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

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 (slides) Lesson 24: Areas and Distances, The Definite Integral (slides) Presentation Transcript

  • Sec on 5.1–5.2 Areas and Distances, The Definite Integral V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 25, 2011.
  • 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
  • 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.
  • 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.
  • 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
  • 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.
  • Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. . b
  • Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. h . b
  • Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. h .
  • Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. h . b
  • Easy Areas: Parallelogram By cu ng and pas ng, a parallelogram can be made into a rectangle. h . So b Fact The area of a parallelogram of base width b and height h is A = bh
  • Easy Areas: Triangle By copying and pas ng, a triangle can be made into a parallelogram. . b
  • Easy Areas: Triangle By copying and pas ng, a triangle can be made into a parallelogram. h . b
  • Easy Areas: Triangle By copying and pas ng, a triangle can be made into a parallelogram. h . So b Fact The area of a triangle of base width b and height h is 1 A = bh 2
  • Easy Areas: Other Polygons Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: . .
  • Hard Areas: Curved Regions . ???
  • Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (a er Euclid) Geometer Weapons engineer
  • Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (a er Euclid) Geometer Weapons engineer
  • Meet the mathematician: Archimedes Greek (Syracuse), 287 BC – 212 BC (a er Euclid) Geometer Weapons engineer
  • Archimedes and the Parabola . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=
  • Archimedes and the Parabola 1 . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1
  • Archimedes and the Parabola 1 1 1 8 8 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8
  • 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 A=1+2· +4· + ··· 8 64
  • 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
  • Summing the series We need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4
  • Summing a geometric series Fact For any number r and any posi ve integer n, (1 − r)(1 + r + r2 + · · · + rn ) = 1 − rn+1 .
  • 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
  • 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 + ··· + r =n 1−r
  • 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+ + + ··· + n = 4 16 4 1 − 1/4
  • 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
  • Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspec ve
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x measure the area of the inscribed rectangles: . 0 1
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x measure the area of the inscribed rectangles: 1 L2 = 8 . 0 1 1 2
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x measure the area of the inscribed rectangles: 1 L2 = 8 L3 = . 0 1 2 1 3 3
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x measure the area of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 . 0 1 2 1 3 3
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x 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
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x 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
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x 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
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x 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
  • Cavalieri’s method Divide up the interval into pieces and 2 y=x 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 =?
  • What is Ln? 1 Divide the interval [0, 1] into n pieces. Then each has width . n
  • 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
  • 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 + · · · + = n n n3 n3
  • 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.
  • 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 + · · · + = n n n3 n3 So n(n − 1)(2n − 1) Ln = 6n3
  • 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 + · · · + = n n n3 n3 So n(n − 1)(2n − 1) 1 Ln = → 6n3 3 as n → ∞.
  • 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
  • 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
  • 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
  • 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)
  • 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 = 4n4
  • 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 → ∞.
  • Cavalieri’s method with different heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 1 + 2 + 3 + ··· + n 3 3 3 3 = n4 1 [ ]2 = 4 1 n(n + 1) n 2 n2 (n + 1)2 1 . = → 4n4 4 as n → ∞.
  • Cavalieri’s method with different heights 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 1 + 2 + 3 + ··· + n 3 3 3 3 = n4 1 [ ]2 = 4 1 n(n + 1) n 2 n2 (n + 1)2 1 . = → 4n4 4 as n → ∞. So even though the rectangles overlap, we s ll get the same answer.
  • 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
  • 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
  • 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
  • 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.
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. le endpoints… ∑ n Ln = f(xi−1 )∆x i=1 . x
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. right endpoints… ∑ n Rn = f(xi )∆x i=1 . x
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. midpoints… ∑ ( xi−1 + xi ) n Mn = f ∆x i=1 2 . x
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. the maximum value on the interval… ∑ n Un = max {f(x)} ∆x xi−1 ≤x≤xi i=1 . x
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. the minimum value on the interval… ∑ n Ln = min {f(x)} ∆x xi−1 ≤x≤xi i=1 . x
  • Forming Riemann sums We have many choices of representa ve points to approximate the area in each subinterval. …even random points! . x
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump 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.
