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.
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 → ∞.
. . . . . .
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:
. .
0
. 1
.
. . . . . .
18. 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
. . . . . .
19. Cavalieri’s method
Divide up the interval into
. = x2
y pieces and measure the area
of the inscribed rectangles:
1
L2 =
8
L3 =
. . . .
0
. 2
1 1
.
.
.
3
3
. . . . . .
20. 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
. 2
1 1
.
.
.
3
3
. . . . . .
21. 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
. 2 3
1 1
.
. .
.
4 4
4
. . . . . .
22. 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
. 2 3
1 1
.
. .
.
4 4
4
. . . . . .
23. 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
. . . . . .
L5 =
4
0
. 2 3
1 1
.
.
. .
.
5
5 5
5
. . . . . .
25. 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
L5
0
. 1
. = + + + =
125 125 125 125 125
.
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 → ∞. . . . . . .
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)
=
n4
. . . . . .
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)
=
n4
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)
=
n4
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.
. . . . . .
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
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .