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V63.0121.041, Calculus I                                           Section 5.4 : The Fundamental Theorem      December 8, 2010



                Section 5.4
                                                                                                      Notes
     The Fundamental Theorem of Calculus

                                   V63.0121.041, Calculus I

                                        New York University


                                      December 8, 2010


 Announcements

       Today: Section 5.4
       Monday, December 13: Section 5.5
       ”Monday,” December 15: Review and Movie Day!
       Monday, December 20, 12:00–1:50pm: Final Exam (location still
       TBD)




 Announcements
                                                                                                      Notes



          Today: Section 5.4
          Monday, December 13:
          Section 5.5
          ”Monday,” December 15:
          Review and Movie Day!
          Monday, December 20,
          12:00–1:50pm: Final Exam
          (location still TBD)




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem    December 8, 2010   2 / 33




 Objectives
                                                                                                      Notes



          State and explain the
          Fundemental Theorems of
          Calculus
          Use the first fundamental
          theorem of calculus to find
          derivatives of functions
          defined as integrals.
          Compute the average value
          of an integrable function
          over a closed interval.




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem    December 8, 2010   3 / 33




                                                                                                                             1
V63.0121.041, Calculus I                                                   Section 5.4 : The Fundamental Theorem         December 8, 2010


 Outline
                                                                                                                 Notes


 Recall: The Evaluation Theorem a/k/a 2nd FTC

 The First Fundamental Theorem of Calculus
   The Area Function
   Statement and proof of 1FTC
   Biographies

 Differentiation of functions defined by integrals
    “Contrived” examples
    Erf
    Other applications




  V63.0121.041, Calculus I (NYU)           Section 5.4 The Fundamental Theorem       December 8, 2010   4 / 33




 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




  V63.0121.041, Calculus I (NYU)           Section 5.4 The Fundamental Theorem       December 8, 2010   5 / 33




 Big time Theorem
                                                                                                                 Notes




 Theorem (The Second Fundamental Theorem of Calculus)
 Suppose f is integrable on [a, b] and f = F for another function F , then
                                               b
                                                   f (x) dx = F (b) − F (a).
                                           a




  V63.0121.041, Calculus I (NYU)           Section 5.4 The Fundamental Theorem       December 8, 2010   6 / 33




                                                                                                                                        2
V63.0121.041, Calculus I                                                      Section 5.4 : The Fundamental Theorem        December 8, 2010


 The Integral as Total Change
                                                                                                                   Notes

 Another way to state this theorem is:
                                         b
                                             F (x) dx = F (b) − F (a),
                                     a

 or the integral of a derivative along an interval is the total change between
 the sides of that interval. This has many ramifications:

 Theorem
 If v (t) represents the velocity of a particle moving rectilinearly, then
                                         t1
                                              v (t) dt = s(t1 ) − s(t0 ).
                                     t0




  V63.0121.041, Calculus I (NYU)         Section 5.4 The Fundamental Theorem           December 8, 2010   7 / 33




 The Integral as Total Change
                                                                                                                   Notes

 Another way to state this theorem is:
                                         b
                                             F (x) dx = F (b) − F (a),
                                     a

 or the integral of a derivative along an interval is the total change between
 the sides of that interval. This has many ramifications:

 Theorem
 If MC (x) represents the marginal cost of making x units of a product, then
                                                                      x
                                   C (x) = C (0) +                        MC (q) dq.
                                                                  0




  V63.0121.041, Calculus I (NYU)         Section 5.4 The Fundamental Theorem           December 8, 2010   7 / 33




 The Integral as Total Change
                                                                                                                   Notes

 Another way to state this theorem is:
                                         b
                                             F (x) dx = F (b) − F (a),
                                     a

 or the integral of a derivative along an interval is the total change between
 the sides of that interval. This has many ramifications:

 Theorem
 If ρ(x) represents the density of a thin rod at a distance of x from its end,
 then the mass of the rod up to x is
                                                              x
                                              m(x) =              ρ(s) ds.
                                                          0


  V63.0121.041, Calculus I (NYU)         Section 5.4 The Fundamental Theorem           December 8, 2010   7 / 33




                                                                                                                                          3
V63.0121.041, Calculus I                                               Section 5.4 : The Fundamental Theorem                 December 8, 2010


