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 just work for areas, any "product law" such as distance=rate x time can be generalized to a similar computation

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Lesson 29: Areas

  1. 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. 2. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Worksheet
  3. 3. Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer
  4. 4. Archimedes found areas of a sequence of triangles inscribed in a parabola. A=
  5. 5. 1 Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1
  6. 6. 1 1 1 8 8 Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8
  7. 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. 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. 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. 10. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Worksheet
  11. 11. Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective
  12. 12. Cavalieri’s method Divide up the interval into pieces and measure the area of the inscribed rectangles:
  13. 13. Cavalieri’s method Divide up the interval into pieces and measure the area of the inscribed rectangles: 1 L2 = 8
  14. 14. Cavalieri’s method Divide up the interval into pieces and measure the area of the inscribed rectangles: 1 L2 = 8 L3 =
  15. 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. 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. 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. 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. 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. 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. 21. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n
  22. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 33. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Worksheet
  34. 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. 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. 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. 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. 38. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Worksheet
  39. 39. Worksheet We will determine the area under y = e x between x = 0 and x = 1.

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