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V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral        December 2, 2010



                          Sections 5.1–5.2                                                                   Notes

                        Areas and Distances
                        The Definite Integral
                                   V63.0121.021, Calculus I

                                           New York University


                                         December 2, 2010


 Announcements

       Final December 20, 12:00–1:50pm




 Announcements
                                                                                                             Notes




          Final December 20,
          12:00–1:50pm
                 cumulative
                 location TBD
                 old exams on common
                 website




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   2 / 56




 Objectives from Section 5.1
                                                                                                             Notes


          Compute the area of a
          region by approximating it
          with rectangles and letting
          the size of the rectangles
          tend to zero.
          Compute the total distance
          traveled by a particle by
          approximating it as distance
          = (rate)(time) and letting
          the time intervals over which
          one approximates tend to
          zero.



  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   3 / 56




                                                                                                                                    1
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral            December 2, 2010


 Objectives from Section 5.2
                                                                                                                 Notes



          Compute the definite
          integral using a limit of
          Riemann sums
          Estimate the definite
          integral using a Riemann
          sum (e.g., Midpoint Rule)
          Reason with the definite
          integral using its elementary
          properties.




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010   4 / 56




 Outline
                                                                                                                 Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010   5 / 56




 Easy Areas: Rectangle
                                                                                                                 Notes

 Definition
 The area of a rectangle with dimensions                          and w is the product A = w .




                                                                                 w




 It may seem strange that this is a definition and not a theorem but we
 have to start somewhere.


  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010   6 / 56




                                                                                                                                        2
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral        December 2, 2010


 Easy Areas: Parallelogram
                                                                                                             Notes
 By cutting and pasting, a parallelogram can be made into a rectangle.




                                          h



                                                  b         b

 So
 Fact
 The area of a parallelogram of base width b and height h is

                                                  A = bh

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   7 / 56




 Easy Areas: Triangle
                                                                                                             Notes
 By copying and pasting, a triangle can be made into a parallelogram.




                                          h



                                                  b

 So
 Fact
 The area of a triangle of base width b and height h is
                                                   1
                                                A = bh
                                                   2

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   8 / 56




 Easy Areas: Other Polygons
                                                                                                             Notes


 Any polygon can be triangulated, so its area can be found by summing the
 areas of the triangles:




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   9 / 56




                                                                                                                                    3
V63.0121.021, Calculus I                                         Sections 5.1–5.2 : Areas, Distances, Integral              December 2, 2010


 Hard Areas: Curved Regions
                                                                                                                    Notes




 ???




  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   10 / 56




 Meet the mathematician: Archimedes
                                                                                                                    Notes




        Greek (Syracuse), 287 BC –
        212 BC (after Euclid)
        Geometer
        Weapons engineer




  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   11 / 56




 Archimedes and the Parabola
                                                                                                                    Notes



                           1                                                      1
                           64                                                     64
                                                        1
                                    1                                        1
                                    8                                        8

                                             1                    1
                                             64                   64



 Archimedes found areas of a sequence of triangles inscribed in a parabola.
                                        1      1
                                A=1+2·    +4·     + ···
                                        8     64
                                      1   1          1
                                   =1+ +    + ··· + n + ···
                                      4 16          4

  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   12 / 56




                                                                                                                                           4
V63.0121.021, Calculus I                                         Sections 5.1–5.2 : Areas, Distances, Integral         December 2, 2010


 Summing the series
                                                                                                               Notes
 We would then need to know the value of the series
                          1   1           1
                      1+ +       + ··· + n + ···
                          4 16            4

 Fact
 For any number r and any positive integer n,

                             (1 − r )(1 + r + · · · + r n ) = 1 − r n+1

 So
                                                                  1 − r n+1
                                   1 + r + · · · + rn =
                                                                    1−r

 Therefore
                1   1         1   1 − (1/4)n+1    1  4
         1+       +   + ··· + n =              → 3 = as n → ∞.
                4 16         4       1 − 1/4      /4 3
  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   13 / 56




 Cavalieri
                                                                                                               Notes




        Italian,
        1598–1647
        Revisited the
        area problem
        with a
        different
        perspective




