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Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
Lesson 24: Area and Distances
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Lesson 24: Area and Distances

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  • 1. Section 5.1 Areas and Distances V63.0121.002.2010Su, Calculus I New York University June 16, 2010 Announcements Quiz Thursday on 4.1–4.4 . . . . . .
  • 2. Announcements Quiz Thursday on 4.1–4.4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 2 / 31
  • 3. Objectives Compute the area of a region by approximating it with rectangles and letting the size of the rectangles tend to zero. Compute the total distance traveled by a particle by approximating it as distance = (rate)(time) and letting the time intervals over which one approximates tend to zero. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 3 / 31
  • 4. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 4 / 31
  • 5. Easy Areas: Rectangle Definition The area of a rectangle with dimensions ℓ and w is the product A = ℓw. w . . . ℓ It may seem strange that this is a definition and not a theorem but we have to start somewhere. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 5 / 31
  • 6. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. . b . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
  • 7. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
  • 8. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
  • 9. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
  • 10. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . So Fact The area of a parallelogram of base width b and height h is A = bh . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 6 / 31
  • 11. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. . b . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
  • 12. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. h . . b . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
  • 13. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. h . . b . So Fact The area of a triangle of base width b and height h is 1 A= bh 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 7 / 31
  • 14. Easy Areas: Other Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: . . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 8 / 31
  • 15. Hard Areas: Curved Regions . ??? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 9 / 31
  • 16. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
  • 17. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
  • 18. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 10 / 31
  • 19. Archimedes and the Parabola . Archimedes found areas of a sequence of triangles inscribed in a parabola. A= . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
  • 20. Archimedes and the Parabola 1 . . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
  • 21. Archimedes and the Parabola 1 . .1 8 .1 8 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
  • 22. Archimedes and the Parabola 1 1 .64 .64 1 . .1 8 .1 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
  • 23. Archimedes and the Parabola 1 1 .64 .64 1 . .1 8 .1 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 =1+ + + ··· + n + ··· 4 16 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 11 / 31
  • 24. Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
  • 25. Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
  • 26. Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 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 1 1 − (1/4)n+1 1+ + + ··· + n = 4 16 4 1 − 1/4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
  • 27. Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 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 1 1 − (1/4)n+1 1 4 1+ + + ··· + n = →3 = 4 16 4 1− 1/4 /4 3 as n → ∞. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 12 / 31
  • 28. Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 13 / 31
  • 29. Cavalieri's method Divide up the interval into 2 y . =x pieces and measure the area of the inscribed rectangles: . . 0 . 1 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 30. Cavalieri's method Divide up the interval into 2 y . =x pieces and measure the area of the inscribed rectangles: 1 L2 = 8 . . . 0 . 1 1 . . 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 31. Cavalieri's method Divide up the interval into 2 y . =x pieces and measure the area of the inscribed rectangles: 1 L2 = 8 L3 = . . . . 0 . 1 2 1 . . . 3 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 32. Cavalieri's method Divide up the interval into 2 y . =x pieces and measure the area of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 . . . . 0 . 1 2 1 . . . 3 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 33. Cavalieri's method Divide up the interval into 2 y . =x pieces and measure the area of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 L4 = . . . . . 0 . 1 2 3 1 . . . . 4 4 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 34. Cavalieri's method Divide up the interval into 2 y . =x 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 . 1 2 3 1 . . . . 4 4 4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 35. Cavalieri's method Divide up the interval into 2 y . =x 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 . 1 2 3 4 1 . L5 = . . . . 5 5 5 5 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 36. Cavalieri's method Divide up the interval into 2 y . =x 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 0 . 1 2 3 4 1 . L5 = + + + = . . . . 125 125 125 125 125 5 5 5 5 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 37. Cavalieri's method Divide up the interval into 2 y . =x 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 0 . 1 . L5 = + + + = . 125 125 125 125 125 Ln =? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 14 / 31
  • 38. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
  • 39. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . The n rectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
  • 40. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . The n rectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = n3 n n3 n3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
  • 41. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . The n rectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
  • 42. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . The n rectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = 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 → ∞. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 15 / 31
  • 43. Cavalieri's method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
  • 44. Cavalieri's method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
  • 45. Cavalieri's method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 3 3 1 + 2 + 3 + · · · + (n − 1)3 = n4 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
  • 46. Cavalieri's method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 3 3 1 + 2 + 3 + · · · + (n − 1)3 = n4 The formula out of the hat is [ ]2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
  • 47. Cavalieri's method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 3 3 1 + 2 + 3 + · · · + (n − 1)3 = n4 The formula out of the hat is [ ]2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) So n2 (n − 1)2 1 Ln = → 4n4 4 as n → ∞. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 16 / 31
  • 48. Cavalieri's method with different heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 3 3 3 1 + 2 + 3 + ··· + n 3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 = → 4n4 4 . as n → ∞. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 17 / 31
  • 49. Cavalieri's method with different heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 3 3 3 1 + 2 + 3 + ··· + n 3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 = → 4n4 4 . as n → ∞. So even though the rectangles overlap, we still get the same answer. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 17 / 31
  • 50. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 18 / 31
  • 51. 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. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step between a and n 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 . . . . . . . . ······ a . . 0 . 1 . 2 . . . . i. n−1..n xx x b b−a x x x xn = a + n · =b n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
