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5.3 The Fundamental Theorem of Calculus

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Title: 5.3 The Fundamental Theorem of Calculus


1
INTEGRALS
5.3The Fundamental Theorem of Calculus
In this section, we will learn about The
Fundamental Theorem of Calculus and its
significance.
2
FUNDAMENTAL THEOREM OF CALCULUS
  • The Fundamental Theorem of Calculus (FTC) is
    appropriately named.
  • It establishes a connection between the two
    branches of calculusdifferential calculus and
    integral calculus.

3
FTC
  • Differential calculus arose from the tangent
    problem.
  • Integral calculus arose from a seemingly
    unrelated problemthe area problem.

4
FTC
  • Newtons mentor at Cambridge, Isaac Barrow
    (16301677), discovered that these two problems
    are actually closely related.
  • In fact, he realized that differentiation and
    integration are inverse processes.

5
FTC
  • The FTC gives the precise inverse relationship
    between the derivative and the integral.

6
FTC
  • It was Newton and Leibniz who exploited this
    relationship and used it to develop calculus into
    a systematic mathematical method.
  • In particular, they saw that the FTC enabled them
    to compute areas and integrals very easily
    without having to compute them as limits of
    sumsas we did in Sections 5.1 and 5.2

7
FTC
Equation 1
  • The first part of the FTC deals with functions
    defined by an equation of the form
  • where f is a continuous function on a, b and x
    varies between a and b.

8
FTC
  • Observe that g depends only on x, which appears
    as the variable upper limit in the integral.
  • If x is a fixed number, then the integral is a
    definite number.
  • If we then let x vary, the number also varies and
    defines a function of x denoted by g(x).

9
FTC
  • If f happens to be a positive function, then g(x)
    can be interpreted as the area under the graph of
    f from a to x, where x can vary from a to b.
  • Think of g as the area so far function, as
    seen here.

10
FTC
Example 1
  • If f is the function whose graph is shown and
    , find the values of
    g(0), g(1), g(2), g(3), g(4), and g(5).
  • Then, sketch a rough graph of g.

11
FTC
Example 1
  • First, we notice that

12
FTC
Example 1
  • From the figure, we see that g(1)
  • is the area of a triangle

13
FTC
Example 1
  • To find g(2), we add to g(1) the area of a
    rectangle

14
FTC
Example 1
  • We estimate that the area under f from 2 to 3 is
    about 1.3.
  • So,

15
FTC
Example 1
  • For t gt 3, f(t) is negative.
  • So, we start subtracting areas, as follows.

16
FTC
Example 1
  • Thus,

17
FTC
Example 1
  • We use these values to sketch the graph of g.
  • Notice that, because f(t) is positive for t lt 3,
    we keep adding area for t lt 3.
  • So, g is increasing up to x 3, where it
    attains a maximum value.
  • For x gt 3, g decreases because f(t) is negative.

18
FTC
  • If we take f(t) t and a 0, then, using
    Exercise 27 in Section 5.2, we have

19
FTC
  • Notice that g(x) x, that is, g f.
  • In other words, if g is defined as the integral
    of f by Equation 1, g turns out to be an
    antiderivative of fat least in this case.

20
FTC
  • If we sketch the derivative of the function g, as
    in the first figure, by estimating slopes of
    tangents, we get a graph like that of f in the
    second figure.
  • So, we suspect that g f in Example 1 too.

21
FTC
  • To see why this might be generally true, we
    consider a continuous function f with f (x) 0.
  • Then, can be
    interpreted as the
  • area under the graph of f from a to x.

22
FTC
  • To compute g(x) from the definition of
    derivative, we first observe that, for h gt 0, g(x
    h) g(x) is obtained by subtracting areas.
  • It is the area under the graph of f from x to x
    h (the gold area).

23
FTC
  • For small h, we see that this area is
    approximately equal to the area of the rectangle
    with height f(x) and width h
  • So,

24
FTC
  • Intuitively, we therefore expect that
  • The fact that this is true, even when f is not
    necessarily positive, is the first part of the
    FTC (FTC1).

25
Fundamental Theorem of Calculus Version 1
  • If f is continuous on a, b, then the function g
    defined by
  • is continuous on a, b and differentiable on
    (a, b), and g(x) f(x).