  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   14 / 56




 Cavalieri’s method
                                                                                                               Notes

                                                         Divide up the interval into pieces
                             y = x2                      and measure the area of the
                                                         inscribed rectangles:
                                                                1
                                                         L2 =
                                                                8
                                                                 1     4     5
                                                         L3   =     +     =
                                                                27 27       27
                                                                 1     4    9    14
                                                         L4   =     +     +    =
                                                                64 64 64         64
                                                                  1      4     9    16   30
  0                                   1                  L5   =      +      +     +    =
                                                                125 125 125 125          125
                                                         Ln   =?



  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   15 / 56




                                                                                                                                      5
V63.0121.021, Calculus I                                         Sections 5.1–5.2 : Areas, Distances, Integral                 December 2, 2010


 What is Ln ?
                                                               1                                                       Notes
 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 =       +    + ··· +          =
                 n3 n3             n3                   n3
 The Arabs knew that
                                                                     n(n − 1)(2n − 1)
                    1 + 22 + 32 + · · · + (n − 1)2 =
                                                                            6
 So
                                           n(n − 1)(2n − 1)   1
                                   Ln =                     →
                                                 6n3          3
 as n → ∞.
  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral           December 2, 2010   16 / 56




 Cavalieri’s method for different functions
                                                                                                                       Notes
 Try the same trick with f (x) = x 3 . We have

                       1      1     1        2           1                           n−1
                 Ln =    ·f      + ·f            + ··· + · f
                       n      n     n        n           n                            n
                       1 1      1 23             1 (n − 1)3
                     = · 3 + · 3 + ··· + ·
                       n n      n n              n     n3
                       1 + 23 + 33 + · · · + (n − 1)3
                     =
                                    n4
 The formula out of the hat is
                                                                                           2
                       1 + 23 + 33 + · · · + (n − 1)3 =                    1
                                                                           2 n(n   − 1)

  So
                                                n2 (n − 1)2   1
                                      Ln =                  →
                                                    4n4       4
 as n → ∞.
  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral           December 2, 2010   17 / 56




 Cavalieri’s method with different heights
                                                                                                                       Notes



                                                                  1 13      1 23            1 n3
                                                         Rn =       · 3 + · 3 + ··· + · 3
                                                                  n n       n n             n n
                                                                  13 + 23 + 33 + · · · + n3
                                                                =
                                                                              n4
                                                                  1 1            2
                                                                = 4 2 n(n + 1)
                                                                  n
                                                                  n2 (n + 1)2    1
                                                                =              →
                                                                      4n4        4
                                    as n → ∞.
 So even though the rectangles overlap, we still get the same answer.




  V63.0121.021, Calculus I (NYU)    Sections 5.1–5.2 Areas, Distances, Integral           December 2, 2010   18 / 56




                                                                                                                                              6
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral               December 2, 2010


 Outline
                                                                                                                    Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010   19 / 56




 Cavalieri’s method in general
                                                                                                                    Notes
 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. Then
        b−a
 ∆x =        . For each i between 1 and n, let xi be the ith step between a
          n
 and b. So

                                                         x0 = a
                                                                              b−a
                                                         x1 = x0 + ∆x = a +
                                                                               n
                                                                                b−a
                                                         x2 = x1 + ∆x = a + 2 ·     ...
                                                                                 n
                                                                      b−a
                                                         xi = a + i ·     ...
                                                                       n
                                                                      b−a
                                                         xn = a + n ·      =b
                                      x                                n
    x0 x1 . . . xi . . .xn−1xn

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010   20 / 56




 Forming Riemann sums
                                                                                                                    Notes
 We have many choices of representative points to approximate the area in
 each subinterval.




       . . . even random points!




