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# Lesson 22: Areas and Distances

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We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.

We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.

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• 1. Section 5.1 Areas and Distances V63.0121.006/016, Calculus I New York University April 13, 2010 Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 12:00noon
• 2. Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 12:00noon V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30
• 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 3 / 30
• 4. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 4 / 30
• 5. Easy Areas: Rectangle Deﬁnition The area of a rectangle with dimensions and w is the product A = w . w It may seem strange that this is a deﬁnition and not a theorem but we have to start somewhere. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 5 / 30
• 6. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. b V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
• 7. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h b V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
• 8. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
• 9. Easy Areas: Parallelogram By cutting and pasting, a parallelogram can be made into a rectangle. h b V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
• 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30
• 11. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. b V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
• 12. Easy Areas: Triangle By copying and pasting, a triangle can be made into a parallelogram. h b V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
• 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30
• 14. Easy Areas: Other Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 8 / 30
• 15. Hard Areas: Curved Regions ??? V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 9 / 30
• 16. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
• 17. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
• 18. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30
• 19. Archimedes found areas of a sequence of triangles inscribed in a parabola. A= V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
• 20. 1 Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
• 21. 1 1 1 8 8 Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
• 22. 1 1 64 64 1 1 1 8 8 1 1 64 64 Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 1 A=1+2· +4· + ··· 8 64 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
• 23. 1 1 64 64 1 1 1 8 8 1 1 64 64 Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 1 A=1+2· +4· + ··· 8 64 1 1 1 =1+ + + ··· + n + ··· 4 16 4 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30
• 24. We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
• 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 + · · · + r n ) = 1 − r n+1 So 1 − r n+1 1 + r + · · · + rn = 1−r V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
• 26. 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 + · · · + r n ) = 1 − r n+1 So 1 − r n+1 1 + r + · · · + rn = 1−r Therefore 1 1 1 1 − (1/4)n+1 1+ + + ··· + n = 4 16 4 1 − 1/4 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
• 27. 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 + · · · + r n ) = 1 − r n+1 So 1 − r n+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 → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30
• 28. Cavalieri Italian, 1598–1647 Revisited the area problem with a diﬀerent perspective V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 13 / 30
• 29. Cavalieri’s method Divide up the interval into pieces y = x2 and measure the area of the inscribed rectangles: 0 1 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 30. Cavalieri’s method Divide up the interval into pieces y = x2 and measure the area of the inscribed rectangles: 1 L2 = 8 0 1 1 2 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 31. Cavalieri’s method Divide up the interval into pieces y = x2 and measure the area of the inscribed rectangles: 1 L2 = 8 L3 = 0 1 2 1 3 3 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 32. Cavalieri’s method Divide up the interval into pieces y = x2 and measure the area of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 0 1 2 1 3 3 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 33. Cavalieri’s method Divide up the interval into pieces y = x2 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 34. Cavalieri’s method Divide up the interval into pieces y = x2 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 35. Cavalieri’s method Divide up the interval into pieces y = x2 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 36. Cavalieri’s method Divide up the interval into pieces y = x2 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 37. Cavalieri’s method Divide up the interval into pieces y = x2 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 =? V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30
• 38. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
• 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 2 1 i −1 (i − 1)2 · = . n n n3 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
• 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 2 1 i −1 (i − 1)2 · = . n n n3 So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = n3 n n3 n3 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
• 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 2 1 i −1 (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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
• 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 2 1 i −1 (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 → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30
• 43. Cavalieri’s method for diﬀerent functions Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + ·f n n n n n n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
• 44. Cavalieri’s method for diﬀerent functions Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 2 3 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
• 45. Cavalieri’s method for diﬀerent functions Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 2 3 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1+2 3 + 33 + · · · + (n − 1)3 = n4 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
• 46. Cavalieri’s method for diﬀerent functions Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 2 3 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1+2 3 + 33 + · · · + (n − 1)3 = n4 The formula out of the hat is 2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
• 47. Cavalieri’s method for diﬀerent functions Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 2 3 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1+2 3 + 33 + · · · + (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 → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30
• 48. Cavalieri’s method with diﬀerent heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 13 + 23 + 33 + · · · + n3 = n4 1 1 2 = 4 2 n(n + 1) n n2 (n + 1)2 1 = → 4n4 4 as n → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
• 49. Cavalieri’s method with diﬀerent heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 13 + 23 + 33 + · · · + n3 = 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30
• 50. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 18 / 30
• 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
• 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
• 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
• 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
• 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30
• 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
• 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30
• 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 b x1 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x1 b x2 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x1 x2 b x3 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x1 x2 x3 b x4 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 b5 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 b6 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 a x x x x x x x matter what choice of ci we 1 2 3 4 5 6 b7 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 matter what choice of ci we 1 2 3 4 5 6 7 b8 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 matter what choice of ci we 1 2 3 4 5 6 7 8b9 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 a x 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x x xb matter what choice of ci we 1 2 3 4 5 6 7 8 9 1011 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 101112 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 ax x x x x x x x xx x x x x xb matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14 11 13 15 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 axxxxxxxxxxxxxxxxb matter what choice of ci we 1 2 3 4 5 6 7 8 910 12 14 16 11 13 15 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 xxxxxxxxxxxxxxxxb x1 2 3 4 5 6 7 8 910 12 14 16 matter what choice of ci we 11 13 15 17 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x1 2 3 4 5 6 7 8x10 12 14 16 18 matter what choice of ci we 9 11 13 15 17 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 x12345678910 12 14 16 18 x 11 13 15 17 19 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 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 a xxxxxxxx xxxxxxxxxxb x123456789101214161820 x 1113151719 matter what choice of ci we made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30
• 78. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 79. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the slope of a curve V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 80. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 81. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the slope of lines V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 82. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope of lines polygons V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 83. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope of lines polygons Approximate curve with a line V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 84. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope of lines polygons Approximate curve with a Approximate region with line polygons V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 85. Analogies The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30
• 86. Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 23 / 30
• 87. Distances Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 24 / 30
• 88. Application: Dead Reckoning V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 25 / 30
• 89. 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 26 / 30
• 90. 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 27 / 30
• 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 28 / 30
• 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 29 / 30
• 93. 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 30 / 30