Lesson 25: Areas and Distances; The Definite Integral

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Thus begins the second "half" of calculus—in which we attempt to find areas of curved regions. Like with derivatives, we use a limiting process starting from things we know (areas of rectangles) and finer and finer approximations.

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Lesson 25: Areas and Distances; The Definite Integral

  1. 1. Section 5.1–5.2 Areas and Distances The Definite Integral V63.0121.034, Calculus I November 30, 2009 Announcements Quiz 5 this week in recitation on 4.1–4.4, 4.7 Final Exam, December 18, 2:00–3:50pm . . . . . .
  2. 2. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Properties of the integral . . . . . .
  3. 3. 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. . . . . . .
  4. 4. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. . b . . . . . . .
  5. 5. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . . . . . . .
  6. 6. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . . . . . . .
  7. 7. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . . . . . . .
  8. 8. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h . . b . So A = bh . . . . . .
  9. 9. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. . b . . . . . . .
  10. 10. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. h . . b . . . . . . .
  11. 11. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. h . . b . So 1 A= bh 2 . . . . . .
  12. 12. Easy Areas: Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: . . . . . . .
  13. 13. Hard Areas: Curved Regions . ??? . . . . . .
  14. 14. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  15. 15. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  16. 16. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  17. 17. . Archimedes found areas of a sequence of triangles inscribed in a parabola. A= . . . . . .
  18. 18. 1 . . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1 . . . . . .
  19. 19. 1 . .1 8 .1 8 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8 . . . . . .
  20. 20. 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 . . . . . .
  21. 21. 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 . . . . . .
  22. 22. We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 . . . . . .
  23. 23. 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 . . . . . .
  24. 24. 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 . . . . . .
  25. 25. 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 = → = 4 16 4 1− 1/4 3/4 3 as n → ∞. . . . . . .
  26. 26. Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective . . . . . .
  27. 27. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: . . 0 . 1 . . . . . . .
  28. 28. 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 . . . . . .
  29. 29. Cavalieri’s method Divide up the interval into . = x2 y pieces and measure the area of the inscribed rectangles: 1 L2 = 8 L3 = . . . . 0 . 1 2 1 . . . 3 3 . . . . . .
  30. 30. 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 . 1 2 1 . . . 3 3 . . . . . .
  31. 31. 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 . 1 2 3 1 . . . . 4 4 4 . . . . . .
  32. 32. 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 . 1 2 3 1 . . . . 4 4 4 . . . . . .
  33. 33. 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 . 1 2 3 4 1 . L5 = . . . . 5 5 5 5 . . . . . .
  34. 34. 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 0 . 1 2 3 4 1 . L5 = + + + = . . . . 125 125 125 125 125 5 5 5 5 . . . . . .
  35. 35. 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 0 . 1 . L5 = + + + = . 125 125 125 125 125 Ln =? . . . . . .
  36. 36. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n . . . . . .
  37. 37. 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 ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 . . . . . .
  38. 38. 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 ( ) 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 . . . . . .
  39. 39. 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 ( ) 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 . . . . . .
  40. 40. 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 ( ) 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 = → 6n3 3 as n → ∞. . . . . . .
  41. 41. 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 . . . . . .
  42. 42. 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 . . . . . .
  43. 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 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 3 3 3 1 + 2 + 3 + · · · + (n − 1 ) = n4 . . . . . .
  44. 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 3 3 3 1 + 2 + 3 + · · · + (n − 1 ) = n4 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = 2 n(n − 1) . . . . . .
  45. 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 3 1 + 2 + 3 + · · · + (n − 1 ) = n4 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = 2 n(n − 1) So n2 (n − 1)2 1 Ln = → 4n4 4 as n → ∞. . . . . . .
  46. 46. Cavalieri’s method with different heights 1 13 1 2 3 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 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 → ∞. . . . . . .
  47. 47. Cavalieri’s method with different heights 1 13 1 2 3 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 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. . . . . . .
