Evaluation Theorem

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  Biocalculus

  Calculus, Probability, and Statistics for the Life Sciences

  • James Stewart, Troy Day, James Stewart(Authors)
  • 2015(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  342 CHAPTER 5 | Integrals 5.3 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calcu-lus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. 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. The Funda-mental Theorem of Calculus gives the precise inverse relationship between the deriva-tive and the integral. It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method. ■ Evaluating Definite Integrals In Section 5.2 we computed integrals from the definition as a limit of Riemann sums and we saw that this procedure is sometimes long and difficult. Newton and Leibniz realized that they could calculate y b a f s x d dx if they happened to know an antiderivative F of f . Their discovery, called the Evaluation Theorem, is part of the Fundamental Theorem of Calculus, which is discussed later in this section. Evaluation Theorem If f is continuous on the interval f a , b g , then y b a f s x d dx -F s b d 2 F s a d where F is any antiderivative of f , that is, F 9 -f . This theorem states that if we know an antiderivative F of f , then we can evaluate y b a f s x d dx simply by subtracting the values of F at the endpoints of the interval f a , b g . It is very surprising that y b a f s x d dx , which was defined by a complicated procedure involv-ing all of the values of f s x d for a < x < b , can be found by knowing the values of F s x d at only two points, a and b .

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  The Fundamental Theorem of Calculus can be applied without modification to definite inte- grals in which the lower limit of integration is greater than or equal to the upper limit of integration. Example 5  1 1 x 2 dx = x 3 3  1 1 = 1 3 − 1 3 = 0  0 4 x dx = x 2 2  0 4 = 0 2 − 16 2 = −8 The latter result is consistent with the result that would be obtained by first reversing the limits of integration in accordance with Definition 4.5.3(b):  0 4 x dx = −  4 0 x dx = − x 2 2  4 0 = −  16 2 − 0 2  = −8 To integrate a continuous function that is defined piecewise on an interval [a, b], split this interval into subintervals at the breakpoints of the function, and integrate separately over each subinterval in accordance with Theorem 4.5.5. 4.6 The Fundamental Theorem of Calculus 243 1 1 2 3 4 2 3 4 5 6 7 x y FIGURE 4.6.5 Example 6 Evaluate  3 0 f (x) dx if f (x) =  x 2 , x < 2 3x − 2, x ≥ 2 Solution See Figure 4.6.5. From Theorem 4.5.5 we can integrate from 0 to 2 and from 2 to 3 separately and add the results. This yields  3 0 f (x) dx =  2 0 f (x) dx +  3 2 f (x) dx =  2 0 x 2 dx +  3 2 (3x − 2) dx = x 3 3  2 0 +  3x 2 2 − 2x  3 2 =  8 3 − 0  +  15 2 − 2  = 49 6 If f is a continuous function on the interval [a, b], then we define the total area between the curve y = f (x) and the interval [a, b] to be total area =  b a | f (x)| dx (7) (Figure 4.6.6). To compute total area using Formula (7), begin by dividing the interval of integra- tion into subintervals on which f (x) does not change sign. On the subintervals for which 0 ≤ f (x) replace | f (x)| by f (x), and on the subintervals for which f (x) ≤ 0 replace | f (x)| by −f (x). Adding the resulting integrals then yields the total area. (a) (b) y = f (x) y =  f (x) A II A III A I a b A I A II A III a b Total area = A I + A II + A III FIGURE 4.6.6 Example 7 Find the total area between the curve y = 1 − x 2 and the x-axis over the interval [0, 2] (Figure 4.6.7).

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  THE FUNDAMENTAL THEOREM OF CALCULUS As in earlier sections, let us begin by assuming that f is nonnegative and continuous on an interval [a, b], in which case the area A under the graph of f over the interval [a, b] is y = f (x) b a x y A Figure 5.6.1 represented by the definite integral A = ∫ b a f (x) dx (1) (Figure 5.6.1). Recall that our discussion of the antiderivative method in Section 5.1 suggested that if A(x) is the area under the graph of f from a to x (Figure 5.6.2), then  A ′ (x) = f (x)  A(a) = 0 The area under the curve from a to a is the area above the single point a, and hence is zero.  A(b) = A The area under the curve from a to b is A. y = f (x) b x a x y A(x) Figure 5.6.2 The formula A ′ (x) = f (x) states that A(x) is an antiderivative of f (x), which implies that every other antiderivative of f (x) on [a, b] can be obtained by adding a constant to A(x). Accordingly, let F(x) = A(x) + C be any antiderivative of f (x), and consider what happens when we subtract F(a) from F(b): F(b) − F(a) = [A(b) + C] − [A(a) + C] = A(b) − A(a) = A − 0 = A Hence (1) can be expressed as ∫ b a f (x) dx = F(b) − F(a) In words, this equation states: The definite integral can be evaluated by finding any antiderivative of the integrand and then subtracting the value of this antiderivative at the lower limit of integration from its value at the upper limit of integration. Although our evidence for this result assumed that f is nonnegative on [a, b], this assump- tion is not essential. 5.6.1 theorem (The Fundamental Theorem of Calculus, Part 1) If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then ∫ b a f (x) dx = F(b) − F(a) (2) 5.6 The Fundamental Theorem of Calculus 295 proof Let x 1 , x 2 , … , x n−1 be any points in [a, b] such that a < x 1 < x 2 < ⋯ < x n−1 < b These values divide [a, b] into n subintervals [a, x 1 ], [x 1 , x 2 ], … , [x n−1 , b] (3) whose lengths, as usual, we denote by Δx 1 , Δx 2 , … , Δx n (see Figure 5.6.3).

