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

6. 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

7.  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

8. 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

9. 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

10. First, we notice that: FTC Example 1

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

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

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

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

15. Thus, FTC Example 1

16. 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

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

18. 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

19. 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

20. 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).

21. 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

22. 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

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

24. 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

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

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

27. 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

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

29. 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].

30. 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.

31. 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.

32. 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.

33. 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

34. 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

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

36. 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

37. 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

38. 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

39. 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

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

41. 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

42. 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

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

44. 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.

45. 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.

46. 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).

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

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