  48. 48. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Properties of the integral . . . . . .
  49. 49. 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 x 1 = x 0 + ∆x = a + n b−a x 2 = x 1 + ∆x = a + 2 · n ······ b−a xi = a + i · n x x x .0 .1 .2 xx x . i . n−1. n ······ . . . . . . . . a . b . b−a xn = a + n · . . . =b . . .
  50. 50. 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 Mn = f ∆x + f ∆x + · · · + f ∆x 2 2 2 . . . . . .
  51. 51. 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 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 . . . . . .
  52. 52. 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. . . . . . .
  53. 53. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) . . . . . .
  54. 54. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the slope of a curve . . . . . .
  55. 55. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve . . . . . .
  56. 56. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve Only know the slope of lines . . . . . .
  57. 57. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve Only know the area of Only know the slope of polygons lines . . . . . .
  58. 58. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve Only know the area of Only know the slope of polygons lines Approximate curve with a line . . . . . .
  59. 59. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve Only know the area of Only know the slope of polygons lines Approximate region Approximate curve with with polygons a line . . . . . .
  60. 60. Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a Want the slope of a curved region curve Only know the area of Only know the slope of polygons lines Approximate region Approximate curve with with polygons a line Take limit over better Take limit over better and better and better approximations approximations . . . . . .
  61. 61. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Properties of the integral . . . . . .
  62. 62. Distances Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. . . . . . .
  63. 63. 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. . . . . . .
  64. 64. 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. . . . . . .
  65. 65. 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. . . . . . .
  66. 66. 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 . . . . . .
  67. 67. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Properties of the integral . . . . . .
  68. 68. The definite integral as a limit Definition If f is a function defined on [a, b], the definite integral of f from a to b is the number ∫ b ∑n f(x) dx = lim f(ci ) ∆x a ∆x→0 i =1 . . . . . .
  69. 69. Notation/Terminology ∫ b f(x) dx a . . . . . .
  70. 70. Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) . . . . . .
  71. 71. Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) f(x) — integrand . . . . . .
  72. 72. Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) . . . . . .
  73. 73. Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) . . . . . .
  74. 74. Notation/Terminology ∫ b f(x) dx a ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of computing an integral is called integration or quadrature . . . . . .
  75. 75. The limit can be simplified Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite ∫ b integral f(x) dx exists. a . . . . . .
  76. 76. The limit can be simplified Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite ∫ b integral f(x) dx exists. a Theorem If f is integrable on [a, b] then ∫ b n ∑ f(x) dx = lim f(xi )∆x, a n→∞ i=1 where b−a ∆x = and xi = a + i ∆x n . . . . . .
  77. 77. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Properties of the integral . . . . . .
  78. 78. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) a . . . . . .
  79. 79. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) a ∫ b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. a a a . . . . . .
  80. 80. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) a ∫ b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. a a a ∫ b ∫ b 3. cf(x) dx = c f(x) dx. a a . . . . . .
  81. 81. Properties of the integral Theorem (Additive Properties of the Integral) Let f and g be integrable functions on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) a ∫ b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. a a a ∫ b ∫ b 3. cf(x) dx = c f(x) dx. a a ∫ b ∫ b ∫ b 4. [f(x) − g(x)] dx = f(x) dx − g(x) dx. a a a . . . . . .
  82. 82. More Properties of the Integral Conventions: ∫ ∫ a b f(x) dx = − f(x) dx b a . . . . . .
  83. 83. More Properties of the Integral Conventions: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 a . . . . . .
  84. 84. More Properties of the Integral Conventions: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 a This allows us to have ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b . . . . . .
  85. 85. Example Suppose f and g are functions with ∫ 4 f(x) dx = 4 0 ∫ 5 f(x) dx = 7 0 ∫ 5 g(x) dx = 3. 0 Find ∫ 5 (a) [2f(x) − g(x)] dx 0 ∫ 5 (b) f(x) dx. 4 . . . . . .
  86. 86. Solution We have (a) ∫ 5 ∫ 5 ∫ 5 [2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx 0 0 0 = 2 · 7 − 3 = 11 . . . . . .
  87. 87. Solution We have (a) ∫ 5 ∫ 5 ∫ 5 [2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx 0 0 0 = 2 · 7 − 3 = 11 (b) ∫ 5 ∫ 5 ∫ 4 f(x) dx = f(x) dx − f(x) dx 4 0 0 =7−4=3 . . . . . .

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