26
FTC1
  • In words, the FTC1 says that the derivative of a
    definite integral with respect to its upper limit
    is the integrand evaluated at the upper limit.

27
FTC1
  • Using Leibniz notation for derivatives, we can
    write the FTC1 as
  • when f is continuous.
  • Roughly speaking, Equation 5 says that, if we
    first integrate f and then differentiate the
    result, we get back to the original function f.

28
FTC1
Example 2
  • Find the derivative of the function
  • As is continuous, the FTC1 gives

29
FTC1
  • A formula of the form may seem like a
    strange way of defining a function.
  • However, books on physics, chemistry, and
    statistics are full of such functions.

30
FRESNEL FUNCTION
Example 3
  • For instance, consider the Fresnel function
  • It is named after the French physicist Augustin
    Fresnel (17881827), famous for his works in
    optics.
  • It first appeared in Fresnels theory of the
    diffraction of light waves.
  • More recently, it has been applied to the design
    of highways.

31
FRESNEL FUNCTION
Example 3
  • The FTC1 tells us how to differentiate the
    Fresnel function
  • This means that we can apply all the methods of
    differential calculus to analyze S.

32
FRESNEL FUNCTION
Example 3
  • The figure shows the graphs of f (x)
    sin(px2/2) and the Fresnel function
  • A computer was used to graph S by computing the
    value of this integral for many values of x.

33
FRESNEL FUNCTION
Example 3
  • It does indeed look as if S(x) is the area under
    the graph of f from 0 to x (until x 1.4, when
    S(x) becomes a difference of areas).

34
FRESNEL FUNCTION
Example 3
  • The other figure shows a larger part of the graph
    of S.

35
FRESNEL FUNCTION
Example 3
  • If we now start with the graph of S here and
    think about what its derivative should look like,
    it seems reasonable that S(x) f(x).
  • For instance, S is increasing when f(x) gt 0 and
    decreasing when f(x) lt 0.

36
FRESNEL FUNCTION
Example 3
  • So, this gives a visual confirmation of the FTC1.

37
FTC1
Example 4
  • Find
  • Here, we have to be careful to use the Chain Rule
    in conjunction with the FTC1.

38
FTC1
Example 4
  • Let u x4. Then,

39
FTC1
  • In Section 5.2, we computed integrals from the
    definition as a limit of Riemann sums and saw
    that this procedure is sometimes long and
    difficult.
  • The second part of the FTC (FTC2), which follows
    easily from the first part, provides us with a
    much simpler method for the evaluation of
    integrals.

40
Fundamental Theorem of Calculus Version 2
  • If f is continuous on a, b, then
  • where F is any antiderivative of f, that is, a
    function such that F f.

41
FTC2
Proof
  • Let
  • We know from the FTC1 that g(x) f(x), that
    is, g is an antiderivative of f.

42
FTC2
ProofEquation 6
  • If F is any other antiderivative of f on a, b,
    then we know from Corollary 7 in Section 4.2 that
    F and g differ by a constant
  • F(x) g(x) C
  • for a lt x lt b.

43
FTC2
Proof
  • However, both F and g are continuous on a, b.
  • Thus, by taking limits of both sides of Equation
    6 (as x ? a and x ? b- ), we see it also holds
    when x a and x b.

44
FTC2
Proof
  • If we put x a in the formula for g(x), we get

45
FTC2
Proof
  • So, using Equation 6 with x b and x a, we
    have

46
FTC2
  • The FTC2 states that, if we know an
    antiderivative F of f, then we can evaluate
    simply by
    subtracting the values of F at the endpoints of
    the interval a, b.

47
FTC2
  • It is very surprising that ,
    which was defined by a complicated procedure
    involving all the values of f(x) for a x b,
    can be found by knowing the values of F(x) at
    only two points, a and b.

48
FTC2
  • At first glance, the theorem may be surprising.
  • However, it becomes plausible if we interpret it
    in physical terms.

49
FTC2
  • If v(t) is the velocity of an object and s(t) is
    its position at time t, then v(t) s(t).
  • So, s is an antiderivative of v.