 My first table of integrals
                                                                                                                     Notes

         [f (x) + g (x)] dx =      f (x) dx +           g (x) dx

                    x n+1
           x n dx =       + C (n = −1)                       cf (x) dx = c      f (x) dx
                    n+1
                                                                   1
                    e x dx = e x + C                                  dx = ln |x| + C
                                                                   x
                                                                     x       ax
               sin x dx = − cos x + C                              a dx =         +C
                                                                            ln a

                 cos x dx = sin x + C                        csc2 x dx = − cot x + C

                sec2 x dx = tan x + C                      csc x cot x dx = − csc x + C
                                                                 1
             sec x tan x dx = sec x + C                    √          dx = arcsin x + C
                                                               1 − x2
                1
                     dx = arctan x + C
              1 + x2

  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem                  December 8, 2010    8 / 33




 Outline
                                                                                                                     Notes


 Recall: The Evaluation Theorem a/k/a 2nd FTC

 The First Fundamental Theorem of Calculus
   The Area Function
   Statement and proof of 1FTC
   Biographies

 Differentiation of functions defined by integrals
    “Contrived” examples
    Erf
    Other applications




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem                  December 8, 2010    9 / 33




 An area function
                                                                                                                     Notes
                                                    x
                   3
 Let f (t) = t and define g (x) =                        f (t) dt. Can we evaluate the integral
                                                0
 in g (x)?
                                         Dividing the interval [0, x] into n pieces
                                                     x                        ix
                                         gives ∆t = and ti = 0 + i∆t = . So
                                                     n                         n
                                                  x x 3 x (2x)3               x (nx)3
                                           Rn =      ·     + ·         + ··· + ·
                                                  n n3        n   n3          n   n3
                                                  x4 3
                                                = 4 1 + 23 + 33 + · · · + n 3
                                                  n
                                                  x4 1            2
                                                = 4 2 n(n + 1)
                                                  n
  0                           x
                                                  x 4 n2 (n + 1)2    x4
                                                =           4
                                                                  →
                                                         4n          4
                                         as n → ∞.
  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem              December 8, 2010       10 / 33




                                                                                                                                            4
V63.0121.041, Calculus I                                                 Section 5.4 : The Fundamental Theorem      December 8, 2010


 An area function, continued
                                                                                                            Notes




 So
                                                             x4
                                               g (x) =          .
                                                             4
 This means that
                                               g (x) = x 3 .




  V63.0121.041, Calculus I (NYU)     Section 5.4 The Fundamental Theorem       December 8, 2010   11 / 33




 The area function
                                                                                                            Notes


 Let f be a function which is integrable (i.e., continuous or with finitely
 many jump discontinuities) on [a, b]. Define
                                                         x
                                        g (x) =              f (t) dt.
                                                     a


       The variable is x; t is a “dummy” variable that’s integrated over.
       Picture changing x and taking more of less of the region under the
       curve.
       Question: What does f tell you about g ?




  V63.0121.041, Calculus I (NYU)     Section 5.4 The Fundamental Theorem       December 8, 2010   12 / 33




 Envisioning the area function
                                                                                                            Notes

 Example
 Suppose f (t) is the function graphed below:
                                       y




                                                                    g
                                                                    x
                                           2    4    6        8 10f

                        x
 Let g (x) =                f (t) dt. What can you say about g ?
                    0



  V63.0121.041, Calculus I (NYU)     Section 5.4 The Fundamental Theorem       December 8, 2010   13 / 33




                                                                                                                                   5
V63.0121.041, Calculus I                                                       Section 5.4 : The Fundamental Theorem              December 8, 2010


 features of g from f
                                                                                                                          Notes



   y
                             Interval sign monotonicity monotonicity concavity
                                      of f    of g         of f        of g
                      g
                               [0, 2]        +
                 x
     2 4 6 8 10f             [2, 4.5]        +
                             [4.5, 6]        −
                               [6, 8]        −
                              [8, 10]        −                                           →                 none

 We see that g is behaving a lot like an antiderivative of f .


