1. INTEGRALS 5

2. 5.3 The Fundamental Theorem of Calculus INTEGRALS In this section, we will learn about: The Fundamental Theorem of Calculus and its significance.

3. The Fundamental Theorem of Calculus (FTC) is appropriately named.  It establishes a connection between the two branches of calculus—differential calculus and integral calculus. FUNDAMENTAL THEOREM OF CALCULUS

4. FTC Differential calculus arose from the tangent problem. 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. FTC

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

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 FTC

8. 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. Equation 1 FTC

9.  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). FTC

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. FTC

11. 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. Example 1 FTC

12. First, we notice that: FTC Example 1

13. From the figure, we see that g(1) is the area of a triangle: Example 1 FTC

14. To find g(2), we add to g(1) the area of a rectangle: Example 1 FTC

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

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

17. Thus, FTC Example 1

18. We use these values to sketch the graph of g.  Notice that, because f(t) is positive for t < 3, we keep adding area 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. Example 1 FTC

19. If we take f(t) = t and a = 0, then, using Exercise 27 in Section 5.2, we have: FTC

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. FTC

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. FTC

22. 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. FTC

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 areas.  It is the area under the graph of f from x to x + h (the gold area). FTC

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: So, FTC

25. 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). FTC

26. 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).

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 limit. FTC1

28. If x and x + h are in (a, b), then Proof FTC1

29. So, for h ≠ 0, Proof—Equation 2 FTC1

30. 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]. Proof FTC1

31. By Property 8 of integrals, we have: That is, Proof FTC1

32. Since h > 0, we can divide this inequality by h: FTC1 Proof

33. 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 < 0. Proof—Equation 3 FTC1

34. 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. Proof FTC1

35. From Equation 3 and the Squeeze Theorem, we conclude that: Proof—Equation 4 FTC1

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]. FTC1

37. 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. Equation 5 FTC1

38. Find the derivative of the function  As is continuous, the FTC1 gives: Example 2 FTC1

39. 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. FTC1 Example 3

40. FRESNEL FUNCTION For instance, consider the Fresnel function  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. Example 3

41. FRESNEL FUNCTION The FTC1 tells us how to differentiate the Fresnel function: S’(x) = sin(πx 2 /2)  This means that we can apply all the methods of differential calculus to analyze S. Example 3

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

43. 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). Example 3 FRESNEL FUNCTION

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

45. 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) > 0 and decreasing when f(x) < 0. Example 3 FRESNEL FUNCTION

46. So, this gives a visual confirmation of the FTC1. Example 3

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

48. Let u = x 4. Then, Example 4 FTC1

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. FTC1

50. FTC2 If f is continuous on [a, b], then where F is any antiderivative of f, that is, a function such that F’ = f.

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

52. FTC2 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. Proof—Equation 6

53. FTC2 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. Proof

54. FTC2 If we put x = a in the formula for g(x), we get: Proof

55. FTC2 So, using Equation 6 with x = b and x = a, we have: Proof

56. 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].

57. FTC2 It’s 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.

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

59. 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.

60. 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.

61. FTC2 Evaluate the integral  The function f(x) = x 3 is continuous on [-2, 1] and we know from Section 4.9 that an antiderivative is F(x) = ¼x 4.  So, the FTC2 gives: Example 5

62. FTC2  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) = ¼x 4, instead of ¼x 4 + 7 or ¼x 4 + C. Example 5

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

64. FTC2 Find the area under the parabola y = x 2 from 0 to 1.  An antiderivative of f(x) = x 2 is F(x) = (1/3)x 3.  The required area is found using the FTC2: Example 6

65. 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.

66. FTC2 Find the area under the cosine curve from 0 to b, where 0 ≤ b ≤ π/2.  Since an antiderivative of f(x) = cos x is F(x) = sin x, we have: Example 7

67. FTC2 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. Example 7

68. 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.

69. FTC2 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

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

71. 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 7.

72. FTC2 What is wrong with this calculation? Example 8

73. FTC2 To start, we notice that the calculation must be wrong because the answer is negative but f(x) = 1/x 2 ≥ 0 and Property 6 of integrals says that when f ≥ 0. Example 9

74. FTC2 The FTC applies to continuous functions.  It can’t be applied here because f(x) = 1/x 2 is not continuous on [-1, 3].  In fact, f has an infinite discontinuity at x = 0.  So, does not exist. Example 9

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

76. 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.

77. 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.

78. 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’s in the form F(b) - F(a).

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

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

81. SUMMARY 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.

82. 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.