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump 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.
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L1 = 3.0 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L2 = 5.25 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L3 = 6.0 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L4 = 6.375 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L5 = 6.59988 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L6 = 6.75 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L7 = 6.85692 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L8 = 6.9375 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L9 = 6.99985 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L10 = 7.04958 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L11 = 7.09064 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L12 = 7.125 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L13 = 7.15332 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L14 = 7.17819 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L15 = 7.19977 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L16 = 7.21875 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L17 = 7.23508 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L18 = 7.24927 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L19 = 7.26228 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L20 = 7.27443 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L21 = 7.28532 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L22 = 7.29448 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L23 = 7.30406 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L24 = 7.3125 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L25 = 7.31944 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L26 = 7.32559 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L27 = 7.33199 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L28 = 7.33798 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L29 = 7.34372 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L30 = 7.34882 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. le endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R1 = 12.0 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R2 = 9.75 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R3 = 9.0 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R4 = 8.625 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R5 = 8.39969 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R6 = 8.25 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R7 = 8.14236 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R8 = 8.0625 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R9 = 7.99974 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R10 = 7.94933 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R11 = 7.90868 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R12 = 7.875 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R13 = 7.84541 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R14 = 7.8209 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R15 = 7.7997 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R16 = 7.78125 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R17 = 7.76443 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R18 = 7.74907 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R19 = 7.73572 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R20 = 7.7243 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R21 = 7.7138 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R22 = 7.70335 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R23 = 7.69531 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R24 = 7.6875 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R25 = 7.67934 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R26 = 7.6715 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R27 = 7.66508 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R28 = 7.6592 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R29 = 7.65388 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump R30 = 7.64864 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. right endpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M1 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M2 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M3 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M4 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M5 = 7.4998 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M6 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M7 = 7.4996 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M8 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M9 = 7.49977 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M10 = 7.49947 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M11 = 7.49966 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M12 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M13 = 7.49937 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M14 = 7.49954 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. midpoints
  • 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M16 = 7.49988 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M17 = 7.49974 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M18 = 7.49916 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M19 = 7.49898 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M20 = 7.4994 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M21 = 7.49951 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M22 = 7.49889 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M23 = 7.49962 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M24 = 7.5 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M25 = 7.49939 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M26 = 7.49847 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M27 = 7.4985 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M28 = 7.4986 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M29 = 7.49878 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump M30 = 7.49872 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. midpoints
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U1 = 12.0 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U2 = 10.55685 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U3 = 10.0379 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U4 = 9.41515 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U5 = 8.96004 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U6 = 8.76895 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U7 = 8.6033 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U8 = 8.45757 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U9 = 8.34564 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U10 = 8.27084 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U11 = 8.20132 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U12 = 8.13838 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U13 = 8.0916 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U14 = 8.05139 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U15 = 8.01364 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U16 = 7.98056 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U17 = 7.9539 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U18 = 7.92815 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U19 = 7.90414 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U20 = 7.88504 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U21 = 7.86737 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U22 = 7.84958 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U23 = 7.83463 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U24 = 7.82187 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U25 = 7.80824 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U26 = 7.79504 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U27 = 7.78429 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U28 = 7.77443 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U29 = 7.76495 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump U30 = 7.7558 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. maximum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L1 = 3.0 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L2 = 4.44312 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L3 = 4.96208 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L4 = 5.58484 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L5 = 6.0395 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L6 = 6.23103 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L7 = 6.39577 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L8 = 6.54242 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L9 = 6.65381 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L10 = 6.72797 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L11 = 6.7979 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L12 = 6.8616 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L13 = 6.90704 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L14 = 6.94762 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L15 = 6.98575 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L16 = 7.01942 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L17 = 7.04536 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L18 = 7.07005 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L19 = 7.09364 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L20 = 7.1136 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L21 = 7.13155 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L22 = 7.14804 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L23 = 7.16441 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L24 = 7.17812 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L25 = 7.19025 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L26 = 7.2019 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L27 = 7.21265 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L28 = 7.22269 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L29 = 7.23251 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. minimum points
  • Theorem of the Day Theorem If f is a con nuous func on on [a, b] or has finitely many jump L30 = 7.24162 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. minimum points
  • Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4)
  • Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the slope of a curve
  • 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
  • Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of lines
  • 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
  • Analogies The Tangent Problem The Area Problem (Ch. 5) (Ch. 2–4) Want the area of a curved Want the slope of a curve region Only know the slope of Only know the area of lines polygons Approximate curve with a line
  • 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
  • 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
  • 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
  • Distances Just like area = length × width, we have distance = rate × me. So here is another use for Riemann sums.