  V63.0121.041, Calculus I (NYU)        Section 5.4 The Fundamental Theorem                  December 8, 2010   14 / 33




 Another Big Time Theorem
                                                                                                                          Notes



 Theorem (The First Fundamental Theorem of Calculus)
 Let f be an integrable function on [a, b] and define
                                                               x
                                           g (x) =                 f (t) dt.
                                                           a

 If f is continuous at x in (a, b), then g is differentiable at x and

                                               g (x) = f (x).




  V63.0121.041, Calculus I (NYU)        Section 5.4 The Fundamental Theorem                  December 8, 2010   15 / 33




 Proving the Fundamental Theorem
                                                                                                                          Notes
 Proof.
 Let h > 0 be given so that x + h < b. We have
                                                                           x+h
                              g (x + h) − g (x)   1
                                                =                                f (t) dt.
                                      h           h                    x

 Let Mh be the maximum value of f on [x, x + h], and let mh the minimum
 value of f on [x, x + h]. From §5.2 we have
                                                     x+h
                                   mh · h ≤                f (t) dt ≤ Mh · h
                                                 x

 So
                         g (x + h) − g (x)
                                   mh ≤    ≤ Mh .
                                 h
 As h → 0, both mh and Mh tend to f (x).

  V63.0121.041, Calculus I (NYU)        Section 5.4 The Fundamental Theorem                  December 8, 2010   16 / 33




                                                                                                                                                 6
V63.0121.041, Calculus I                                           Section 5.4 : The Fundamental Theorem      December 8, 2010


 Meet the Mathematician: James Gregory
                                                                                                      Notes




        Scottish, 1638-1675
        Astronomer and Geometer
        Conceived transcendental
        numbers and found evidence
        that π was transcendental
        Proved a geometric version
        of 1FTC as a lemma but
        didn’t take it further




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem   December 8, 2010   17 / 33




 Meet the Mathematician: Isaac Barrow
                                                                                                      Notes




        English, 1630-1677
        Professor of Greek, theology,
        and mathematics at
        Cambridge
        Had a famous student




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem   December 8, 2010   18 / 33




 Meet the Mathematician: Isaac Newton
                                                                                                      Notes




        English, 1643–1727
        Professor at Cambridge
        (England)
        Philosophiae Naturalis
        Principia Mathematica
        published 1687




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem   December 8, 2010   19 / 33




                                                                                                                             7
V63.0121.041, Calculus I                                            Section 5.4 : The Fundamental Theorem       December 8, 2010


 Meet the Mathematician: Gottfried Leibniz
                                                                                                        Notes




        German, 1646–1716
        Eminent philosopher as well
        as mathematician
        Contemporarily disgraced by
        the calculus priority dispute




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem     December 8, 2010   20 / 33




 Differentiation and Integration as reverse processes
                                                                                                        Notes
 Putting together 1FTC and 2FTC, we get a beautiful relationship between
 the two fundamental concepts in calculus.
 Theorem (The Fundamental Theorem(s) of Calculus)

   I. If f is a continuous function, then
                                                    x
                                         d
                                                        f (t) dt = f (x)
                                         dx     a

       So the derivative of the integral is the original function.
  II. If f is a differentiable function, then
                                         b
                                             f (x) dx = f (b) − f (a).
                                     a

       So the integral of the derivative of is (an evaluation of) the original
       function.
  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem     December 8, 2010   21 / 33




 Outline
                                                                                                        Notes


 Recall: The Evaluation Theorem a/k/a 2nd FTC

 The First Fundamental Theorem of Calculus
   The Area Function
   Statement and proof of 1FTC
   Biographies

 Differentiation of functions defined by integrals
    “Contrived” examples
    Erf
    Other applications




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem     December 8, 2010   22 / 33




                                                                                                                               8
V63.0121.041, Calculus I                                                          Section 5.4 : The Fundamental Theorem                   December 8, 2010


 Differentiation of area functions
                                                                                                                                  Notes
 Example
                         3x
 Let h(x) =                   t 3 dt. What is h (x)?
                     0


 Solution (Using 2FTC)
                 3x
            t4            1
 h(x) =                  = (3x)4 =           1
                                             4   · 81x 4 , so h (x) = 81x 3 .
            4    0        4

 Solution (Using 1FTC)
                                                                                                           u
 We can think of h as the composition g ◦ k, where g (u) =                                                     t 3 dt and
                                                                                                       0
 k(x) = 3x. Then h (x) = g (u) · k (x), or

             h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 .

