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# Lesson 24: Areas, Distances, the Integral (Section 021 slides

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We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.

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### Lesson 24: Areas, Distances, the Integral (Section 021 slides

1. 1. . Sections 5.1–5.2 Areas and Distances The Definite Integral V63.0121.021, Calculus I New York University December 2, 2010 Announcements Final December 20, 12:00–1:50pm . . . . . .
2. 2. Announcements Final December 20, 12:00–1:50pm cumulative location TBD old exams on common website . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56
3. 3. Objectives from Section 5.1 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. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 3 / 56
4. 4. Objectives from Section 5.2 Compute the definite integral using a limit of Riemann sums Estimate the definite integral using a Riemann sum (e.g., Midpoint Rule) Reason with the definite integral using its elementary properties. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 4 / 56
5. 5. OutlineArea through the Centuries Euclid Archimedes CavalieriGeneralizing Cavalieri’s method AnalogiesDistances Other applicationsThe definite integral as a limitEstimating the Definite IntegralProperties of the integralComparison Properties of the Integral . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 5 / 56
6. 6. Easy Areas: RectangleDefinitionThe area of a rectangle with dimensions ℓ and w is the product A = ℓw. w . ℓIt may seem strange that this is a definition and not a theorem but wehave to start somewhere. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 6 / 56
7. 7. Easy Areas: ParallelogramBy cutting and pasting, a parallelogram can be made into a rectangle. . b . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
8. 8. Easy Areas: ParallelogramBy cutting and pasting, a parallelogram can be made into a rectangle. h . b . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
9. 9. Easy Areas: ParallelogramBy cutting and pasting, a parallelogram can be made into a rectangle. h . . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
10. 10. Easy Areas: ParallelogramBy cutting and pasting, a parallelogram can be made into a rectangle. h . b . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
11. 11. Easy Areas: ParallelogramBy cutting and pasting, a parallelogram can be made into a rectangle. h . bSoFactThe area of a parallelogram of base width b and height h is A = bh . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
12. 12. Easy Areas: TriangleBy copying and pasting, a triangle can be made into a parallelogram. . b . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
13. 13. Easy Areas: TriangleBy copying and pasting, a triangle can be made into a parallelogram. h . b . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
14. 14. Easy Areas: TriangleBy copying and pasting, a triangle can be made into a parallelogram. h . bSoFactThe area of a triangle of base width b and height h is 1 A= bh 2 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
15. 15. Easy Areas: Other PolygonsAny polygon can be triangulated, so its area can be found by summingthe areas of the triangles: . . . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 9 / 56
16. 16. Hard Areas: Curved Regions .??? . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 10 / 56
17. 17. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
18. 18. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
19. 19. Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
20. 20. Archimedes and the Parabola .Archimedes found areas of a sequence of triangles inscribed in aparabola. A= . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
21. 21. Archimedes and the Parabola 1 .Archimedes found areas of a sequence of triangles inscribed in aparabola. A=1 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
22. 22. Archimedes and the Parabola 1 1 1 8 8 .Archimedes found areas of a sequence of triangles inscribed in aparabola. 1 A=1+2· 8 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
23. 23. Archimedes and the Parabola 1 1 64 64 1 1 1 8 8 1 1 64 64 .Archimedes found areas of a sequence of triangles inscribed in aparabola. 1 1 A=1+2· +4· + ··· 8 64 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
24. 24. Archimedes and the Parabola 1 1 64 64 1 1 1 8 8 1 1 64 64 .Archimedes found areas of a sequence of triangles inscribed in aparabola. 1 1 A=1+2· +4· + ··· 8 64 1 1 1 =1+ + + ··· + n + ··· 4 16 4 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
25. 25. Summing the seriesWe would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
26. 26. Summing the seriesWe would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4FactFor any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1So 1 − rn+1 1 + r + · · · + rn = 1−r . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
27. 27. Summing the seriesWe would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4FactFor any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1So 1 − rn+1 1 + r + · · · + rn = 1−rTherefore 1 1 1 1 − (1/4)n+1 1+ + + ··· + n = 4 16 4 1 − 1/4 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
28. 28. Summing the seriesWe would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4FactFor any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1So 1 − rn+1 1 + r + · · · + rn = 1−rTherefore 1 1 1 1 − (1/4)n+1 1 4 1+ + + ··· + n = → 3 = as n → ∞. 4 16 4 1− 1/4 /4 3 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
29. 29. Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 14 / 56
30. 30. Cavalieris method Divide up the interval into 2 y=x pieces and measure the area of the inscribed rectangles: . 0 1 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
31. 31. 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 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
32. 32. 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 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
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 . 0 1 2 1 3 3 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
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 L4 = . 0 1 2 3 1 4 4 4 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
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 1 4 4 4 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
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 0 1 2 3 4 1 L5 = 5 5 5 5 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