                                                                           x
 In general, 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

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010   21 / 56




                                                                                                                                           7
V63.0121.021, Calculus I                                              Sections 5.1–5.2 : Areas, Distances, Integral                   December 2, 2010


 Theorem of the Day
                                                                                                                              Notes


 Theorem

 If f is a continuous function on [a, b]
                                                                                       M15 = 7.49968
 or has finitely many jump
 discontinuities, then
                                    n
      lim Sn = lim                       f (ci )∆x
     n→∞             n→∞
                                   i=1

 exists and is the same value no
 matter what choice of ci we make.                                                                              x




  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010       22 / 56




 Analogies
                                                                                                                              Notes


                                                                   The Area Problem (Ch. 5)
 The Tangent Problem
 (Ch. 2–4)                                                                 Want the area of a curved
                                                                           region
        Want the slope of a curve
                                                                           Only know the area of
        Only know the slope of lines
                                                                           polygons
        Approximate curve with a
                                                                           Approximate region with
        line
                                                                           polygons
        Take limit over better and
                                                                           Take limit over better and
        better approximations
                                                                           better approximations




  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010       23 / 56




 Outline
                                                                                                                              Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral         December 2, 2010       24 / 56




                                                                                                                                                     8
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral                December 2, 2010


 Distances
                                                                                                                     Notes




 Just like area = length × width, we have

                                    distance = rate × time.

 So here is another use for Riemann sums.




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   25 / 56




 Application: Dead Reckoning
                                                                                                                     Notes




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   26 / 56




 Computing position by Dead Reckoning
                                                                                                                     Notes
 Example
 A sailing ship is cruising back and forth along a channel (in a straight
 line). At noon the ship’s position and velocity are recorded, but shortly
 thereafter a storm blows in and position is impossible to measure. The
 velocity continues to be recorded at thirty-minute intervals.

                  Time                   12:00         12:30         1:00        1:30    2:00
                  Speed (knots)            4             8            12           6       4
                  Direction                E             E            E           E       W
                  Time                    2:30         3:00          3:30        4:00
                  Speed                    3             3             5           9
                  Direction                W             E            E           E

 Estimate the ship’s position at 4:00pm.

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   27 / 56




                                                                                                                                            9
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral         December 2, 2010


 Solution
                                                                                                              Notes

 Solution
 We estimate that the speed of 4 knots (nautical miles per hour) is
 maintained from 12:00 until 12:30. So over this time interval the ship
 travels
                          4 nmi     1
                                      hr = 2 nmi
                            hr      2
 We can continue for each additional half hour and get

   distance = 4 × 1/2 + 8 × 1/2 + 12 × 1/2
                     + 6 × 1/2 − 4 × 1/2 − 3 × 1/2 + 3 × 1/2 + 5 × 1/2
                                                                                             = 15.5

 So the ship is 15.5 nmi east of its original position.


  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   28 / 56




 Analysis
                                                                                                              Notes




       This method of measuring position by recording velocity was necessary
       until global-positioning satellite technology became widespread
       If we had velocity estimates at finer intervals, we’d get better
       estimates.
       If we had velocity at every instant, a limit would tell us our exact
       position relative to the last time we measured it.




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   29 / 56




 Other uses of Riemann sums
                                                                                                              Notes




 Anything with a product!
       Area, volume
       Anything with a density: Population, mass
       Anything with a “speed:” distance, throughput, power
       Consumer surplus
       Expected value of a random variable




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   30 / 56




                                                                                                                                    10
V63.0121.021, Calculus I                                                Sections 5.1–5.2 : Areas, Distances, Integral          December 2, 2010


 Outline
                                                                                                                       Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)           Sections 5.1–5.2 Areas, Distances, Integral    December 2, 2010   31 / 56




 The definite integral as a limit
                                                                                                                       Notes




 Definition
 If f is a function 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




  V63.0121.021, Calculus I (NYU)           Sections 5.1–5.2 Areas, Distances, Integral    December 2, 2010   32 / 56




 Notation/Terminology
                                                                                                                       Notes


                                       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 integration (a is the lower limit and b the
       upper limit)
       dx — ??? (a parenthesis? an infinitesimal? a variable?)
       The process of computing an integral is called integration or
       quadrature



  V63.0121.021, Calculus I (NYU)           Sections 5.1–5.2 Areas, Distances, Integral    December 2, 2010   33 / 56




                                                                                                                                             11
V63.0121.021, Calculus I                                                       Sections 5.1–5.2 : Areas, Distances, Integral                  December 2, 2010