  V63.0121.041, Calculus I (NYU)          Section 5.4 The Fundamental Theorem                   December 8, 2010        23 / 33




 Differentiation of area functions, in general
                                                                                                                                  Notes

       by 1FTC
                                                     k(x)
                                        d
                                                            f (t) dt = f (k(x))k (x)
                                        dx       a
       by reversing the order of integration:
                                  b                                    h(x)
                     d                                      d
                                      f (t) dt = −                            f (t) dt = −f (h(x))h (x)
                     dx       h(x)                          dx     b

       by combining the two above:

                     k(x)                                       k(x)                  0
           d                                 d
                              f (t) dt =                               f (t) dt +           f (t) dt
           dx     h(x)                       dx             0                        h(x)

                                                                          = f (k(x))k (x) − f (h(x))h (x)


  V63.0121.041, Calculus I (NYU)          Section 5.4 The Fundamental Theorem                   December 8, 2010        24 / 33




 Another Example
                                                                                                                                  Notes

 Example
                         sin2 x
 Let h(x) =                       (17t 2 + 4t − 4) dt. What is h (x)?
                     0


 Solution




  V63.0121.041, Calculus I (NYU)          Section 5.4 The Fundamental Theorem                   December 8, 2010        25 / 33




                                                                                                                                                         9
V63.0121.041, Calculus I                                                        Section 5.4 : The Fundamental Theorem                     December 8, 2010


 A Similar Example
                                                                                                                                  Notes

 Example
                            sin2 x
 Let h(x) =                          (17t 2 + 4t − 4) dt. What is h (x)?
                        3


 Solution




  V63.0121.041, Calculus I (NYU)            Section 5.4 The Fundamental Theorem                      December 8, 2010   26 / 33




 Compare
                                                                                                                                  Notes
 Question
 Why is
                        sin2 x                                              sin2 x
               d                                                   d
                                 (17t 2 + 4t − 4) dt =                               (17t 2 + 4t − 4) dt?
               dx   0                                              dx   3

 Or, why doesn’t the lower limit appear in the derivative?

 Answer
 Because
      sin2 x                                        3                                       sin2 x
               (17t 2 + 4t − 4) dt =                    (17t 2 + 4t − 4) dt +                        (17t 2 + 4t − 4) dt
  0                                             0                                       3

 So the two functions differ by a constant.

  V63.0121.041, Calculus I (NYU)            Section 5.4 The Fundamental Theorem                      December 8, 2010   27 / 33




 The Full Nasty
                                                                                                                                  Notes


 Example
                                                          ex
 Find the derivative of F (x) =                                sin4 t dt.
                                                        x3


 Solution




 Notice here it’s much easier than finding an antiderivative for sin4 .




  V63.0121.041, Calculus I (NYU)            Section 5.4 The Fundamental Theorem                      December 8, 2010   28 / 33




                                                                                                                                                        10
V63.0121.041, Calculus I                                            Section 5.4 : The Fundamental Theorem            December 8, 2010


 Why use 1FTC?
                                                                                                             Notes



 Question
 Why would we use 1FTC to find the derivative of an integral? It seems
 like confusion for its own sake.

 Answer

       Some functions are difficult or impossible to integrate in elementary
       terms.
       Some functions are naturally defined in terms of other integrals.




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem          December 8, 2010   29 / 33




 Erf
                                                                                                             Notes
 Here’s a function with a funny name but an important role:
                                         x
                                    2          2
                          erf(x) = √       e −t dt.
                                     π 0
 It turns out erf is the shape of the bell curve. We can’t find erf(x),
                                                       2     2
 explicitly, but we do know its derivative: erf (x) = √ e −x .
                                                        π
 Example
      d
 Find    erf(x 2 ).
      dx

 Solution




  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem          December 8, 2010   30 / 33




 Other functions defined by integrals
                                                                                                             Notes


       The future value of an asset:
                                                         ∞
                                     FV (t) =                π(s)e −rs ds
                                                     t

       where π(s) is the profitability at time s and r is the discount rate.
       The consumer surplus of a good:
                                                       q∗
                                   CS(q ∗ ) =               (f (q) − p ∗ ) dq
                                                   0

       where f (q) is the demand function and p ∗ and q ∗ the equilibrium
       price and quantity.



