Section	5.1
              Areas	and	Distances

                   V63.0121, Calculus	I



                      April	13, ...
Outline


  Archimedes


  Cavalieri


  Generalizing	Cavalieri’s	method


  Distances


  Other	applications




        ...
Meet	the	mathematician: Archimedes




     287	BC –	212	BC (after
     Euclid)
     Geometer
     Weapons	engineer




  ...
Meet	the	mathematician: Archimedes




     287	BC –	212	BC (after
     Euclid)
     Geometer
     Weapons	engineer




  ...
Meet	the	mathematician: Archimedes




     287	BC –	212	BC (after
     Euclid)
     Geometer
     Weapons	engineer




  ...
.

Archimedes	found	areas	of	a	sequence	of	triangles	inscribed	in	a
parabola.

               A=




                     ...
1
                               .



                                .

Archimedes	found	areas	of	a	sequence	of	triangles...
1
                               .
                   .1                      .1
                    8                    ...
1                                            1
             .64                                          .64
             ...
1                                      1
             .64                                    .64
                         ...
We	would	then	need	to	know	the	value	of	the	series
                      1   1         1
                            + ···...
We	would	then	need	to	know	the	value	of	the	series
                       1   1         1
                             + ·...
We	would	then	need	to	know	the	value	of	the	series
                       1   1         1
                             + ·...
We	would	then	need	to	know	the	value	of	the	series
                       1   1         1
                             + ·...
Outline


  Archimedes


  Cavalieri


  Generalizing	Cavalieri’s	method


  Distances


  Other	applications




        ...
Cavalieri



     Italian,
     1598–1647
     Revisited
     the	area
     problem
     with	a
     different
     perspe...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
Cavalieri’s	method

                        Divide	up	the	interval	into
                        pieces	and	measure	the	are...
What	is Ln ?
                                                                   1
   Divide	the	interval [0, 1] into n pie...
What	is Ln ?
                                                                1
   Divide	the	interval [0, 1] into n pieces...
What	is Ln ?
                                                                1
   Divide	the	interval [0, 1] into n pieces...
What	is Ln ?
                                                                1
   Divide	the	interval [0, 1] into n pieces...
What	is Ln ?
                                                                1
   Divide	the	interval [0, 1] into n pieces...
Cavalieri’s	method	for	different	functions
   Try	the	same	trick	with f(x) = x3 . We	have
                     ()         ...
Cavalieri’s	method	for	different	functions
   Try	the	same	trick	with f(x) = x3 . We	have
                     ()         ...
Cavalieri’s	method	for	different	functions
   Try	the	same	trick	with f(x) = x3 . We	have
                     ()         ...
Cavalieri’s	method	for	different	functions
   Try	the	same	trick	with f(x) = x3 . We	have
                     ()         ...
Cavalieri’s	method	for	different	functions
   Try	the	same	trick	with f(x) = x3 . We	have
                     ()         ...
Cavalieri’s	method	with	different	heights



                                 1 13 1 2 3                 1 n3
            ...
Cavalieri’s	method	with	different	heights



                                         1 13 1 2 3                 1 n3
    ...
Outline


  Archimedes


  Cavalieri


  Generalizing	Cavalieri’s	method


  Distances


  Other	applications




        ...
Cavalieri’s	method	in	general
   Let f be	a	positive	function	defined	on	the	interval [a, b]. We	want
   to	find	the	area	be...
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	on [a, b] or	has	finitely	many	jump
  discontinuities, the...
Outline


  Archimedes


  Cavalieri


  Generalizing	Cavalieri’s	method


  Distances


  Other	applications




        ...
Distances




  Just	like area = length × width, we	have

                       distance = rate × time.

  So	here	is	ano...
Example
A sailing	ship	is	cruising	back	and	forth	along	a	channel	(in	a
straight	line). At	noon	the	ship’s	position	and	ve...
Solution
We	estimate	that	the	speed	of	4	knots	(nautical	miles	per	hour)	is
maintained	from	12:00	until	12:30. So	over	thi...
Analysis




      This	method	of	measuring	position	by	recording	velocity	is
      known	as dead	reckoning.
      If	we	h...
Outline


  Archimedes


  Cavalieri


  Generalizing	Cavalieri’s	method


  Distances


  Other	applications




        ...
Other	uses	of	Riemann	sums




  Anything	with	a	product!
      Area, volume
      Anything	with	a	density: Population, ma...
Summary




     Riemann	sums	can	be	used	to	estimate	areas, distance, and
     other	quantities
     The	limit	of	Riemann...
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Lesson 23: Areas and Distances (Section 10 version)

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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.

