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.

1. Section 5.1
Areas and Distances
V63.0121, Calculus I
April 13, 2009
Announcements
Moving to 624 today (no OH)
Quiz 5 this week on §§4.1–4.4
Back HW returned in recitation this week
. . . . . .

2. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Distances
Other applications
. . . . . .

3. Meet the mathematician: Archimedes
287 BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

4. Meet the mathematician: Archimedes
287 BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

5. Meet the mathematician: Archimedes
287 BC – 212 BC (after
Euclid)
Geometer
Weapons engineer
. . . . . .

6. .
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=
. . . . . .

7. 1
.
.
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=1
. . . . . .

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

9. 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
. . . . . .

10. 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
+ ··· + n + ···
=1+ +
4 16 4
. . . . . .

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

12. We would then need to know the value of the series
1 1 1
+ ··· + n + ···
1+ +
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
. . . . . .

13. We would then need to know the value of the series
1 1 1
+ ··· + n + ···
1+ +
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 − (1/4)n+1
1 1 1
+ ··· + n =
1+ +
1 − 1/4
4 16 4
. . . . . .

14. We would then need to know the value of the series
1 1 1
+ ··· + n + ···
1+ +
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 − (1/4)n+1
1 1 1 1 4
+ ··· + n = →
1+ + =
1−
4 16 4 3
1/4 3/4
as n → ∞.
. . . . . .

15. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Distances
Other applications
. . . . . .

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

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

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

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

20. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y of the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
.
. . . . . .

21. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y of the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
L4 =
.
. . . . . .

22. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y of the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
64 64 64 64
.
. . . . . .

23. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y of the inscribed rectangles:
1
L2 =
8
1 4 5
L3 = + =
27 27 27
1 4 9 14
L4 = + + =
64 64 64 64
. L5 =
. . . . . .

24. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y 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
. L5 = + + + =
125 125 125 25 125
. . . . . .

25. Cavalieri’s method
Divide up the interval into
pieces and measure the area
. = x2
y 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
. L5 = + + + =
125 125 125 25 125
Ln =?
. . . . . .

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

27. 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 ( )
i − 1 2 (i − 1)2
1
· = .
n3
n n
. . . . . .

28. 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 ( )
i − 1 2 (i − 1)2
1
· = .
n3
n n
So
22 1 + 22 + 32 + · · · + (n − 1)2
(n − 1)2
1
+ 3 + ··· +
Ln = =
n3 n n3 n3
. . . . . .

29. 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 ( )
i − 1 2 (i − 1)2
1
· = .
n3
n n
So
22 1 + 22 + 32 + · · · + (n − 1)2
(n − 1)2
1
+ 3 + ··· +
Ln = =
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
. . . . . .

30. 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 ( )
i − 1 2 (i − 1)2
1
· = .
n3
n n
So
22 1 + 22 + 32 + · · · + (n − 1)2
(n − 1)2
1
+ 3 + ··· +
Ln = =
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 → ∞. . . . . . .

31. Cavalieri’s method for different functions
Try the same trick with f(x) = x3 . We have
() () ( )
n−1
1 1 1 2 1
Ln = · f + ·f + ··· + · f
n n n n n n
. . . . . .

32. Cavalieri’s method for different functions
Try the same trick with f(x) = x3 . We have
() () ( )
n−1
1 1 1 2 1
Ln = · f + ·f + ··· + · f
n n n n n n
1 23 1 (n − 1)3
11
· 3 + · 3 + ··· + ·
=
n3
nn nn n
. . . . . .

33. Cavalieri’s method for different functions
Try the same trick with f(x) = x3 . We have
() () ( )
n−1
1 1 1 2 1
Ln = · f + ·f + ··· + · f
n n n n n n
1 23 1 (n − 1)3
11
· 3 + · 3 + ··· + ·
=
n3
nn nn n
3 3 3
1 + 2 + 3 + · · · + (n − 1)
=
n3
. . . . . .

34. Cavalieri’s method for different functions
Try the same trick with f(x) = x3 . We have
() () ( )
n−1
1 1 1 2 1
Ln = · f + ·f + ··· + · f
n n n n n n
1 23 1 (n − 1)3
11
· 3 + · 3 + ··· + ·
=
n3
nn nn n
3 3 3
1 + 2 + 3 + · · · + (n − 1)
=
n3
The formula out of the hat is
[1 ]2
1 + 23 + 33 + · · · + (n − 1)3 = − 1)
2 n(n
. . . . . .

35. Cavalieri’s method for different functions
Try the same trick with f(x) = x3 . We have
() () ( )
n−1
1 1 1 2 1
Ln = · f + ·f + ··· + · f
n n n n n n
1 23 1 (n − 1)3
11
· 3 + · 3 + ··· + ·
=
n3
nn nn n
3 3 3
1 + 2 + 3 + · · · + (n − 1)
=
n3
The formula out of the hat is
[1 ]2
1 + 23 + 33 + · · · + (n − 1)3 = − 1)
2 n(n
So
n2 (n − 1)2 1
→
Ln =
4n4 4
as n → ∞.
. . . . . .

36. Cavalieri’s method with different heights
1 13 1 2 3 1 n3
· 3 + · 3 + ··· + · 3
Rn =
nn nn nn
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 → ∞.
. . . . . .

37. Cavalieri’s method with different heights
1 13 1 2 3 1 n3
· 3 + · 3 + ··· + · 3
Rn =
nn nn nn
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.
. . . . . .

38. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Distances
Other applications
. . . . . .

39. 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
. For each i between 1 and n, let xi be the nth
Then ∆x =
n
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
xx x
. i . n−1. n
xxx
.0 .1 .2 ······
. .......
b−a
a
. b
. xn = a + n · =b
. . . . . .

40. 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
∆x + · · · + f
Mn = f ∆x + f ∆x
2 2 2
. . . . . .

41. 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
∆x + · · · + f
Mn = 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
. . . . . .

42. 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.
. . . . . .

43. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Distances
Other applications
. . . . . .

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

45. 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.
. . . . . .

46. 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.
. . . . . .

47. 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.
. . . . . .

48. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Distances
Other applications
. . . . . .

49. 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
. . . . . .

50. Summary
Riemann sums can be used to estimate areas, distance, and
other quantities
The limit of Riemann sums can get the exact value
Coming up: giving this limit a name and working with it.
. . . . . .