 The limit can be simplified
                                                                                                                                      Notes
 Theorem
 If f is continuous on [a, b] or if f has only finitely many jump
 discontinuities, then f is integrable on [a, b]; that is, the definite integral
      b
          f (x) dx exists.
  a


 Theorem
 If f is integrable on [a, b] then
                                              b                                n
                                                  f (x) dx = lim                    f (xi )∆x,
                                          a                         n→∞
                                                                             i=1

 where
                                              b−a
                                 ∆x =                              and             xi = a + i ∆x
                                               n

  V63.0121.021, Calculus I (NYU)                  Sections 5.1–5.2 Areas, Distances, Integral            December 2, 2010   34 / 56




 Example: Integral of x
                                                                                                                                      Notes
 Example
                  3
 Find                 x dx
              0


 Solution
                                                    3         3i
 For any n we have ∆x =                               and xi = . So
                                                    n         n
                                     n                               n                                    n
                                                                             3i         3           9
                             Rn =         f (xi ) ∆x =                                          =              i
                                                                             n          n           n2
                                    i=1                            i=1                                   i=1
                                    9 n(n + 1)    9
                                =      ·       −→
                                    n2   2        2
              3
                               9
 So               x dx =         = 4.5
          0                    2

  V63.0121.021, Calculus I (NYU)                  Sections 5.1–5.2 Areas, Distances, Integral            December 2, 2010   35 / 56




 Example: Integral of x 2
                                                                                                                                      Notes
 Example
                  3
 Find                 x 2 dx
              0


 Solution




  V63.0121.021, Calculus I (NYU)                  Sections 5.1–5.2 Areas, Distances, Integral            December 2, 2010   36 / 56




                                                                                                                                                            12
V63.0121.021, Calculus I                                           Sections 5.1–5.2 : Areas, Distances, Integral         December 2, 2010


 Example: Integral of x 3
                                                                                                                 Notes
 Example
              3
 Find             x 3 dx
          0


 Solution




  V63.0121.021, Calculus I (NYU)      Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   37 / 56




 Outline
                                                                                                                 Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)      Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   38 / 56




 Estimating the Definite Integral
                                                                                                                 Notes
 Example
                        1
                              4
 Estimate                          dx using M4 .
                    0       1 + x2

 Solution
                                      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
                    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

  V63.0121.021, Calculus I (NYU)      Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   39 / 56




                                                                                                                                       13
V63.0121.021, Calculus I                                             Sections 5.1–5.2 : Areas, Distances, Integral                December 2, 2010


 Estimating the Definite Integral (Continued)
                                                                                                                          Notes
 Example
                       1
                             4
 Estimate                         dx using L4 and R4
                   0       1 + x2

 Answer




  V63.0121.021, Calculus I (NYU)        Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   40 / 56




 Outline
                                                                                                                          Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)        Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   41 / 56




 Properties of the integral
                                                                                                                          Notes


 Theorem (Additive Properties of the Integral)
 Let f and g be integrable functions 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
             b                                      b                        b
  4.             [f (x) − g (x)] dx =                   f (x) dx −               g (x) dx.
         a                                      a                        a




  V63.0121.021, Calculus I (NYU)        Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010   42 / 56




                                                                                                                                                14
V63.0121.021, Calculus I                                                        Sections 5.1–5.2 : Areas, Distances, Integral       December 2, 2010


 Proofs
                                                                                                                            Notes


 Proofs.

        When integrating a constant function 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.
        Ditto for constant multiples
        Ditto for differences




  V63.0121.021, Calculus I (NYU)            Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   43 / 56




 Example                                                                                                                    Notes
              3
 Find              x 3 − 4.5x 2 + 5.5x + 1 dx
          0


 Solution




  V63.0121.021, Calculus I (NYU)            Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   44 / 56




 More Properties of the Integral
                                                                                                                            Notes



 Conventions:
                                                a                               b
                                                    f (x) dx = −                    f (x) dx
                                            b                               a
                                                           a
                                                               f (x) dx = 0
                                                       a
 This allows us to have
             c                         b                            c
  5.              f (x) dx =               f (x) dx +                   f (x) dx for all a, b, and c.
         a                         a                            b