50
FTC2
  • In Section 5.1, we considered an object that
    always moves in the positive direction.
  • Then, we guessed that the area under the velocity
    curve equals the distance traveled.
  • In symbols,
  • That is exactly what the FTC2 says in this
    context.

51
FTC2
Example 5
  • Evaluate the integral
  • The function f(x) ex is continuous everywhere
    and we know that an antiderivative is F(x) ex.
  • So, the FTC2 gives

52
FTC2
Example 5
  • Notice that the FTC2 says that we can use any
    antiderivative F of f.
  • So, we may as well use the simplest one, namely
    F(x) ex, instead of ex 7 or ex C.

53
FTC2
  • We often use the notation
  • So, the equation of the FTC2 can be written as

54
FTC2
Example 6
  • Find the area under the parabola y x2 from 0 to
    1.
  • An antiderivative of f(x) x2 is F(x) (1/3)x3.
  • The required area is found using the FTC2

55
FTC2
  • If you compare the calculation in Example 6 with
    the one in Example 2 in Section 5.1, you will see
    the FTC gives a much shorter method.

56
FTC2
Example 7
  • Evaluate
  • An antiderivative of f(x) 1/x is F(x) ln x.
  • As 3 x 6, we can write F(x) ln x.

57
FTC2
Example 7
  • Therefore,

58
FTC2
Example 8
  • Find the area under the cosine curve from 0 to b,
    where 0 b p/2.
  • As an antiderivative of f(x) cos x is F(x)
    sin x, we have

59
FTC2
Example 8
  • In particular, taking b p/2, we have proved
    that the area under the cosine curve from 0 to
    p/2 is sin(p/2) 1

60
FTC2
  • When the French mathematician Gilles de Roberval
    first found the area under the sine and cosine
    curves in 1635, this was a very challenging
    problem that required a great deal of ingenuity.

61
FTC2
  • If we didnt have the benefit of the FTC, we
    would have to compute a difficult limit of sums
    using either
  • Obscure trigonometric identities
  • A computer algebra system (CAS), as in Section 5.1

62
FTC2
  • It was even more difficult for Roberval.
  • The apparatus of limits had not been invented in
    1635.

63
FTC2
  • However, in the 1660s and 1670s, when the FTC was
    discovered by Barrow and exploited by Newton and
    Leibniz, such problems became very easy.
  • You can see this from Example 8.

64
FTC2
Example 9
  • What is wrong with this calculation?

65
FTC2
Example 9
  • To start, we notice that the calculation must be
    wrong because the answer is negative but
    f (x) 1/x2 0 and Property 6 of integrals
    says that when f
    0.

66
FTC2
Example 9
  • The FTC applies to continuous functions.
  • It cannot be applied here because f(x) 1/x2 is
    not continuous on -1, 3.
  • In fact, f has an infinite discontinuity at x
    0.
  • So, does not exist.

67
INVERSE PROCESSES
  • We end this section by bringing together the two
    parts of the FTC.

68
FTC
  • Suppose f is continuous on a, b.
  • 1.If , then g(x) f(x).
  • 2. , where F is any
    antiderivative of f, that is, F f.

69
INVERSE PROCESSES
  • We noted that the FTC1 can be rewritten as
  • This says that, if f is integrated and then the
    result is differentiated, we arrive back at the
    original function f.

70
INVERSE PROCESSES
  • As F(x) f(x), the FTC2 can be rewritten as
  • This version says that, if we take a function F,
    first differentiate it, and then integrate the
    result, we arrive back at the original function
    F.
  • However, it is in the form F(b) - F(a).

71
INVERSE PROCESSES
  • Taken together, the two parts of the FTC say that
    differentiation and integration are inverse
    processes.
  • Each undoes what the other does.

72
SUMMARY
  • The FTC is unquestionably the most important
    theorem in calculus.
  • Indeed, it ranks as one of the great
    accomplishments of the human mind.

73
SUMMARY
  • Before it was discoveredfrom the time of Eudoxus
    and Archimedes to that of Galileo and
    Fermatproblems of finding areas, volumes, and
    lengths of curves were so difficult that only a
    genius could meet the challenge.

74
SUMMARY
  • Now, armed with the systematic method that Newton
    and Leibniz fashioned out of the theorem, we will
    see in the chapters to come that these
    challenging problems are accessible to all of
    us.
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