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  Calculus

  Concepts and Contexts, Enhanced Edition

  • James Stewart(Author)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 371 The Fundamental Theorem of Calculus Suppose is continuous on . 1. If , then . 2. , where is any antiderivative of , that is, We noted that Part 1 can be rewritten as which says that if is integrated and then the result is differentiated, we arrive back at the original function . In Section 5.3 we reformulated Part 2 as the Net Change Theorem: This version says that if we take a function , first differentiate it, and then integrate the result, we arrive back at the original function , but in the form . Taken togeth-er, the two parts of the Fundamental Theorem of Calculus say that differentiation and inte-gration are inverse processes. Each undoes what the other does. The Fundamental Theorem of Calculus is unquestionably the most important theorem in calculus and, indeed, it ranks as one of the great accomplishments of the human mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo and Fermat, problems of finding areas, volumes, and lengths of curves were so difficult that only a genius could meet the challenge. But now, armed with the systematic method that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the chapters to come that these challenging problems are accessible to all of us. Proof of FTC1 Here we give a proof of Part 1 of the Fundamental Theorem of Calculus without assuming the existence of an antiderivative of . Let . If and are in the open interval , then and so, for , For now let’s assume that .

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  Advanced Calculus

  Theory and Practice

  • John Petrovic(Author)
  • 2013(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  15 Fundamental Theorems of Multivariable Calculus The Fundamental Theorem of Calculus expresses a relationship between the derivative and the definite integral. When a function depends on more than one variable, the connection is still there, although it is less transparent. In this chapter we will look at some multivariable generalizations of the Fundamental Theorem of Calculus. 15.1 Curves in R n Let us start with the following question. Suppose that C is a curve in the xy -plane, and that we are interested in its length. The first order of business is to define what we mean by a curve and by its length . Let f : [ a,b ] → R n . When f is continuous, we call it a path in R n . The image of f is called a curve in R n . The function f is a parameterization of the curve C . Example 15.1.1. f ( t ) = (cos t, sin t ), 0 ≤ t ≤ 2 π . This is a parameterization of the unit circle in R 2 . Example 15.1.2. g ( t ) = (cos 2 t, sin 2 t ), 0 ≤ t ≤ π . The path g traces the same curve as in Example 15.1.1. We say that two paths f : [ a,b ] → R n and g : [ c,d ] → R n are equivalent if there exists a C 1 bijection ϕ : [ a,b ] → [ c,d ] such that ϕ ′ ( t ) > 0 for all t ∈ [ a,b ] and f = g ◦ ϕ. The paths f and g in the examples above are equivalent. The bijection ϕ : [0 , 2 π ] → [0 ,π ] is given by ϕ ( t ) = t/ 2. The relation of equivalence of paths is an equivalence relation (Problem 15.1.1). Example 15.1.3. h ( t ) = (cos t, sin t ), 0 ≤ t ≤ 4 π . The curve is again the unit circle, but the path is not equivalent to those in Examples 15.1.1 and 15.1.2 (Problem 15.1.2). The obvious difference is that f traces the curve only once, whereas h loops twice around the origin. Throughout this chapter we will assume that the paths are simple , meaning that the function f is injective, with a possible exception at the endpoints. Namely, we can have f ( a ) = f ( b ), in which case we say that a path is closed . Example 15.1.4. k ( t ) = (sin t, cos t ), 0 ≤ t ≤ 2 π .

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  Calculus: An Applied Approach, Brief, International Metric Edition

