Lesson 24: Areas and Distances, The Definite Integral (handout)
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Lesson 24: Areas and Distances, The Definite Integral (handout)

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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.

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 and Distances, The Definite Integral (handout) Lesson 24: Areas and Distances, The Definite Integral (handout) Document Transcript

  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Sec on 5.1–5.2 Areas and Distances, The Definite Integral V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 25, 2011 . . Notes Announcements Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm cumula ve loca on TBD old exams on common website . . Notes Objectives from Section 5.1 Compute the area of a region by approxima ng it with rectangles and le ng the size of the rectangles tend to zero. Compute the total distance traveled by a par cle by approxima ng it as distance = (rate)( me) and le ng the me intervals over which one approximates tend to zero. . . . 1.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Objectives from Section 5.2 Compute the definite integral using a limit of Riemann sums Es mate the definite integral using a Riemann sum (e.g., Midpoint Rule) Reason with the definite integral using its elementary proper es. . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . Notes Easy Areas: Rectangle Defini on 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 defini on and not a theorem but . we have to start somewhere. . . 2.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. h . So b b Fact The area of a parallelogram of base width b and height h is A = bh . . Notes Easy Areas: Triangle By copying and pas ng, 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 . Notes Easy Areas: Other Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: . . . . . 3.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Hard Areas: Curved Regions . ??? . . . Notes Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (a er Euclid) Geometer Weapons engineer . . Notes 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 a parabola. 1 1 1 1 1 A=1+2· +4· + ··· = 1 + + + ··· + n + ··· 8 64 4 16 4 . . . 4.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Summing a geometric series Fact For any number r and any posi ve integer n, (1 − r)(1 + r + r2 + · · · + rn ) = 1 − rn+1 . Proof. (1 − r)(1 + r + r2 + · · · + rn ) = (1 + r + r2 + · · · + rn ) − r(1 + r + r2 + · · · + rn ) = (1 + r + r2 + · · · + rn ) − (r + r2 + r3 · · · + rn + rn+1 ) = 1 − rn+1 . . Notes Summing a geometric series Fact For any number r and any posi ve integer n, (1 − r)(1 + r + r2 + · · · + rn ) = 1 − rn+1 . Corollary 1 − rn+1 1 + r + · · · + rn = 1−r . . Notes Summing the series We need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 Using the corollary, 1 1 1 1 − (1/4)n+1 1 4 1+ + + ··· + n = → 3 = as n → ∞. 4 16 4 1 − 1/4 /4 3 . . . 5.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspec ve . . Notes Cavalieri’s method Divide up the interval into pieces and y = x2 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 =? . . Notes The Square Pyramidial Numbers Fact Let n be a posi ve integer. Then n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6 This formula was known to the Arabs and discussed by Fibonacci in his book Liber Abaci. . . . 6.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes 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 + 3 + ··· + 3 = n n n n3 So n(n − 1)(2n − 1) 1 Ln = → 6n3 3 . as n → ∞. . Notes Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have ( ) ( ) ( ) 1 1 1 2 1 n−1 Ln = · f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1 + 23 + 33 + · · · + (n − 1)3 = n4 n2 (n − 1)2 1 = → 4n4 4 as n → ∞. . . Notes Nicomachus’s Theorem Fact (Nicomachus 1st c. CE, Aryabhata 5th c., Al-Karaji 11th c.) 1 + 23 + 33 + · · · + (n − 1)3 = [1 + 2 + · · · + (n − 1)]2 [1 ]2 = 2 n(n − 1) . . . 7.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes 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 s ll get the same answer. . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . Notes Cavalieri’s method in general Problem Let f be a posi ve func on defined on the interval [a, b]. Find the area between x = a, x = b, y = 0, and y = f(x). . . x x x0 x1. . . xi . . xn−1 n . . . 