  V63.0121.021, Calculus I (NYU)            Sections 5.1–5.2 Areas, Distances, Integral        December 2, 2010   46 / 56




                                                                                                                                                  15
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral         December 2, 2010



 Example                                                                                                      Notes
 Suppose f and g are functions with
             4
                 f (x) dx = 4
         0
             5
                 f (x) dx = 7
         0
             5
                 g (x) dx = 3.
         0
 Find
            5
 (a)            [2f (x) − g (x)] dx
        0
            5
 (b)            f (x) dx.
        4




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   47 / 56




 Solution                                                                                                     Notes




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   48 / 56




 Outline
                                                                                                              Notes
 Area through the Centuries
    Euclid
    Archimedes
    Cavalieri
 Generalizing Cavalieri’s method
   Analogies
 Distances
    Other applications
 The definite integral as a limit
 Estimating the Definite Integral
 Properties of the integral
 Comparison Properties of the Integral

  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   49 / 56




                                                                                                                                    16
V63.0121.021, Calculus I                                                     Sections 5.1–5.2 : Areas, Distances, Integral                  December 2, 2010


 Comparison Properties of the Integral
                                                                                                                                    Notes
 Theorem
 Let f and g be integrable functions on [a, b].
  6. If f (x) ≥ 0 for all x in [a, b], then
                                                                 b
                                                                     f (x) dx ≥ 0
                                                             a

  7. If f (x) ≥ g (x) for all x in [a, b], then
                                                       b                         b
                                                           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

  V63.0121.021, Calculus I (NYU)              Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010       50 / 56




 The integral of a nonnegative function is
 nonnegative                                                                                                                        Notes

 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


 Since Sn ≥ 0 for all n, the limit
 of {Sn } is nonnegative, too:
           b
                                                                                                                      x
               f (x) dx = lim Sn ≥ 0
       a                       n→∞
                                          ≥0


  V63.0121.021, Calculus I (NYU)              Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010       51 / 56




 The definite integral is “increasing”
                                                                                                                                    Notes
 Proof.

 Let h(x) = f (x) − g (x). If                                                                                   f (x)
 f (x) ≥ g (x) for all x in [a, b],
 then h(x) ≥ 0 for all x in [a, b].                                                             h(x)            g (x)
 So by the previous property
                       b
                           h(x) dx ≥ 0
                   a                                                                                                  x
 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

           b                        b
 So            f (x) dx ≥               g (x) dx.
       a                        a

  V63.0121.021, Calculus I (NYU)              Sections 5.1–5.2 Areas, Distances, Integral          December 2, 2010       52 / 56




                                                                                                                                                          17
V63.0121.021, Calculus I                                              Sections 5.1–5.2 : Areas, Distances, Integral                 December 2, 2010


 Bounding the integral using bounds of the function
                                                                                                                            Notes
 Proof.
 If m ≤ f (x) ≤ M on for all x in
 [a, b], then by the previous                                             y
 property
                                                                         M
       b                     b                     b
           m dx ≤                f (x) dx ≤            M dx
   a                     a                     a
                                                                                                         f (x)
 By Property 8, the integral of a
 constant function is the product
 of the constant and the width of                                        m
 the interval. So:
                             b
                                                                                                              x
 m(b−a) ≤                        f (x) dx ≤ M(b−a)                                     a
                         a                                                                           b


  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010       53 / 56




                                                                                                                            Notes
 Example
                     2
                         1
 Estimate                  dx using the comparison properties.
                 1       x

 Solution




  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010       54 / 56




 Summary
                                                                                                                            Notes




           We can compute the area of a curved region with a limit of Riemann
           sums
           We can compute the distance traveled from the velocity with a limit
           of Riemann sums
           Many other important uses of this process.




