The Fundamental Theorem of Calculus

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In this section, we will learn about:
The Fundamental Theorem of Calculus and its significance.
1. 5
2. 5.3
The Fundamental
Theorem of Calculus
In this section, we will learn about:
The Fundamental Theorem of Calculus
and its significance.
3. FUNDAMENTAL THEOREM OF CALCULUS
The Fundamental Theorem of Calculus
(FTC) is appropriately named.
 It establishes a connection between the two
branches of calculus—differential calculus and
integral calculus.
4. Differential calculus arose from the tangent
Integral calculus arose from a seemingly
unrelated problem—the area problem.
5. Newton’s mentor at Cambridge, Isaac Barrow
(1630–1677), discovered that these two
problems are actually closely related.
 In fact, he realized that differentiation and
integration are inverse processes.
6. The FTC gives the precise inverse
relationship between the derivative
and the integral.
7. 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 sums—as we did
in Sections 5.1 and 5.2
8. FTC Equation 1
The first part of the FTC deals with functions
defined by an equation of the form
x
g ( x)  f (t ) dt
a
where f is a continuous function on [a, b]
and x varies between a and b.
9. x
g ( x)  f (t ) dt
a
 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 x
is a definite number.  f (t ) dt
a
x
 If we then let x vary, the number f (t ) dt

also varies and defines a function aof x denoted by g(x).
10. 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.
11. FTC Example 1
If f is the function
whose graph is shown
x
and g ( x)   f (t ) dt ,
0
find the values of:
g(0), g(1), g(2), g(3),
g(4), and g(5).
 Then, sketch a rough graph of g.
12. FTC Example 1
First, we notice that:
0
g (0)  f (t ) dt 0
0
13. FTC Example 1
From the figure, we see that g(1) is
the area of a triangle:
1
g (1)  f (t ) dt
0
 (12)
1
2
1
14. FTC Example 1
To find g(2), we add to g(1) the area of
a rectangle:
2
g (2)  f (t ) dt
0
1 2
 f (t ) dt   f (t ) dt
0 1
1  (1 2)
3
15. FTC Example 1
We estimate that the area under f from 2 to 3
3
So, g (3)  g (2)  
2
f (t ) dt
3  1.3
4.3
16. FTC Example 1
For t > 3, f(t) is negative.
So, we start subtracting areas, as
17. FTC Example 1
4
g (4) g (3)   f (t ) dt 4.3  ( 1.3) 3.0
3
5
g (5)  g (4)   f (t ) dt 3  ( 1.3) 1.7
4
18. FTC Example 1
We use these values to sketch the graph
of g.
 Notice that, because f(t)
is positive for t < 3,
for t < 3.
 So, g is increasing up to
x = 3, where it attains
a maximum value.
 For x > 3, g decreases
because f(t) is negative.
19. If we take f(t) = t and a = 0, then,
using Exercise 27 in Section 5.2,
we have: 2
x x
g ( x)  tdt 
0 2
20. 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 f—at least in this case.
21. 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.
22. To see why this might be generally true, we
consider a continuous function f with f(x) ≥ 0.
x
 Then, g ( x )  f (t )dt can be interpreted as
a
the area under the graph of f from a to x.
23. To compute g’(x) from the definition of
derivative, we first observe that, for h > 0,
g(x + h) – g(x) is obtained by subtracting
 It is the area
under the graph
of f from x to x + h
(the gold area).
24. For small h, you can see that this area is
approximately equal to the area of the
rectangle with height f(x) and width h:
g ( x  h)  g ( x) hf ( x)
So, g ( x  h)  g ( x)
h
 f ( x)
25. Intuitively, we therefore expect that:
g ( x  h)  g ( x )
g '( x) lim  f ( x)
h 0 h
 The fact that this is true, even when f is not
necessarily positive, is the first part of the FTC
(FTC1).
26. If f is continuous on [a, b], then the function g
defined by x
g ( x)  f (t )dt a x b
a
is continuous on [a, b] and differentiable on
(a, b), and g’(x) = f(x).
27. 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
28. FTC1 Proof
If x and x + h are in (a, b), then
g ( x  h)  g ( x )
x h x
 f (t )dt   f (t )dt
a a
x
  f (t)dt  
a
x h
x 
f (t )dt 
x
 f (t )dt
a
(Property 5)
x h
 f (t )dt
x
29. FTC1 Proof—Equation 2
So, for h ≠ 0,
g ( x  h)  g ( x ) 1 x  h
  f (t )dt
h h x
30. FTC1 Proof
For now, let us assume that h > 0.