8.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Cavalieri’s method in general For each posi ve 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 n between a and b. 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 x x0 x1. . . xi . . xn−1 n xn = a + n · =b n . . Notes Forming Riemann Sums Choose ci to be a point in the ith interval [xi−1 , xi ]. Form the Riemann sum ∑ n Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x = f(ci )∆x i=1 Thus we approximate area under a curve by a sum of areas of rectangles. . . Notes Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. …even random points! . x . . . 9.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M15 = 7.49968 discon nui es, then { n } ∑ lim Sn = lim f(ci )∆x n→∞ n→∞ i=1 exists and is the same value no . x ma er what choice of ci we make. . . Notes 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 be er and Take limit over be er and be er approxima ons be er approxima ons . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . . 10.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Distances Just like area = length × width, we have distance = rate × me. So here is another use for Riemann sums. . . Notes Application: Dead Reckoning . . Notes 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 posi on and velocity are recorded, but shortly therea er a storm blows in and posi on is impossible to measure. The velocity con nues to be recorded at thirty-minute intervals. . . . 11.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Computing position by Dead Reckoning Example Time 12:00 12:30 1:00 1:30 2:00 Speed (knots) 4 8 12 6 4 Direc on E E E E W Time 2:30 3:00 3:30 4:00 Speed 3 3 5 9 Direc on W E E E Es mate the ship’s posi on at 4:00pm. . . Notes Solution Solu on We es mate that the speed of 4 knots (nau cal miles per hour) is maintained from 12:00 un l 12:30. So over this me interval the ship travels ( )( ) 4 nmi 1 hr = 2 nmi hr 2 We can con nue for each addi onal 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 posi on. . Notes Analysis This method of measuring posi on by recording velocity was necessary un l global-posi oning satellite technology became widespread If we had velocity es mates at finer intervals, we’d get be er es mates. If we had velocity at every instant, a limit would tell us our exact posi on rela ve to the last me we measured it. . . . 12.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Other uses of Riemann sums Anything with a product! Area, volume Anything with a density: Popula on, mass Anything with a “speed:” distance, throughput, power Consumer surplus Expected value of a random variable . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . Notes The definite integral as a limit Defini on If f is a func on defined on [a, b], the definite integral of f from a to b is the number ∫ b ∑n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1 . . . 13.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Notation/Terminology ∫ b ∑ n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1 ∫ — integral sign (swoopy S) f(x) — integrand a and b — limits of integra on (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of compu ng an integral is called integra on or quadrature . . Notes The limit can be simplified Theorem If f is con nuous on [a, b] or if f has only finitely many jump discon nui es, then f is integrable on [a, b]; that is, the definite ∫ b integral f(x) dx exists. a So we can find the integral by compu ng the limit of any sequence of Riemann sums that we like, . . Example ∫ 3 Notes Find x dx 0 Solu on 3 3i For any n we have ∆x = and for each i between 0 and n, xi = . n n For each i, take xi to represent the func on on the ith interval. So ∫ 3 ∑n ∑ ( 3i ) ( 3 ) n x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n 9 ∑ n 9 n(n + 1) 9 = lim i = lim 2 · = ·1 n→∞ n2 n→∞ n 2 2 i=1 . . . 14.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Example ∫ 3 Notes Find x2 dx 0 Solu on . . Example ∫ 3 Notes Find x3 dx 0 Solu on . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . . 15.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Estimating the Definite Integral Example ∫ 1 4 Es mate dx using M4 . 