Section 5.1           Areas and Distances                V63.0121.002.2010Su, Calculus I                        New York U...
Announcements           Quiz Thursday on 4.1–4.4                                                                          ...
Objectives           Compute the area of a           region by approximating it           with rectangles and letting     ...
Outline Area through the Centuries    Euclid    Archimedes    Cavalieri Generalizing Cavalieri’s method   Analogies Distan...
Easy Areas: Rectangle Definition The area of a rectangle with dimensions ℓ and w is the product A = ℓw.                   ...
Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle.                           ...
Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle.                           ...
Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle.                           ...
Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle.                           ...
Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle.                           ...
Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram.                                 ...
Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram.                                 ...
Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram.                                 ...
Easy Areas: Other Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles...
Hard Areas: Curved Regions                             . ???                                                              ...
Meet the mathematician: Archimedes          Greek (Syracuse), 287 BC          – 212 BC (after Euclid)          Geometer   ...
Meet the mathematician: Archimedes          Greek (Syracuse), 287 BC          – 212 BC (after Euclid)          Geometer   ...
Meet the mathematician: Archimedes          Greek (Syracuse), 287 BC          – 212 BC (after Euclid)          Geometer   ...
Archimedes and the Parabola                                                       . Archimedes found areas of a sequence o...
Archimedes and the Parabola                                                      1                                        ...
Archimedes and the Parabola                                                           1                                   ...
Archimedes and the Parabola                              1                                                           1    ...
Archimedes and the Parabola                              1                                                           1    ...
Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series                     ...
Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series                     ...
Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series                     ...
Summing the series [label=archimedes-parabola-sum] We would then need to know the value of the series                     ...
Cavalieri          Italian,          1598–1647          Revisited the          area          problem with          a diffe...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
Cavalieris method                                                              Divide up the interval into                ...
What is Ln ?                                                                                             1 Divide the inte...
What is Ln ?                                                              1 Divide the interval [0, 1] into n pieces. Then...
What is Ln ?                                                              1 Divide the interval [0, 1] into n pieces. Then...
What is Ln ?                                                              1 Divide the interval [0, 1] into n pieces. Then...
What is Ln ?                                                              1 Divide the interval [0, 1] into n pieces. Then...
Cavalieris method for different functions Try the same trick with f(x) = x3 . We have                      ( )            ...
Cavalieris method for different functions Try the same trick with f(x) = x3 . We have                      ( )            ...
Cavalieris method for different functions Try the same trick with f(x) = x3 . We have                      ( )            ...
Cavalieris method for different functions Try the same trick with f(x) = x3 . We have                      ( )            ...
Cavalieris method for different functions Try the same trick with f(x) = x3 . We have                      ( )            ...
Cavalieris method with different heights                                                              1 13 1 23           ...
Cavalieris method with different heights                                                              1 13 1 23           ...
Outline Area through the Centuries    Euclid    Archimedes    Cavalieri Generalizing Cavalieri’s method   Analogies Distan...
Cavalieris method in general . Let f be a positive function defined on the interval [a, b]. We want to find the area betwe...
Cavalieris method in general . Let f be a positive function defined on the interval [a, b]. We want to find the area betwe...
Cavalieris method in general . Let f be a positive function defined on the interval [a, b]. We want to find the area betwe...
Cavalieris method in general . Let f be a positive function defined on the interval [a, b]. We want to find the area betwe...
Cavalieris method in general . Let f be a positive function defined on the interval [a, b]. We want to find the area betwe...
Forming Riemann sums We have many choices of how to approximate the area:       Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 ...
Forming Riemann sums We have many choices of how to approximate the area:       Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 ...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Theorem of the Day   Theorem   If f is a continuous function or   has finitely many jump   discontinuities on [a, b], then...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)          Want the...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Analogies  The Tangent Problem                                       The Area Problem (Ch. 5)  (Ch. 2–4)                  ...
Outline Area through the Centuries    Euclid    Archimedes    Cavalieri Generalizing Cavalieri’s method   Analogies Distan...
Distances Just like area = length × width, we have                                        distance = rate × time. So here ...
Application: Dead Reckoning                                                                          .   .   .      .     ...
Computing position by Dead Reckoning Example A sailing ship is cruising back and forth along a channel (in a straight line...
Solution Solution We estimate that the speed of 4 knots (nautical miles per hour) is maintained from 12:00 until 12:30. So...
Analysis         This method of measuring position by recording velocity was         necessary until global-positioning sa...
Other uses of Riemann sums Anything with a product!         Area, volume         Anything with a density: Population, mass...
Surplus by picture                                           c                                           . onsumer surplus...
Summary         We can compute the area of a curved region with a limit of         Riemann sums         We can compute the...
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Lesson 24: Area and Distances

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Lesson 24: Area and Distances

  1. 1. Section 5.1 Areas and Distances V63.0121.002.2010Su, Calculus I New York University June 16, 2010Announcements Quiz Thursday on 4.1–4.4 . . . . . .
  2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 15. Hard Areas: Curved Regions . ??? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 9 / 31
  16. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 29. Cavalieris 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. 30. Cavalieris 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. 31. Cavalieris 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. 32. Cavalieris 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. 33. Cavalieris 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. 34. Cavalieris 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. 35. Cavalieris 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. 36. Cavalieris 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. 37. Cavalieris 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. 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. 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. 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. 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. 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. 43. Cavalieris 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. 44. Cavalieris 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. 45. Cavalieris 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. 46. Cavalieris 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. 47. Cavalieris 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. 48. Cavalieris 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. 49. Cavalieris 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. 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. 51. Cavalieris 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. 52. Cavalieris 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. 53. Cavalieris 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. 54. Cavalieris 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. 55. Cavalieris 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 88. Application: Dead Reckoning . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.1 Areas and Distances June 16, 2010 25 / 31
  89. 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. 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. 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. 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. 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. 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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