  • Application: Dead Reckoning
  • 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.
  • 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.
  • 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.
  • 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.
  • 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
  • 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
  • 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
  • Notation/Terminology ∫ b ∑ n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1
  • Notation/Terminology ∫ b ∑ n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1 ∫ — integral sign (swoopy S)
  • Notation/Terminology ∫ b ∑ n f(x) dx = lim f(ci ) ∆x a ∆x→0 i=1 ∫ — integral sign (swoopy S) f(x) — integrand
  • 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)
  • 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?)
  • 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
  • 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
  • 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 ∫ 3Find x dx 0
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n n
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor each i, take xi to represent the func on on the ith interval.
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor each i, take xi to represent the func on on the ith interval. So ∫ 3 x dx = lim Rn 0 n→∞
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor each i, take xi to represent the func on on the ith interval. So ∫ 3 ∑ n x dx = lim Rn = lim f(xi ) ∆x 0 n→∞ n→∞ i=1
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor 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
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor 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
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor 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 = lim 2 i n→∞ n i=1
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor 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) = lim 2 i = lim 2 · n→∞ n n→∞ n 2 i=1
  • Example ∫ 3Find x dx 0Solu on 3 3iFor any n we have ∆x = and for each i between 0 and n, xi = . n nFor 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 2 i = lim 2 · = ·1 n→∞ n n→∞ n 2 2 i=1
  • Example ∫ 3Find x2 dx 0
  • Example ∫ 3Find x2 dx 0Solu on
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . n n
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 x2 dx = lim Rn 0 n→∞
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n x2 dx = lim Rn = lim f(xi ) ∆x 0 n→∞ n→∞ i=1
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n 27 ∑ 2 n = lim 3 i n→∞ n i=1
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n 27 ∑ 2 n 27 n(n + 1)(2n + 1) = lim 3 i = lim 3 · n→∞ n n→∞ n 6 i=1
  • Example ∫ 3Find x2 dx 0Solu on 3 3iFor any n and i we have ∆x = and xi = . So n n ∫ 3 ∑n ∑ ( 3i )2 ( 3 ) n 2 x dx = lim Rn = lim f(xi ) ∆x = lim 0 n→∞ n→∞ i=1 n→∞ i=1 n n 27 ∑ 2 n 27 n(n + 1)(2n + 1) = lim 3 i = lim 3 · n→∞ n n→∞ n 6 i=1 27 = =9 3
  • Example ∫ 3Find x3 dx 0
  • Example ∫ 3Find x3 dx 0Solu on 3 3iFor any n we have ∆x = and xi = . So n n ∑n ∑ ( 3i )3 ( 3 ) 81 ∑ n n Rn = f(xi ) ∆x = = 4 i3 i=1 i=1 n n n i=1 81 n2 (n + 1)2 81 = 4· −→ n 4 4 ∫ 3 81So x3 dx = = 20.25 0 4
  • 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
  • Estimating the Definite Integral Example ∫ 1 4 Es mate dx using M4 . 0 1 + x2