0 1 + x2 Solu on 1 1 3 We have x0 = 0, x1 = , x2 = , x3 = , x4 = 1. 4 2 4 1 3 5 7 So c1 = , c2 = , c3 = , c4 = . 8 8 8 8 . . Notes Estimating the Definite Integral Example ∫ 1 4 Es mate dx using M4 . 0 1 + x2 Solu on ( ) 1 4 4 4 4 M4 = + + + 4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2 ( ) 1 4 4 4 4 = + + + 4 65/64 73/64 89/64 113/64 64 64 64 64 = + + + ≈ 3.1468 65 73 89 113 . . Notes Estimating the Definite Integral Example ∫ 1 4 Es mate dx using L4 and R4 0 1 + x2 Answer . . . 16.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Estimating the Definite Integral Example ∫ 1 4 Es mate dx using L4 and R4 0 1 + x2 Answer . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . Notes Properties of the integral Theorem (Addi ve Proper es of the Integral) Let f and g be integrable func ons on [a, b] and c a constant. Then ∫ b 1. c dx = c(b − a) ∫a b ∫ b ∫ b 2. [f(x) + g(x)] dx = f(x) dx + g(x) dx. ∫a b ∫ b a a 3. cf(x) dx = c f(x) dx. ∫a b a ∫ b ∫ b 4. [f(x) − g(x)] dx = f(x) dx − g(x) dx. a a a . . . 17.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Proofs Proofs. When integra ng a constant func on c, each Riemann sum equals c(b − a). A Riemann sum for f + g equals a Riemann sum for f plus a Riemann sum for g. Using the sum rule for limits, the integral of a sum is the sum of the integrals. Di o for constant mul ples Di o for differences . . Example ∫ 3 Notes ( 3 ) Find x − 4.5x2 + 5.5x + 1 dx 0 Solu on . . Notes More Properties of the Integral Conven ons: ∫ ∫ a b f(x) dx = − f(x) dx b a ∫ a f(x) dx = 0 a This allows us to have Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b . . . 18.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Illustrating Property 5 Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ b ∫ c f(x) dx f(x) dx a b . a c x b . . Notes Illustrating Property 5 Theorem ∫ c ∫ b ∫ c 5. f(x) dx = f(x) dx + f(x) dx for all a, b, and c. a a b y ∫ ∫ c c f(x) dx f(x) dx = b∫ a b − f(x) dx . c a c x b . . Notes Using the Properties Example Suppose f and g are func ons with ∫ 4 Find∫ f(x) dx = 4 5 (a) [2f(x) − g(x)] dx ∫0 5 ∫0 5 f(x) dx = 7 (b) f(x) dx. ∫0 5 4 g(x) dx = 3. 0 . . . 19.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Solu on . . Notes Outline Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applica ons The definite integral as a limit Es ma ng the Definite Integral Proper es of the integral Comparison Proper es of the Integral . . Notes Comparison Properties of the Integral Theorem Let f and g be integrable func ons on [a, b]. ∫ b 6. If f(x) ≥ 0 for all x in [a, b], then f(x) dx ≥ 0 a ∫ b ∫ b 7. If f(x) ≥ g(x) for all x in [a, b], then f(x) dx ≥ g(x) dx a a 8. If m ≤ f(x) ≤ M for all x in [a, b], then ∫ b m(b − a) ≤ f(x) dx ≤ M(b − a) a . . . 20.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Notes Integral of a nonnegative function is nonnegative Proof. If f(x) ≥ 0 for all x in [a, b], then for any number of divisions n and choice of sample points {ci }: ∑ n ∑ n Sn = f(ci ) ∆x ≥ 0 · ∆x = 0 i=1 ≥0 i=1 . x Since Sn ≥ 0 for all n, the limit of {Sn } is nonnega ve, too: ∫ b f(x) dx = lim Sn ≥ 0 a n→∞ ≥0 . . Notes The integral is “increasing” Proof. Let h(x) = f(x) − g(x). If f(x) ≥ g(x) for all x in [a, b], then h(x) ≥ 0 for all f(x) x in [a, b]. So by the previous h(x) g(x) property ∫ b h(x) dx ≥ 0 . x a This means that ∫ b ∫ b ∫ b ∫ b f(x) dx − g(x) dx = (f(x) − g(x)) dx = h(x) dx ≥ 0 a a a a . . Notes Bounding the integral Proof. If m ≤ f(x) ≤ M on for all x in [a, b], then by y the previous property ∫ b ∫ b ∫ b M m dx ≤ f(x) dx ≤ M dx a a a f(x) By Property 8, the integral of a constant func on is the product of the constant and m the width of the interval. So: ∫ b . x m(b − a) ≤ f(x) dx ≤ M(b − a) a b a . . . 21.
  • . V63.0121.001: Calculus I Sec on 5.1–5.2: Areas, Distances, Integral . . April 25, 2011 Example Notes ∫ 2 1 Es mate dx using the comparison proper es. 1 x Solu on . . Notes 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. . . Notes Summary The definite integral is a limit of Riemann Sums The definite integral can be es mated with Riemann Sums The definite integral can be distributed across sums and constant mul ples of func ons The definite integral can be bounded using bounds for the func on . . . 22.