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Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
Lesson 29: Areas
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Lesson 29: Areas

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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 …

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

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  • 1. Section 5.1 Areas and Distances Math 1a December 5, 2007 Announcements my next office hours: Today 1–3 (SC 323) MT II is nearly graded. You’ll get it back Friday Final seview sessions: Wed 1/9 and Thu 1/10 in Hall D, Sun 1/13 in Hall C, all 7–8:30pm Final tentatively scheduled for January 17
  • 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.

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