  V63.0121.041, Calculus I (NYU)   Section 5.4 The Fundamental Theorem          December 8, 2010   31 / 33




                                                                                                                                   11
V63.0121.041, Calculus I                                             Section 5.4 : The Fundamental Theorem           December 8, 2010


 Surplus by picture
                                                                                                             Notes
                                            consumer surplus
                         price (p)
                                             producer surplus
                                                                       supply


                          p∗                     equilibrium

                                                       market revenue

                                                                  demand f (q)


                                              q∗                   quantity (q)


  V63.0121.041, Calculus I (NYU)     Section 5.4 The Fundamental Theorem        December 8, 2010   32 / 33




 Summary
                                                                                                             Notes



       Functions defined as integrals can be differentiated using the first
       FTC:                         x
                               d
                                      f (t) dt = f (x)
                              dx a
       The two FTCs link the two major processes in calculus: differentiation
       and integration
                                            F (x) dx = F (x) + C

       Follow the calculus wars on twitter: #calcwars




  V63.0121.041, Calculus I (NYU)     Section 5.4 The Fundamental Theorem        December 8, 2010   33 / 33




                                                                                                             Notes




                                                                                                                                   12

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Lesson 26: The Fundamental Theorem of Calculus (Section 041 handout)

  • 1. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Section 5.4 Notes The Fundamental Theorem of Calculus V63.0121.041, Calculus I New York University December 8, 2010 Announcements Today: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD) Announcements Notes Today: Section 5.4 Monday, December 13: Section 5.5 ”Monday,” December 15: Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD) V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 2 / 33 Objectives Notes State and explain the Fundemental Theorems of Calculus Use the first fundamental theorem of calculus to find derivatives of functions defined as integrals. Compute the average value of an integrable function over a closed interval. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 3 / 33 1
  • 2. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 4 / 33 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 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 5 / 33 Big time Theorem Notes Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F for another function F , then b f (x) dx = F (b) − F (a). a V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 6 / 33 2
  • 3. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If v (t) represents the velocity of a particle moving rectilinearly, then t1 v (t) dt = s(t1 ) − s(t0 ). t0 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 33 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If MC (x) represents the marginal cost of making x units of a product, then x C (x) = C (0) + MC (q) dq. 0 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 33 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If ρ(x) represents the density of a thin rod at a distance of x from its end, then the mass of the rod up to x is x m(x) = ρ(s) ds. 0 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 7 / 33 3
  • 4. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 My first table of integrals Notes [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 1 e x dx = e x + C dx = ln |x| + C x x ax sin x dx = − cos x + C a dx = +C ln a cos x dx = sin x + C csc2 x dx = − cot x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 sec x tan x dx = sec x + C √ dx = arcsin x + C 1 − x2 1 dx = arctan x + C 1 + x2 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 8 / 33 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 9 / 33 An area function Notes x 3 Let f (t) = t and define g (x) = f (t) dt. Can we evaluate the integral 0 in g (x)? Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x x 3 x (2x)3 x (nx)3 Rn = · + · + ··· + · n n3 n n3 n n3 x4 3 = 4 1 + 23 + 33 + · · · + n 3 n x4 1 2 = 4 2 n(n + 1) n 0 x x 4 n2 (n + 1)2 x4 = 4 → 4n 4 as n → ∞. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 10 / 33 4
  • 5. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 An area function, continued Notes So x4 g (x) = . 4 This means that g (x) = x 3 . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 11 / 33 The area function Notes Let f be a function which is integrable (i.e., continuous or with finitely many jump discontinuities) on [a, b]. Define x g (x) = f (t) dt. a The variable is x; t is a “dummy” variable that’s integrated over. Picture changing x and taking more of less of the region under the curve. Question: What does f tell you about g ? V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 12 / 33 Envisioning the area function Notes Example Suppose f (t) is the function graphed below: y g x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 13 / 33 5