  V63.0121.021, Calculus I (NYU)         Sections 5.1–5.2 Areas, Distances, Integral       December 2, 2010       55 / 56




                                                                                                                                                  18
V63.0121.021, Calculus I                                        Sections 5.1–5.2 : Areas, Distances, Integral         December 2, 2010


 Summary
                                                                                                              Notes




       The definite integral is a limit of Riemann Sums
       The definite integral can be estimated with Riemann Sums
       The definite integral can be distributed across sums and constant
       multiples of functions
       The definite integral can be bounded using bounds for the function




  V63.0121.021, Calculus I (NYU)   Sections 5.1–5.2 Areas, Distances, Integral   December 2, 2010   56 / 56




                                                                                                              Notes




                                                                                                              Notes




                                                                                                                                    19

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Lesson 22: Optimization Problems (handout)
 
Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)Lesson 21: Curve Sketching (slides)
Lesson 21: Curve Sketching (slides)
 
Lesson 20: Derivatives and the Shapes of Curves (slides)
Lesson 20: Derivatives and the Shapes of Curves (slides)Lesson 20: Derivatives and the Shapes of Curves (slides)
Lesson 20: Derivatives and the Shapes of Curves (slides)
 
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 20: Derivatives and the Shapes of Curves (handout)Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 20: Derivatives and the Shapes of Curves (handout)
 
Lesson 19: The Mean Value Theorem (slides)
Lesson 19: The Mean Value Theorem (slides)Lesson 19: The Mean Value Theorem (slides)
Lesson 19: The Mean Value Theorem (slides)
 
Lesson 18: Maximum and Minimum Values (slides)
Lesson 18: Maximum and Minimum Values (slides)Lesson 18: Maximum and Minimum Values (slides)
Lesson 18: Maximum and Minimum Values (slides)
 

Lesson 24: Areas, Distances, the Integral (Section 021 handout)

