INTEGRALS

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INTEGRALS
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INTEGRALS
5.3The Fundamental Theorem of Calculus
In this section, we will learn about The
Fundamental Theorem of Calculus and its
significance.
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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.

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FTC

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

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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.

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FTC

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

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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

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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.

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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).

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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.

Figure 5.3.1, p. 314
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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.

Figure 5.3.2, p. 314
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FTC
Example 1

• First, we notice that

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FTC
Example 1

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

Figure 5.3.3a, p. 314
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FTC
Example 1

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

Figure 5.3.3b, p. 314
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FTC
Example 1

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

Figure 5.3.3c, p. 314
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FTC
Example 1

• Thus,

Figure 5.3.3d, p. 314
Figure 5.3.3e, p. 314
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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.

Figure 5.3.4, p. 314
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FTC

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

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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.

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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.

Figure 5.3.4, p. 314
Figure 5.3.2, p. 314
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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.

Figure 5.3.1, p. 314
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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).

Figure 5.3.5, p. 315
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FTC

• 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
• So,

Figure 5.3.5, p. 315
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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).

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FTC1

• 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).

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FTC1
Proof

• If x and x h are in (a, b), then

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FTC1
ProofEquation 2

• So, for h ? 0,

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FTC1
Proof

• For now, let us assume that h gt 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.

Figure 5.3.6, p. 316
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FTC1
Proof

• By Property 8 of integrals, we have
• That is,

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FTC1
Proof

• Since h gt 0, we can divide this inequality by h

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FTC1
ProofEquation 3

• Now, we use Equation 2 to replace the middle part
  of this inequality
• Inequality 3 can be proved in a similar manner
  for the case h lt 0.

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FTC1
Proof

• Now, we let h ? 0.
• Then, u ? x and v ? x, since u and v lie between
  x and x h.
• Therefore, and because f is continuous at x.

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FTC1
ProofEquation 4

• From Equation 3 and the Squeeze Theorem, we
  conclude that

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FTC1
Equation 5

• 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.

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FTC1
Example 2

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

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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.

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FRESNEL FUNCTION
Example 3

• The FTC1 tells us how to differentiate the
  Fresnel function S(x) sin(px2/2)
• This means that we can apply all the methods of
  differential calculus to analyze S.

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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.

Figure 5.3.7, p. 317
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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).

Figure 5.3.7, p. 317
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FRESNEL FUNCTION
Example 3

• The other figure shows a larger part of the
  graph of S.

Figure 5.3.7, p. 317
Figure 5.3.8, p. 317
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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.

Figure 5.3.7, p. 317
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FTC1
Example 4

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

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FTC1
Example 4

• Let u x4.
• Then,

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FTC2

• If f is continuous on a, b, then
• where F is any antiderivative of f, that is, a
  function such that F f.

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FTC2
Proof

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

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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.

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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.

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FTC2
Proof

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

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FTC2
Proof

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

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FTC2
Example 5

• Evaluate the integral
• The function f(x) x3 is continuous on -2, 1
  and we know from Section 4.9 that an
  antiderivative is F(x) ¼x4.
• So, the FTC2 gives

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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) ¼x4, instead of ¼x4 7 or ¼x4 C.

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FTC2

• We often use the notation
• So, the equation of the FTC2 can be written as
• Other common notations are and
  .

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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

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FTC2
Example 7

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

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FTC2
Example 7

• 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.

Figure 5.3.9, p. 319
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FTC2
Example 8

• What is wrong with this calculation?

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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.

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FTC2
Example 9

• The FTC applies to continuous functions.
• It cant 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.

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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.

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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.

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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, its in the form F(b) - F(a).

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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.

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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.