  • 52. 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. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step between a and n 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 . . . . . . . . ······ a . . 0 . 1 . 2 . . . . i. n−1..n xx x b b−a x x x xn = a + n · =b n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
  • 53. 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. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step between a and n 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 . . . . . . . . ······ a . . 0 . 1 . 2 . . . . i. n−1..n xx x b b−a x x x xn = a + n · =b n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
  • 54. 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. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step between a and n 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 . . . . . . . . ······ a . . 0 . 1 . 2 . . . . i. n−1..n xx x b b−a x x x xn = a + n · =b n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
  • 55. 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. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step between a and n 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 . . . . . . . . ······ a . . 0 . 1 . 2 . . . . i. n−1..n xx x b b−a x x x xn = a + n · =b n . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 19 / 31
  • 56. 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 Mn = f ∆x + f ∆x + · · · + f ∆x 2 2 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
  • 57. 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 Mn = f ∆x + 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 20 / 31
  • 58. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . a . ..1 xb matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 59. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . a . x .1 ..2 xb matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 60. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . a . x .1 x .2 ..3 xb matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 61. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . a . x .1 x .2 x .3 ..4 xb matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 62. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . a x x x x x . . . . . .. matter what choice of ci we 1 2 3 4 b 5 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 63. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . a x x x x x x . . . . . . .. matter what choice of ci we 1 2 3 4 5 b 6 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 64. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . . ax x x x x x x . . . . . . . .. matter what choice of ci we 1 2 3 4 5 6 b 7 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 65. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . . . ax x x x x x x x . . . . . . . . .. b matter what choice of ci we 1 2 3 4 5 6 7 8 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 66. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . . . . ax x x x x x x x x . . . . . . . . . .. b matter what choice of ci we 1 2 3 4 5 6 7 8 9 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 67. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . . . . . ax x x x x x x x x xb . . . . . . . . . . .. matter what choice of ci we 1 2 3 4 5 6 7 8 9 10 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 68. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . . . . . . . . . . . . ax x x x x x x x xx xb . . . . . . . . . . . .. matter what choice of ci we 1 2 3 4 5 6 7 8 9 1011 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 69. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ............ ax x x x x x x x xx x xb . . . . . . . . . . . . .. matter what choice of ci we 1 2 3 4 5 6 7 8 9 10 12 11 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 70. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ............. ax x x x x x x x xx x x xb .. . . . . . . . .. . . .. matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 11 13 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 71. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . .............. ax x x x x x x x xx x x x xb .. . . . . . . . .. . . . . . matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14 11 13 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 72. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ............... a xxxxxxxxxxxxxxb .. . . . . . . . .. . . . . . . x1 2 3 4 5 6 7 8 910 12 14 matter what choice of ci we 11 13 15 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 73. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ................ a xxxxxxxx xxxxxxb . . . . . . . . . .. . . . . . . . x1 2 3 4 5 6 7 8x10 12 14 16 matter what choice of ci we 9 11 13 15 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 74. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ................. a xxxxxxxx xxxxxxxb .. . . . . . . . .. . . . . . . . . x1 2 3 4 5 6 7 8x10 12 14 16 matter what choice of ci we 9 11 13 15 17 made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 75. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . .................. a xxxxxxxx xxxxxxxxb .. . . . . . . . .. . . . . . . . . . x12345678910 12 14 16 18 x 11 13 15 17 matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 76. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . ................... a xxxxxxxx xxxxxxxxxb .. . . . . . . . .. . . . . . . . . . . x1234567891012141618 x 1113151719 matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 77. Theorem of the Day Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then ∑ n lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . .................... axxxxxxxx xxxxxxxxxxb .. . . . . . . . .. . . . . . . . . . . . x123456789 1113151719 x101214161820 matter what choice of ci we made. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 21 / 31
  • 78. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 79. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the slope of a curve . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 80. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 81. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of lines . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 82. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of Only know the area of lines polygons . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 83. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of Only know the area of lines polygons Approximate curve with a line . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 84. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of Only know the area of lines polygons Approximate curve with a Approximate region with line polygons . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 85. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of Only know the area of lines polygons Approximate curve with a Approximate region with line polygons Take limit over better and Take limit over better and better approximations better approximations . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 22 / 31
  • 86. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 23 / 31
  • 87. Distances Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 24 / 31
  • 88. Application: Dead Reckoning . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 25 / 31
  • 89. Computing position by Dead Reckoning 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 26 / 31
  • 90. Solution 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 27 / 31
  • 91. Analysis This method of measuring position by recording velocity was necessary until global-positioning satellite technology became widespread 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 28 / 31
  • 92. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 29 / 31
  • 93. Surplus by picture c . onsumer surplus p . rice (p) s . upply .∗ . p . . quilibrium e d . emand f(q) . . .∗ q q . uantity (q) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 30 / 31
  • 94. Summary We can compute the area of a curved region with a limit of Riemann sums We can compute the distance traveled from the velocity with a limit of Riemann sums Many other important uses of this process. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 31 / 31

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