  • 1. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Sections 5.1–5.2 Notes Areas and Distances The Definite Integral V63.0121.021, Calculus I New York University December 2, 2010 Announcements Final December 20, 12:00–1:50pm Announcements Notes Final December 20, 12:00–1:50pm cumulative location TBD old exams on common website V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56 Objectives from Section 5.1 Notes Compute the area of a region by approximating it with rectangles and letting the size of the rectangles tend to zero. Compute the total distance traveled by a particle by approximating it as distance = (rate)(time) and letting the time intervals over which one approximates tend to zero. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 3 / 56 1
  • 2. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Objectives from Section 5.2 Notes Compute the definite integral using a limit of Riemann sums Estimate the definite integral using a Riemann sum (e.g., Midpoint Rule) Reason with the definite integral using its elementary properties. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 4 / 56 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 5 / 56 Easy Areas: Rectangle Notes Definition The area of a rectangle with dimensions and w is the product A = w . w It may seem strange that this is a definition and not a theorem but we have to start somewhere. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 6 / 56 2
  • 3. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Easy Areas: Parallelogram Notes By cutting and pasting, a parallelogram can be made into a rectangle. h b b So Fact The area of a parallelogram of base width b and height h is A = bh V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56 Easy Areas: Triangle Notes By copying and pasting, a triangle can be made into a parallelogram. h b So Fact The area of a triangle of base width b and height h is 1 A = bh 2 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56 Easy Areas: Other Polygons Notes Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 9 / 56 3
  • 4. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Hard Areas: Curved Regions Notes ??? V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 10 / 56 Meet the mathematician: Archimedes Notes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56 Archimedes and the Parabola Notes 1 1 64 64 1 1 1 8 8 1 1 64 64 Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 1 A=1+2· +4· + ··· 8 64 1 1 1 =1+ + + ··· + n + ··· 4 16 4 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56 4
  • 5. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Summing the series Notes We would then need to know the value of the series 1 1 1 1+ + + ··· + n + ··· 4 16 4 Fact For any number r and any positive integer n, (1 − r )(1 + r + · · · + r n ) = 1 − r n+1 So 1 − r n+1 1 + r + · · · + rn = 1−r Therefore 1 1 1 1 − (1/4)n+1 1 4 1+ + + ··· + n = → 3 = as n → ∞. 4 16 4 1 − 1/4 /4 3 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56 Cavalieri Notes Italian, 1598–1647 Revisited the area problem with a different perspective V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 14 / 56 Cavalieri’s method Notes Divide up the interval into pieces y = x2 and measure the area of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 1 4 9 16 30 0 1 L5 = + + + = 125 125 125 125 125 Ln =? V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56 5
  • 6. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 What is Ln ? 1 Notes 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 = + + ··· + = n3 n3 n3 n3 The Arabs knew that n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6 So n(n − 1)(2n − 1) 1 Ln = → 6n3 3 as n → ∞. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56 Cavalieri’s method for different functions Notes Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1 + 23 + 33 + · · · + (n − 1)3 = n4 The formula out of the hat is 2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) So n2 (n − 1)2 1 Ln = → 4n4 4 as n → ∞. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56 Cavalieri’s method with different heights Notes 1 13 1 23 1 n3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 13 + 23 + 33 + · · · + n3 = n4 1 1 2 = 4 2 n(n + 1) n n2 (n + 1)2 1 = → 4n4 4 as n → ∞. So even though the rectangles overlap, we still get the same answer. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56 6
  • 7. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 19 / 56 Cavalieri’s method in general Notes 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. Then b−a ∆x = . For each i between 1 and n, let xi be the ith step between a n and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · ... n b−a xi = a + i · ... n b−a xn = a + n · =b x n x0 x1 . . . xi . . .xn−1xn V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 20 / 56 Forming Riemann sums Notes We have many choices of representative points to approximate the area in each subinterval. . . . even random points! x In general, 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 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56 7
  • 8. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Theorem of the Day Notes Theorem If f is a continuous function on [a, b] M15 = 7.49968 or has finitely many jump discontinuities, then n lim Sn = lim f (ci )∆x n→∞ n→∞ i=1 exists and is the same value no matter what choice of ci we make. x V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56 Analogies Notes The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope of lines polygons Approximate curve with a Approximate region with line polygons Take limit over better and Take limit over better and better approximations better approximations V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 24 / 56 8
  • 9. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Distances Notes Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 25 / 56 Application: Dead Reckoning Notes V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 26 / 56 Computing position by Dead Reckoning Notes Example A sailing ship is cruising back and forth along a channel (in a straight line). At noon the ship’s position and velocity are recorded, but shortly thereafter a storm blows in and position is impossible to measure. The velocity continues to be recorded at thirty-minute intervals. Time 12:00 12:30 1:00 1:30 2:00 Speed (knots) 4 8 12 6 4 Direction E E E E W Time 2:30 3:00 3:30 4:00 Speed 3 3 5 9 Direction W E E E Estimate the ship’s position at 4:00pm. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 27 / 56 9