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Lesson 23: Areas and Distances (Section 10 version)

  1. 1. Section 5.1 Areas and Distances V63.0121, Calculus I April 13, 2009 Announcements Moving to 624 today (no OH) Quiz 5 this week on §§4.1–4.4 Back HW returned in recitation this week . . . . . .
  2. 2. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  3. 3. Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  4. 4. Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  5. 5. Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  6. 6. . Archimedes found areas of a sequence of triangles inscribed in a parabola. A= . . . . . .
  7. 7. 1 . . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1 . . . . . .
  8. 8. 1 . .1 .1 8 8 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8 . . . . . .
  9. 9. 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 . . . . . .
  10. 10. 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 + ··· + n + ··· =1+ + 4 16 4 . . . . . .
  11. 11. We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 4 16 4 . . . . . .
  12. 12. We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 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 . . . . . .
  13. 13. We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 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 − (1/4)n+1 1 1 1 + ··· + n = 1+ + 1 − 1/4 4 16 4 . . . . . .
  14. 14. We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 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 − (1/4)n+1 1 1 1 1 4 + ··· + n = → 1+ + = 1− 4 16 4 3 1/4 3/4 as n → ∞. . . . . . .
  15. 15. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  16. 16. Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective . . . . . .
  17. 17. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: . . . . . . .
  18. 18. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 . . . . . . .
  19. 19. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 L3 = . . . . . . .
  20. 20. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 . . . . . . .
  21. 21. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 L4 = . . . . . . .
  22. 22. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 . . . . . . .
  23. 23. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 . L5 = . . . . . .
  24. 24. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y 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 . L5 = + + + = 125 125 125 25 125 . . . . . .
  25. 25. Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y 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 . L5 = + + + = 125 125 125 25 125 Ln =? . . . . . .
  26. 26. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n . . . . . .
  27. 27. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n . . . . . .
  28. 28. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = n3 n n3 n3 . . . . . .
  29. 29. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = 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 . . . . . .
  30. 30. What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = 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 → ∞. . . . . . .
  31. 31. Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n . . . . . .
  32. 32. Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n . . . . . .
  33. 33. Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 . . . . . .
  34. 34. Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = − 1) 2 n(n . . . . . .
  35. 35. Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = − 1) 2 n(n So n2 (n − 1)2 1 → Ln = 4n4 4 as n → ∞. . . . . . .
  36. 36. Cavalieri’s method with different heights 1 13 1 2 3 1 n3 · 3 + · 3 + ··· + · 3 Rn = nn nn nn 1 3 + 23 + 33 + · · · + n3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 → = 4 4n 4 . as n → ∞. . . . . . .
  37. 37. Cavalieri’s method with different heights 1 13 1 2 3 1 n3 · 3 + · 3 + ··· + · 3 Rn = nn nn nn 1 3 + 23 + 33 + · · · + n3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 → = 4 4n 4 . as n → ∞. So even though the rectangles overlap, we still get the same answer. . . . . . .
  38. 38. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  39. 39. Cavalieri’s method in general Let f be a positive function defined on the interval [a, b]. We want to find the area between x = a, x = b, y = 0, and y = f(x). For each positive integer n, divide up the interval into n pieces. b−a . For each i between 1 and n, let xi be the nth Then ∆x = n 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 xx x . i . n−1. n xxx .0 .1 .2 ······ . ....... b−a a . b . xn = a + n · =b . . . . . .
  40. 40. Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x 1 + x2 xn−1 + xn ∆x + · · · + f Mn = f ∆x + f ∆x 2 2 2 . . . . . .
  41. 41. Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x 1 + x2 xn−1 + xn ∆x + · · · + f Mn = 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 . . . . . .
  42. 42. Theorem of the Day Theorem If f is a continuous function on [a, b] or has finitely many jump discontinuities, then lim Sn = lim {f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x} n→∞ n→∞ exists and is the same value no matter what choice of ci we made. . . . . . .
  43. 43. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  44. 44. Distances Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. . . . . . .
  45. 45. 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. . . . . . .
  46. 46. 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. . . . . . .
  47. 47. Analysis This method of measuring position by recording velocity is known as dead reckoning. If we had velocity estimates at finer intervals, we’d get better estimates. If we had velocity at every instant, a limit would tell us our exact position relative to the last time we measured it. . . . . . .
  48. 48. Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  49. 49. 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 . . . . . .
  50. 50. Summary Riemann sums can be used to estimate areas, distance, and other quantities The limit of Riemann sums can get the exact value Coming up: giving this limit a name and working with it. . . . . . .
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