
=
n4
. . . . . .

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
3 3 3
1 + 2 + 3 + · · · + (n − 1 )
=
n4
The formula out of the hat is
[1 ]2
1 + 23 + 33 + · · · + (n − 1)3 = 2 n(n − 1)
. . . . . .

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 3
1 + 2 + 3 + · · · + (n − 1 )
=
n4
The formula out of the hat is
[1 ]2
1 + 23 + 33 + · · · + (n − 1)3 = 2 n(n − 1)
So
n2 (n − 1)2 1
Ln = →
4n4 4
as n → ∞.
. . . . . .

46. Cavalieri’s method with different heights
1 13 1 2 3 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
1 3 + 23 + 33 + · · · + n3
=
n4
1 [1 ]2
= 4 2 n (n + 1 )
n
n2 (n + 1)2 1
= 4
→
4n 4
.
as n → ∞.
. . . . . .

47. Cavalieri’s method with different heights
1 13 1 2 3 1 n3
Rn = · 3 + · 3 + ··· + · 3
n n n n n n
1 3 + 23 + 33 + · · · + n3
=
n4
1 [1 ]2
= 4 2 n (n + 1 )
n
n2 (n + 1)2 1
= 4
→
4n 4
.
as n → ∞.
So even though the rectangles overlap, we still get the same
answer.
. . . . . .

48. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Properties of the integral
. . . . . .

49. 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.
b−a
Then ∆x = . For each i between 1 and n, let xi be the nth
n
step between a and b. So
x0 = a
b−a
x 1 = x 0 + ∆x = a +
n
b−a
x 2 = x 1 + ∆x = a + 2 ·
n
······
b−a
xi = a + i ·
n
x x x
.0 .1 .2 xx x
. i . n−1. n ······
. . . . . . . .
a
. b
. b−a
xn = a + n ·
. . .
=b . . .

50. 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 + x 1 x 1 + x2 xn−1 + xn
Mn = f ∆x + f ∆x + · · · + f ∆x
2 2 2
. . . . . .

51. 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 + x 1 x 1 + 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
. . . . . .

52. Theorem of the Day
Theorem
If f is a continuous function on [a, b] or has finitely many jump
discontinuities, then
lim Sn = lim {f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x}
n→∞ n→∞
exists and is the same value no matter what choice of ci we made.
. . . . . .

53. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
. . . . . .

54. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the slope of a
curve
. . . . . .

55. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
. . . . . .

56. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
Only know the slope of
lines
. . . . . .

57. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
Only know the area of
Only know the slope of
polygons
lines
. . . . . .

58. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
Only know the area of
Only know the slope of
polygons
lines
Approximate curve with
a line
. . . . . .

59. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
Only know the area of
Only know the slope of
polygons
lines
Approximate region
Approximate curve with
with polygons
a line
. . . . . .

60. Analogies
The Tangent Problem
The Area Problem (Ch. 5)
(Ch. 2–4)
Want the area of a
Want the slope of a
curved region
curve
Only know the area of
Only know the slope of
polygons
lines
Approximate region
Approximate curve with
with polygons
a line
Take limit over better
Take limit over better
and better
and better
approximations
approximations
. . . . . .

61. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Properties of the integral
. . . . . .

62. Distances
Just like area = length × width, we have
distance = rate × time.
So here is another use for Riemann sums.
. . . . . .

63. 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.
. . . . . .

64. 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.
. . . . . .

65. Analysis
This method of measuring position by recording velocity is
known as dead reckoning.
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.
. . . . . .

66. 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
. . . . . .

67. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Properties of the integral
. . . . . .

68. The definite integral as a limit
Definition
If f is a function defined on [a, b], the definite integral of f from a
to b is the number
∫ b ∑n
f(x) dx = lim f(ci ) ∆x
a ∆x→0
i =1
. . . . . .

69. Notation/Terminology
∫ b
f(x) dx
a
. . . . . .

70. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
. . . . . .

71. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
. . . . . .

72. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b
the upper limit)
. . . . . .

73. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b
the upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
. . . . . .

74. Notation/Terminology
∫ b
f(x) dx
a
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integration (a is the lower limit and b
the upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration or
quadrature
. . . . . .

75. The limit can be simplified
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists.
a
. . . . . .

76. The limit can be simplified
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite
∫ b
integral f(x) dx exists.
a
Theorem
If f is integrable on [a, b] then
∫ b n
∑
f(x) dx = lim f(xi )∆x,
a n→∞
i=1
where
b−a
∆x = and xi = a + i ∆x
n
. . . . . .

77. Outline
Area through the Centuries
Euclid
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Analogies
Distances
Other applications
The definite integral as a limit
Properties of the integral
. . . . . .

78. Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
∫ b
1. c dx = c(b − a)
a
. . . . . .

79. Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
∫ b
1. c dx = c(b − a)
a
∫ b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
a a a
. . . . . .

80. Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
∫ b
1. c dx = c(b − a)
a
∫ b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
a a a
∫ b ∫ b
3. cf(x) dx = c f(x) dx.
a a
. . . . . .

81. Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant.
Then
∫ b
1. c dx = c(b − a)
a
∫ b ∫ b ∫ b
2. [f(x) + g(x)] dx = f(x) dx + g(x) dx.
a a a
∫ b ∫ b
3. cf(x) dx = c f(x) dx.
a a
∫ b ∫ b ∫ b
4. [f(x) − g(x)] dx = f(x) dx − g(x) dx.
a a a
. . . . . .

82. More Properties of the Integral
Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
. . . . . .

83. More Properties of the Integral
Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
. . . . . .

84. More Properties of the Integral
Conventions: ∫ ∫
a b
f(x) dx = − f(x) dx
b a
∫ a
f(x) dx = 0
a
This allows us to have
∫ c ∫ b ∫ c
5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c.
a a b
. . . . . .

85. Example
Suppose f and g are functions with
∫ 4
f(x) dx = 4
0
∫ 5
f(x) dx = 7
0
∫ 5
g(x) dx = 3.
0
Find
∫ 5
(a) [2f(x) − g(x)] dx
0
∫ 5
(b) f(x) dx.
4
. . . . . .

86. Solution
We have
(a)
∫ 5 ∫ 5 ∫ 5
[2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx
0 0 0
= 2 · 7 − 3 = 11
. . . . . .

87. Solution
We have
(a)
∫ 5 ∫ 5 ∫ 5
[2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx
0 0 0
= 2 · 7 − 3 = 11
(b)
∫ 5 ∫ 5 ∫ 4
f(x) dx = f(x) dx − f(x) dx
4 0 0
=7−4=3
. . . . . .