We look at the area problem of finding areas of curved regions. Archimedes had a method for parabolas, Cavalieri had a method for other graphs, and Riemann generalized the whole thing. It doesn't just work for areas, any "product law" such as distance=rate x time can be generalized to a similar computation

1. Section 5.1
Areas and Distances
Math 1a
December 5, 2007
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2. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Worksheet

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

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

5. 1
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
A=1

6. 1
1 1
8 8
Archimedes found areas of a sequence of triangles inscribed in a
parabola.
1
A=1+2·
8

7. 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

8. 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

9. 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 − x)(1 + r + · · · + r n ) = 1 − r n+1
So
1 − r n+1
1 + r + · · · + rn =
1−r
Therefore
1 − (1/4)n+1
1 1 1 1 4
+ ··· + n = →3 =
1+ +
1 − 1/4
4 16 4 /4 3
as n → ∞.

10. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Worksheet

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

12. Cavalieri’s method
Divide up the interval into
pieces and measure the area
of the inscribed rectangles:

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

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

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

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

17. Cavalieri’s method
Divide up the interval into
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

18. Cavalieri’s method
Divide up the interval into
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 =

19. Cavalieri’s method
Divide up the interval into
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 = + + + =
125 125 125 25 125

20. Cavalieri’s method
Divide up the interval into
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 = + + + =
125 125 125 25 125
Ln =?

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

22. 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
2
(i − 1)2
i −1
1
· = .
n3
n n

23. 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
2
(i − 1)2
i −1
1
· = .
n3
n n
So
22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2
1
+ 3 + ··· +
Ln = =
n3 n n3 n3

24. 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
2
(i − 1)2
i −1
1
· = .
n3
n n
So
22 (n − 1)2 1 + 22 + 32 + · · · + (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

25. 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
2
(i − 1)2
i −1
1
· = .
n3
n n
So
22 (n − 1)2 1 + 22 + 32 + · · · + (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 = 3
6n 3
as n → ∞.

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

27. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
n−1
1 1 1 2 1
·f + ·f + ··· + · f
Ln =
n n n n n n
3 1 (n − 1)3
11 12
= · 3 + · 3 + ··· + ·
n3
nn nn n

28. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
n−1
1 1 1 2 1
·f + ·f + ··· + · f
Ln =
n n n n n n
3 1 (n − 1)3
11 12
= · 3 + · 3 + ··· + ·
n3
nn nn n
3 + 33 + · · · + (n − 1)3
1+2
=
n3

29. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
n−1
1 1 1 2 1
·f + ·f + ··· + · f
Ln =
n n n n n n
3 1 (n − 1)3
11 12
= · 3 + · 3 + ··· + ·
n3
nn nn n
3 + 33 + · · · + (n − 1)3
1+2
=
n3
The formula out of the hat is
2
1 + 23 + 33 + · · · + (n − 1)3 = 1
− 1)
2 n(n

30. Cavalieri’s method for different functions
Try the same trick with f (x) = x 3 . We have
n−1
1 1 1 2 1
·f + ·f + ··· + · f
Ln =
n n n n n n
3 1 (n − 1)3
11 12
= · 3 + · 3 + ··· + ·
n3
nn nn n
3 + 33 + · · · + (n − 1)3
1+2
=
n3
The formula out of the hat is
2
1 + 23 + 33 + · · · + (n − 1)3 = 1
− 1)
2 n(n
So
n2 (n − 1)2 1
→
Ln = 4
4n 4
as n → ∞.

31. Cavalieri’s method with different heights
1 13 1 23 1 n3
· 3 + · 3 + ··· + · 3
Rn =
nn nn nn
3 + 23 + 3 3 + · · · + n 3
1
=
n4
11 2
= 4 2 n(n + 1)
n
n2 (n + 1)2 1
→
= 4
4n 4
as n → ∞.

32. Cavalieri’s method with different heights
1 13 1 23 1 n3
· 3 + · 3 + ··· + · 3
Rn =
nn nn nn
3 + 23 + 3 3 + · · · + n 3
1
=
n4
11 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.

33. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Worksheet

34. 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
x1 = x0 + ∆x = a +
n
b−a
x2 = x1 + ∆x = a + 2 ·
n
······
b−a
xi = a + i ·
n
xi xn−1xn
x0 x1 x2 ······
b−a
a b xn = a + n · =b
n

35. 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
∆x + · · · + f
Mn = f ∆x + f ∆x
2 2 2

36. 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
∆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

37. 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.

38. Outline
Archimedes
Cavalieri
Generalizing Cavalieri’s method
Worksheet

39. Worksheet
We will determine the area under y = e x between x = 0 and x = 1.