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 2 3 4 1 L5 = + + + = 125 125 125 125 125 5 5 5 5 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
38. 38. 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 =? . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
39. 39. What is Ln ? 1Divide the interval [0, 1] into n pieces. Then each has width . n . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
40. 40. What is Ln ? 1Divide the interval [0, 1] into n pieces. Then each has width . The nrectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
41. 41. What is Ln ? 1Divide the interval [0, 1] into n pieces. Then each has width . The nrectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = n3 n n3 n3 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
42. 42. What is Ln ? 1Divide the interval [0, 1] into n pieces. Then each has width . The nrectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = n3 n n3 n3The Arabs knew that n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6So n(n − 1)(2n − 1) Ln = 6n3 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
43. 43. What is Ln ? 1Divide the interval [0, 1] into n pieces. Then each has width . The nrectangle over the ith interval and under the parabola has area ( ) 1 i − 1 2 (i − 1)2 · = . n n n3So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + 3 + ··· + = n3 n n3 n3The Arabs knew that n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6So n(n − 1)(2n − 1) 1 Ln = 3 → 6n 3as n → ∞. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
44. 44. Cavalieris method for different functionsTry 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 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
45. 45. Cavalieris method for different functionsTry 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 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
46. 46. Cavalieris method for different functionsTry 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 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
47. 47. Cavalieris method for different functionsTry 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 = n4The formula out of the hat is [ ]2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
48. 48. Cavalieris method for different functionsTry 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 = n4The 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 4as n → ∞. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
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 → ∞. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56
50. 50. 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. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56
51. 51. OutlineArea through the Centuries Euclid Archimedes CavalieriGeneralizing Cavalieri’s method AnalogiesDistances Other applicationsThe definite integral as a limitEstimating the Definite IntegralProperties of the integralComparison Properties of the Integral . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 19 / 56
52. 52. Cavalieris method in generalLet f be a positive function defined on the interval [a, b]. We want tofind 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 ith step between na 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 b−a . x xn = a + n · =b x0 x1 . . . xi . . .xn−1xn n . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 20 / 56
53. 53. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval.left endpoints… ∑ n Ln = f(xi−1 )∆x i=1 . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
54. 54. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. right endpoints… ∑ n Rn = f(xi )∆x i=1 . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
55. 55. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. midpoints… ∑ ( xi−1 + xi ) n Mn = f ∆x 2 i=1 . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
56. 56. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. the minimum value on theinterval… . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
57. 57. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. the maximum value on theinterval… . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
58. 58. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. …even random points! . x . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
59. 59. Forming Riemann sumsWe have many choices of representative points to approximate thearea in each subinterval. …even random points! . xIn general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form theRiemann sum ∑ n Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x = f(ci )∆x i=1 . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
60. 60. Theorem of the DayTheoremIf f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
61. 61. Theorem of the DayTheoremIf f is a continuous function on [a, b]or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
62. 62. Theorem of the DayTheoremIf f is a continuous function on [a, b] L1 = 3.0or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
63. 63. Theorem of the DayTheoremIf f is a continuous function on [a, b] L2 = 5.25or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
64. 64. Theorem of the DayTheoremIf f is a continuous function on [a, b] L3 = 6.0or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
65. 65. Theorem of the DayTheoremIf f is a continuous function on [a, b] L4 = 6.375or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
66. 66. Theorem of the DayTheoremIf f is a continuous function on [a, b] L5 = 6.59988or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
67. 67. Theorem of the DayTheoremIf f is a continuous function on [a, b] L6 = 6.75or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
68. 68. Theorem of the DayTheoremIf f is a continuous function on [a, b] L7 = 6.85692or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
69. 69. Theorem of the DayTheoremIf f is a continuous function on [a, b] L8 = 6.9375or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
70. 70. Theorem of the DayTheoremIf f is a continuous function on [a, b] L9 = 6.99985or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
71. 71. Theorem of the DayTheoremIf f is a continuous function on [a, b] L10 = 7.04958or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
72. 72. Theorem of the DayTheoremIf f is a continuous function on [a, b] L11 = 7.09064or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
73. 73. Theorem of the DayTheoremIf f is a continuous function on [a, b] L12 = 7.125or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
74. 74. Theorem of the DayTheoremIf f is a continuous function on [a, b] L13 = 7.15332or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