  • 10. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Solution Notes Solution We estimate that the speed of 4 knots (nautical miles per hour) is maintained from 12:00 until 12:30. So over this time interval the ship travels 4 nmi 1 hr = 2 nmi hr 2 We can continue for each additional half hour and get distance = 4 × 1/2 + 8 × 1/2 + 12 × 1/2 + 6 × 1/2 − 4 × 1/2 − 3 × 1/2 + 3 × 1/2 + 5 × 1/2 = 15.5 So the ship is 15.5 nmi east of its original position. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 28 / 56 Analysis Notes This method of measuring position by recording velocity was necessary until global-positioning satellite technology became widespread If we had velocity estimates at finer intervals, we’d get better estimates. If we had velocity at every instant, a limit would tell us our exact position relative to the last time we measured it. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 29 / 56 Other uses of Riemann sums Notes Anything with a product! Area, volume Anything with a density: Population, mass Anything with a “speed:” distance, throughput, power Consumer surplus Expected value of a random variable V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 30 / 56 10
  • 11. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 31 / 56 The definite integral as a limit Notes Definition If f is a function 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 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 32 / 56 Notation/Terminology Notes 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 integration (a is the lower limit and b the upper limit) dx — ??? (a parenthesis? an infinitesimal? a variable?) The process of computing an integral is called integration or quadrature V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56 11
  • 12. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 The limit can be simplified Notes Theorem If f is continuous on [a, b] or if f has only finitely many jump discontinuities, then f is integrable on [a, b]; that is, the definite integral b f (x) dx exists. a Theorem If f is integrable on [a, b] then b n f (x) dx = lim f (xi )∆x, a n→∞ i=1 where b−a ∆x = and xi = a + i ∆x n V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 34 / 56 Example: Integral of x Notes Example 3 Find x dx 0 Solution 3 3i For any n we have ∆x = and xi = . So n n n n n 3i 3 9 Rn = f (xi ) ∆x = = i n n n2 i=1 i=1 i=1 9 n(n + 1) 9 = · −→ n2 2 2 3 9 So x dx = = 4.5 0 2 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56 Example: Integral of x 2 Notes Example 3 Find x 2 dx 0 Solution V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56 12
  • 13. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Example: Integral of x 3 Notes Example 3 Find x 3 dx 0 Solution V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 37 / 56 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 38 / 56 Estimating the Definite Integral Notes Example 1 4 Estimate dx using M4 . 0 1 + x2 Solution 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 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 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56 13
  • 14. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Estimating the Definite Integral (Continued) Notes Example 1 4 Estimate dx using L4 and R4 0 1 + x2 Answer V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 40 / 56 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 41 / 56 Properties of the integral Notes Theorem (Additive Properties of the Integral) Let f and g be integrable functions 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 b b b 4. [f (x) − g (x)] dx = f (x) dx − g (x) dx. a a a V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56 14
  • 15. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Proofs Notes Proofs. When integrating a constant function 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. Ditto for constant multiples Ditto for differences V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 43 / 56 Example Notes 3 Find x 3 − 4.5x 2 + 5.5x + 1 dx 0 Solution V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 44 / 56 More Properties of the Integral Notes Conventions: a b f (x) dx = − f (x) dx b a a f (x) dx = 0 a This allows us to have c b c 5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c. a a b V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 46 / 56 15
  • 16. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Example Notes Suppose f and g are functions with 4 f (x) dx = 4 0 5 f (x) dx = 7 0 5 g (x) dx = 3. 0 Find 5 (a) [2f (x) − g (x)] dx 0 5 (b) f (x) dx. 4 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 47 / 56 Solution Notes V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 48 / 56 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications The definite integral as a limit Estimating the Definite Integral Properties of the integral Comparison Properties of the Integral V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 49 / 56 16
  • 17. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Comparison Properties of the Integral Notes Theorem Let f and g be integrable functions on [a, b]. 6. If f (x) ≥ 0 for all x in [a, b], then b f (x) dx ≥ 0 a 7. If f (x) ≥ g (x) for all x in [a, b], then b b 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 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56 The integral of a nonnegative function is nonnegative Notes 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 Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too: b x f (x) dx = lim Sn ≥ 0 a n→∞ ≥0 V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 51 / 56 The definite integral is “increasing” Notes Proof. Let h(x) = f (x) − g (x). If f (x) f (x) ≥ g (x) for all x in [a, b], then h(x) ≥ 0 for all x in [a, b]. h(x) g (x) So by the previous property b h(x) dx ≥ 0 a x 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 b b So f (x) dx ≥ g (x) dx. a a V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 52 / 56 17
  • 18. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Bounding the integral using bounds of the function Notes Proof. If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous y property M b b b m dx ≤ f (x) dx ≤ M dx a a a f (x) By Property 8, the integral of a constant function is the product of the constant and the width of m the interval. So: b x m(b−a) ≤ f (x) dx ≤ M(b−a) a a b V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 53 / 56 Notes Example 2 1 Estimate dx using the comparison properties. 1 x Solution V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 54 / 56 Summary Notes We can compute the area of a curved region with a limit of Riemann sums We can compute the distance traveled from the velocity with a limit of Riemann sums Many other important uses of this process. V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 55 / 56 18
  • 19. V63.0121.021, Calculus I Sections 5.1–5.2 : Areas, Distances, Integral December 2, 2010 Summary Notes The definite integral is a limit of Riemann Sums The definite integral can be estimated with Riemann Sums The definite integral can be distributed across sums and constant multiples of functions The definite integral can be bounded using bounds for the function V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 56 / 56 Notes Notes 19