  • 6. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 features of g from f Notes y Interval sign monotonicity monotonicity concavity of f of g of f of g g [0, 2] + x 2 4 6 8 10f [2, 4.5] + [4.5, 6] − [6, 8] − [8, 10] − → none We see that g is behaving a lot like an antiderivative of f . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 14 / 33 Another Big Time Theorem Notes Theorem (The First Fundamental Theorem of Calculus) Let f be an integrable function on [a, b] and define x g (x) = f (t) dt. a If f is continuous at x in (a, b), then g is differentiable at x and g (x) = f (x). V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 15 / 33 Proving the Fundamental Theorem Notes Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h mh · h ≤ f (t) dt ≤ Mh · h x So g (x + h) − g (x) mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f (x). V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 16 / 33 6
  • 7. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Meet the Mathematician: James Gregory Notes Scottish, 1638-1675 Astronomer and Geometer Conceived transcendental numbers and found evidence that π was transcendental Proved a geometric version of 1FTC as a lemma but didn’t take it further V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 17 / 33 Meet the Mathematician: Isaac Barrow Notes English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 18 / 33 Meet the Mathematician: Isaac Newton Notes English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 19 / 33 7
  • 8. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Meet the Mathematician: Gottfried Leibniz Notes German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 20 / 33 Differentiation and Integration as reverse processes Notes Putting together 1FTC and 2FTC, we get a beautiful relationship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) I. If f is a continuous function, then x d f (t) dt = f (x) dx a So the derivative of the integral is the original function. II. If f is a differentiable function, then b f (x) dx = f (b) − f (a). a So the integral of the derivative of is (an evaluation of) the original function. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 21 / 33 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus The Area Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 22 / 33 8
  • 9. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Differentiation of area functions Notes Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 Solution (Using 1FTC) u We can think of h as the composition g ◦ k, where g (u) = t 3 dt and 0 k(x) = 3x. Then h (x) = g (u) · k (x), or h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 23 / 33 Differentiation of area functions, in general Notes by 1FTC k(x) d f (t) dt = f (k(x))k (x) dx a by reversing the order of integration: b h(x) d d f (t) dt = − f (t) dt = −f (h(x))h (x) dx h(x) dx b by combining the two above: k(x) k(x) 0 d d f (t) dt = f (t) dt + f (t) dt dx h(x) dx 0 h(x) = f (k(x))k (x) − f (h(x))h (x) V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 24 / 33 Another Example Notes Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 0 Solution V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 25 / 33 9
  • 10. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 A Similar Example Notes Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 3 Solution V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 26 / 33 Compare Notes Question Why is sin2 x sin2 x d d (17t 2 + 4t − 4) dt = (17t 2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the derivative? Answer Because sin2 x 3 sin2 x (17t 2 + 4t − 4) dt = (17t 2 + 4t − 4) dt + (17t 2 + 4t − 4) dt 0 0 3 So the two functions differ by a constant. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 27 / 33 The Full Nasty Notes Example ex Find the derivative of F (x) = sin4 t dt. x3 Solution Notice here it’s much easier than finding an antiderivative for sin4 . V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 28 / 33 10
  • 11. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Why use 1FTC? Notes Question Why would we use 1FTC to find the derivative of an integral? It seems like confusion for its own sake. Answer Some functions are difficult or impossible to integrate in elementary terms. Some functions are naturally defined in terms of other integrals. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 29 / 33 Erf Notes Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t find erf(x), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π Example d Find erf(x 2 ). dx Solution V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 30 / 33 Other functions defined by integrals Notes The future value of an asset: ∞ FV (t) = π(s)e −rs ds t where π(s) is the profitability at time s and r is the discount rate. The consumer surplus of a good: q∗ CS(q ∗ ) = (f (q) − p ∗ ) dq 0 where f (q) is the demand function and p ∗ and q ∗ the equilibrium price and quantity. V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 31 / 33 11
  • 12. V63.0121.041, Calculus I Section 5.4 : The Fundamental Theorem December 8, 2010 Surplus by picture Notes consumer surplus price (p) producer surplus supply p∗ equilibrium market revenue demand f (q) q∗ quantity (q) V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 32 / 33 Summary Notes Functions defined as integrals can be differentiated using the first FTC: x d f (t) dt = f (x) dx a The two FTCs link the two major processes in calculus: differentiation and integration F (x) dx = F (x) + C Follow the calculus wars on twitter: #calcwars V63.0121.041, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 8, 2010 33 / 33 Notes 12