 Since f is continuous on [x, x + h], the Extreme Value
Theorem says that there are numbers u and v in
[x, x + h] such that f(u) = m and f(v) = M.
 m and M are the absolute
minimum and maximum
values of f on [x, x + h].
31. FTC1 Proof
By Property 8 of integrals, we have:
x h
mh  f (t ) dt Mh
x
x h
That is, f (u )h  
x
f (t ) dt  f (v)h
32. FTC1 Proof
Since h > 0, we can divide this inequality
by h:
1 x h
f (u )   f (t )dt  f (v )
h x
33. FTC1 Proof—Equation 3
Now, we use Equation 2 to replace the middle
part of this inequality:
g ( x  h)  g ( x )
f (u )   f (v )
h
 Inequality 3 can be proved in a similar manner
for the case h < 0.
34. FTC1 Proof
Now, we let h → 0.
Then, u → x and v → x, since u and v lie
between x and x + h.
 Therefore, lim f (u ) lim f (u )  f ( x)
h 0 u x
and lim f (v) lim f (v)  f ( x)
h 0 v x
because f is continuous at x.
35. FTC1 Proof—Equation 4
From Equation 3 and the Squeeze
Theorem, we conclude that:
g ( x  h)  g ( x )
g '( x) lim  f ( x)
h 0 h
36. If x = a or b, then Equation 4 can be
interpreted as a one-sided limit.
 Then, Theorem 4 in Section 2.8 (modified
for one-sided limits) shows that g is continuous
on [a, b].
37. FTC1 Proof—Equation 5
Using Leibniz notation for derivatives, we can
write the FTC1 as d x
 f (t )dt  f ( x)
dx a
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.
38. FTC1 Example 2
Find the derivative of the function
x
2
g ( x)  1  t dt
0
 As f (t )  1  t 2 is continuous, the FTC1 gives:
2
g '( x)  1  x
39. A formula of the form x
g ( x)  f (t )dt
a
may seem like a strange way of defining
a function.
 However, books on physics, chemistry, and
statistics are full of such functions.
40. FRESNEL FUNCTION Example 3
For instance, consider the Fresnel function
x
2
S ( x)  sin( t / 2) dt
0
 It is named after the French physicist Augustin Fresnel
(1788–1827), famous for his works in optics.
 It first appeared in Fresnel’s theory of the diffraction
of light waves.
 More recently, it has been applied to the design
of highways.
41. FRESNEL FUNCTION Example 3
The FTC1 tells us how to differentiate
the Fresnel function:
S’(x) = sin(πx2/2)
 This means that we can apply all the methods
of differential calculus to analyze S.
42. FRESNEL FUNCTION Example 3
The figure shows the graphs of
f(x) = sin(πx2/2) and the Fresnel function
x
S ( x)  f (t )dt
0
 A computer was used
to graph S by computing
the value of this integral
for many values of x.
43. 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).
44. FRESNEL FUNCTION Example 3
The other figure shows a larger part
of the graph of S.
45. FRESNEL FUNCTION Example 3
think about what its derivative should look like,
it seems reasonable that S’(x) = f(x).
 For instance, S is
increasing when f(x) > 0
and decreasing when
f(x) < 0.
46. FRESNEL FUNCTION Example 3
So, this gives a visual confirmation
of the FTC1.
47. FTC1 Example 4
d x4
Find  sec t dt
dx 1
 Here, we have to be careful to use the Chain Rule
in conjunction with the FTC1.
48. FTC1 Example 4
Let u = x4.
d x 4
d u
 sec t dt   sec t dt
dx 1 dx 1
d u du
  
du 1 
sec t dt
dx
(Chain Rule)
du
sec u (FTC1)
dx
4 3
sec( x ) 4 x
49. 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.
50. If f is continuous on [a, b], then
b
 f ( x)dx F (b)  F (a)
a
where F is any antiderivative of f,
that is, a function such that F’ = f.
51. FTC2 Proof
x
Let g ( x )   f (t ) dt
a
We know from the FTC1 that g’(x) = f(x),
that is, g is an antiderivative of f.
52. FTC2 Proof—Equation 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 < x < b.
53. 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.