75. 75. Theorem of the DayTheoremIf f is a continuous function on [a, b] L14 = 7.17819or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
76. 76. Theorem of the DayTheoremIf f is a continuous function on [a, b] L15 = 7.19977or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
77. 77. Theorem of the DayTheoremIf f is a continuous function on [a, b] L16 = 7.21875or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
78. 78. Theorem of the DayTheoremIf f is a continuous function on [a, b] L17 = 7.23508or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
79. 79. Theorem of the DayTheoremIf f is a continuous function on [a, b] L18 = 7.24927or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
80. 80. Theorem of the DayTheoremIf f is a continuous function on [a, b] L19 = 7.26228or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
81. 81. Theorem of the DayTheoremIf f is a continuous function on [a, b] L20 = 7.27443or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
82. 82. Theorem of the DayTheoremIf f is a continuous function on [a, b] L21 = 7.28532or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
83. 83. Theorem of the DayTheoremIf f is a continuous function on [a, b] L22 = 7.29448or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
84. 84. Theorem of the DayTheoremIf f is a continuous function on [a, b] L23 = 7.30406or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
85. 85. Theorem of the DayTheoremIf f is a continuous function on [a, b] L24 = 7.3125or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
86. 86. Theorem of the DayTheoremIf f is a continuous function on [a, b] L25 = 7.31944or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
87. 87. Theorem of the DayTheoremIf f is a continuous function on [a, b] L26 = 7.32559or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
88. 88. Theorem of the DayTheoremIf f is a continuous function on [a, b] L27 = 7.33199or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
89. 89. Theorem of the DayTheoremIf f is a continuous function on [a, b] L28 = 7.33798or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
90. 90. Theorem of the DayTheoremIf f is a continuous function on [a, b] L29 = 7.34372or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
91. 91. Theorem of the DayTheoremIf f is a continuous function on [a, b] L30 = 7.34882or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. left endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
92. 92. Theorem of the DayTheoremIf f is a continuous function on [a, b] R1 = 12.0or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
93. 93. Theorem of the DayTheoremIf f is a continuous function on [a, b] R2 = 9.75or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
94. 94. Theorem of the DayTheoremIf f is a continuous function on [a, b] R3 = 9.0or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
95. 95. Theorem of the DayTheoremIf f is a continuous function on [a, b] R4 = 8.625or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
96. 96. Theorem of the DayTheoremIf f is a continuous function on [a, b] R5 = 8.39969or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
97. 97. Theorem of the DayTheoremIf f is a continuous function on [a, b] R6 = 8.25or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
98. 98. Theorem of the DayTheoremIf f is a continuous function on [a, b] R7 = 8.14236or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
99. 99. Theorem of the DayTheoremIf f is a continuous function on [a, b] R8 = 8.0625or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
100. 100. Theorem of the DayTheoremIf f is a continuous function on [a, b] R9 = 7.99974or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
101. 101. Theorem of the DayTheoremIf f is a continuous function on [a, b] R10 = 7.94933or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
102. 102. Theorem of the DayTheoremIf f is a continuous function on [a, b] R11 = 7.90868or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
103. 103. Theorem of the DayTheoremIf f is a continuous function on [a, b] R12 = 7.875or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
104. 104. Theorem of the DayTheoremIf f is a continuous function on [a, b] R13 = 7.84541or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
105. 105. Theorem of the DayTheoremIf f is a continuous function on [a, b] R14 = 7.8209or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
106. 106. Theorem of the DayTheoremIf f is a continuous function on [a, b] R15 = 7.7997or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
107. 107. Theorem of the DayTheoremIf f is a continuous function on [a, b] R16 = 7.78125or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
108. 108. Theorem of the DayTheoremIf f is a continuous function on [a, b] R17 = 7.76443or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
109. 109. Theorem of the DayTheoremIf f is a continuous function on [a, b] R18 = 7.74907or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
110. 110. Theorem of the DayTheoremIf f is a continuous function on [a, b] R19 = 7.73572or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
111. 111. Theorem of the DayTheoremIf f is a continuous function on [a, b] R20 = 7.7243or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
112. 112. Theorem of the DayTheoremIf f is a continuous function on [a, b] R21 = 7.7138or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
113. 113. Theorem of the DayTheoremIf f is a continuous function on [a, b] R22 = 7.70335or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
114. 114. Theorem of the DayTheoremIf f is a continuous function on [a, b] R23 = 7.69531or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
115. 115. Theorem of the DayTheoremIf f is a continuous function on [a, b] R24 = 7.6875or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
116. 116. Theorem of the DayTheoremIf f is a continuous function on [a, b] R25 = 7.67934or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
117. 117. Theorem of the DayTheoremIf f is a continuous function on [a, b] R26 = 7.6715or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
118. 118. Theorem of the DayTheoremIf f is a continuous function on [a, b] R27 = 7.66508or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
119. 119. Theorem of the DayTheoremIf f is a continuous function on [a, b] R28 = 7.6592or has finitely many jumpdiscontinuities, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1exists and is the same value no . xmatter what choice of ci we make. right endpoints . . . . . . V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56