  • 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
  • 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
  • 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
  • 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
  • Estimating the Definite Integral Example ∫ 1 4 Es mate dx using L4 and R4 0 1 + x2
  • Estimating the Definite Integral Example ∫ 1 4 Es mate dx using L4 and R4 0 1 + x2 Answer ( ) 1 4 4 4 4 L4 = + + + 41 + (0)2 1 + (1/4)2 1 + (1/2)2 1 + (3/4)2 16 4 16 =1+ + + ≈ 3.38118 17 5 25
  • Estimating the Definite Integral Example ∫ 1 4 Es mate dx using L4 and R4 0 1 + x2 Answer ( ) 1 4 4 4 4 R4 = + + + 4 1 + (1/4)2 1 + (1/2)2 1 + (3/4)2 1 + (1)2 16 4 16 1 = + + + ≈ 2.88118 17 5 25 2
  • 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
  • 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
  • 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 a a
  • 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 a a ∫ b ∫ b 3. cf(x) dx = c f(x) dx. a a
  • 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 a a ∫ b ∫ b 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
  • 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 ( 3 )Find x − 4.5x2 + 5.5x + 1 dx 0
  • Example ∫ 3 ( 3 )Find x − 4.5x2 + 5.5x + 1 dx 0Solu on ∫ 3 (x3 −4.5x2 + 5.5x + 1) dx 0 ∫ 3 ∫ 3 ∫ 3 ∫ 3 = x3 dx − 4.5 x2 dx + 5.5 x dx + 1 dx 0 0 0 0 = 20.25 − 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5
  • Example ∫ 3 ( 3 )Find x − 4.5x2 + 5.5x + 1 dx 0Solu on ∫ 3 (x3 −4.5x2 + 5.5x + 1) dx 0 ∫ 3 ∫ 3 ∫ 3 ∫ 3 = x3 dx − 4.5 x2 dx + 5.5 x dx + 1 dx 0 0 0 0 = 20.25 − 4.5 · 9 + 5.5 · 4.5 + 3 · 1 = 7.5(This is the func on we were es ma ng the integral of before)
  • 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. midpoints
  • 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
  • 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 . a c x b
  • 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 f(x) dx a . a c x b
  • 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
  • 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 ∫ c f(x) dx f(x) dx f(x) dx a a b . a c x b
  • 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 . a c x b
  • 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 f(x) dx a . a c x b
  • 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 f(x) dx = b∫ b − f(x) dx . c a c x b
  • 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
  • Using the Properties Example Suppose f and g are func ons with ∫ 4 Find∫ f(x) dx = 4 5 0 (a) [2f(x) − g(x)] dx ∫ 5 0 ∫ 5 f(x) dx = 7 0 (b) f(x) dx. ∫ 5 4 g(x) dx = 3. 0
  • Solu onWe have ∫ ∫ ∫ 5 5 5 [2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx(a) 0 0 0 = 2 · 7 − 3 = 11
  • Solu onWe have ∫ ∫ ∫ 5 5 5 [2f(x) − g(x)] dx = 2 f(x) dx − g(x) dx(a) 0 0 0 = 2 · 7 − 3 = 11 ∫ 5 ∫ 5 ∫ 4 f(x) dx = f(x) dx − f(x) dx(b) 4 0 0 =7−4=3
  • 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
  • Comparison Properties of the Integral Theorem Let f and g be integrable func ons on [a, b].
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Example ∫ 2 1Es mate dx using the comparison proper es. 1 x
  • Example ∫ 2 1Es mate dx using the comparison proper es. 1 xSolu onSince 1 1 ≤x≤ 2 1for all x in [1, 2], we have ∫ 2 1 1 ·1≤ dx ≤ 1 · 1 2 1 x
  • 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.
  • 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