54. FTC2 Proof
If we put x = a in the formula for g(x),
we get:
a
g (a )  f (t ) dt 0
a
55. FTC2 Proof
So, using Equation 6 with x = b and x = a,
we have:
F (b)  F (a ) [ g (b)  C ]  [ g (a)  C ]
 g (b)  g (a )
 g (b)
b
 f (t )dt
a
56. The FTC2 states that, if we know an
antiderivative F of f, then we can evaluate
b
 f ( x)dx
a
simply by subtracting the
of F at the endpoints of the interval [a, b].
57. b
It’s very surprising that  f ( x ) dx , which
a
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.
58. At first glance, the theorem may be
 However, it becomes plausible if we interpret it
in physical terms.
59. 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.
60. 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.
b
 In symbols,
 v(t ) dt s(b)  s(a)
a
 That is exactly what the FTC2 says in this context.
61. FTC2 Example 5
3
x
Evaluate the integral  e dx
1
 The function f(x) = ex is continuous everywhere
and we know that an antiderivative is F(x) = ex.
3
x
 So, the FTC2 gives:
 dx F (3)  F (1)
1
e
e3  e
62. 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.
63. We often use the notation
b
F ( x)] F (b)  F (a )
a
So, the equation of the FTC2 can be written
as: b
b
 f ( x)dx F ( x)]
a a where F '  f
b
 Other common notations are F ( x ) | and [ F ( x)]ba .
a
64. 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:
3 3 3 1
2
1 x  1 0 1
A x dx     
0 3 0 3 3 3
65. 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
66. FTC2 Example 7
dx 6
3 x
1 6
 The given integral is an abbreviation for  dx.
3 x
 An antiderivative of f(x) = 1/x is F(x) = ln |x|.
 As 3 ≤ x ≤ 6, we can write F(x) = ln x.
67. FTC2 Example 7
 Therefore, 6 1 6
3 x dx ln x]3
ln 6  ln 3
6
ln
3
ln 2
68. FTC2 Example 8
Find the area under the cosine curve
from 0 to b, where 0 ≤ b ≤ π/2.
 As an antiderivative of f(x) = cos x is F(x) = sin x,
we have:
b
A  cos x dx sin x]b0 sin b  sin 0 sin b
0
69. FTC2 Example 8
 In particular, taking b = π/2, we have
proved that the area under the cosine curve
from 0 to π/2 is:
sin(π/2) = 1
70. 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.
71. If we didn’t 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
72. It was even more difficult for
 The apparatus of limits had not been invented
in 1635.
73. 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.
74. FTC2 Example 9
What is wrong with this calculation?
1 3
3 1 x  1 4
 1 x 2
dx     1 
 1 1 3 3
75. 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
b
says that a f ( x ) dx 0 when f ≥ 0.
76. FTC2 Example 9
The FTC applies to continuous functions.
 It can’t be applied here because f(x) = 1/x2
is not continuous on [-1, 3].
 In fact, f has an infinite discontinuity at x = 0.
31
 So,  2 dx does not exist.
1 x
77. INVERSE PROCESSES
We end this section by
bringing together the two parts
of the FTC.
78. Suppose f is continuous on [a, b].
x
1.If g ( x)   f (t ) dt , then g’(x) = f(x).
a
b
2.  f ( x) dx F (b)  F (a) , where F is
a
any antiderivative of f, that is, F’ = f.
79. INVERSE PROCESSES
We noted that the FTC1 can be rewritten
as: d x
 f (t ) dt  f ( x )
dx a
 This says that, if f is integrated and then
the result is differentiated, we arrive back
at the original function f.
80. INVERSE PROCESSES
As F’(x) = f(x), the FTC2 can be rewritten
as: b
 F '( x ) dx  F (b )  F ( a )
a
 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’s in the form F(b) - F(a).
81. INVERSE PROCESSES
Taken together, the two parts of the FTC
say that differentiation and integration are
inverse processes.
 Each undoes what the other does.
82. The FTC is unquestionably the most
important theorem in calculus.
 Indeed, it ranks as one of the great
accomplishments of the human mind.
83. Before it was discovered—from the time
of Eudoxus and Archimedes to that of Galileo
and Fermat—problems of finding areas,
volumes, and lengths of curves were so
difficult that only a genius could meet
the challenge.
84. 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.