  • Ron Larson(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  For an example of a definite integral, see Example 1. 2. State the Fundamental Theorem of Calculus (page 340) . For examples of the Fundamental Theorem of Calculus, see Examples 2 and 3. 3. State the properties of definite integrals (page 341) . For examples of the properties of definite integrals, see Examples 4 and 5. 4. State the definition of the average value of a function (page 345) . For an example of finding the average value of a function, see Example 7. 5. State the rules for integrating even and odd functions (page 346) . For an example of integrating even and odd functions, see Example 8. Dusan Zidar/Shutterstock.com Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 348 Chapter 5 Integration and Its Applications Exercises 5.4 See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises. Evaluating a Definite Integral Using a Geometric Formula In Exercises 1–6, sketch the region whose area is represented by the definite integral. Then use a geometric formula to evaluate the integral. See Example 1. 1. integral.alt1 3 1 5 d x 2. integral.alt1 5 2 4 d x 3. integral.alt1 4 0 x d x 4. integral.alt1 3 0 x 3 d x 5. integral.alt1 3 -3 radical.alt2 9 -x 2 d x 6. integral.alt1 4 -4 radical.alt2 16 -x 2 d x Using Properties of Definite Integrals In Exercises 7 and 8, use the values ∫ 5 0 f ( x ) dx = 6 and ∫ 5 0 g ( x ) dx = 2 to evaluate each definite integral. 7. (a) integral.alt1 5 0 [ f ( x ) + g ( x )] d x (b) integral.alt1 5 0 [ f ( x ) -g ( x )] d x (c) integral.alt1 5 0 -4 f ( x ) d x (d) integral.alt1 5 0 [ f ( x ) -3 g ( x )] d x 8. (a) integral.alt1 5 0 2 g ( x ) d x (b) integral.alt1 0 5 f ( x ) d x (c) integral.alt1 5 5 f ( x ) d x (d) integral.alt1 5 0 [ f ( x ) -f ( x )] d x Finding Area by the Fundamental Theorem In Exercises 9–16, find the area of the region. See Example 2. 9. y = x -x 2 10. y = 1 -x 4 1 1 4 x y 1 -1 x 2 y 11.

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  Theoretical Introduction to Mathematical Theorems, A

  • (Author)
  • 2014(Publication Date)
  • Learning Press

    (Publisher)

  The first part of the theorem, sometimes called the first fundamental theorem of calculus , shows that an indefinite integration can be reversed by a differentiation. The first part is also important because it guarantees the existence of antiderivatives for continuous functions. The second part, sometimes called the second fundamental theorem of calculus , allows one to compute the definite integral of a function by using any one of its infinitely many antiderivatives. This part of the theorem has invaluable practical applications, because it markedly simplifies the computation of definite integrals. The first published statement and proof of a restricted version of the fundamental theorem was by James Gregory (1638–1675). Isaac Barrow (1630–1677) proved the first completely general version of the theorem, while Barrow's student Isaac Newton (1643– 1727) completed the development of the surrounding mathematical theory. Gottfried Leibniz (1646–1716) systematized the knowledge into a calculus for infinitesimal quantities. Physical intuition Intuitively, the theorem simply states that the sum of infinitesimal changes in a quantity over time (or over some other quantity) adds up to the net change in the quantity. In the case of a particle traveling in a straight line, its position, x , is given by x ( t ) where t is time and x ( t ) means that x is a function of t . The derivative of this function is equal to the infinitesimal change in quantity, d x , per infinitesimal change in time, d t (of course, the derivative itself is dependent on time). This change in displacement per change in time is the velocity v of the particle. In Leibniz's notation: Rearranging this equation, it follows that: By the logic above, a change in x (or Δ x ) is the sum of the infinitesimal changes d x . It is also equal to the sum of the infinitesimal products of the derivative and time.

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  Mathematical Proofs and Theorems

  • (Author)
  • 2014(Publication Date)
  • Library Press

    (Publisher)

  Computing the derivative of a function and “finding the area” under its curve are opposite operations. This is the crux of the Fundamental Theorem of Calculus. Most of the theorem's proof is devoted to showing that the area function A ( x ) exists in the first place. Formal statements There are two parts to the Fundamental Theorem of Calculus. Loosely put, the first part deals with the derivative of an antiderivative, while the second part deals with the relationship between antiderivatives and definite integrals. First part This part is sometimes referred to as the First Fundamental Theorem of Calculus . Let ƒ be a continuous real-valued function defined on a closed interval [ a , b ]. Let F be the function defined, for all x in [ a , b ], by Then, F is continuous on [ a , b ], differentiable on the open interval ( a , b ), and for all x in ( a , b ). Corollary The fundamental theorem is often employed to compute the definite integral of a function ƒ for which an antiderivative g is known. Specifically, if ƒ is a real-valued continuous function on [ a , b ], and g is an antiderivative of ƒ in [ a , b ], then ______________________________ WORLD TECHNOLOGIES ______________________________ The corollary assumes continuity on the whole interval. This result is strengthened slightly in the following theorem. Second part This part is sometimes referred to as the Second Fundamental Theorem of Calculus or the Newton-Leibniz Axiom . Let ƒ be a real-valued function defined on a closed interval [ a , b ] that admits an antiderivative g on [ a , b ]. That is, ƒ and g are functions such that for all x in [ a , b ], If ƒ is integrable on [ a , b ] then Notice that the Second part is somewhat stronger than the Corollary because it does not assume that ƒ is continuous. Note that when an antiderivative g exists, then there are infinitely many antiderivatives for ƒ , obtained by adding to g an arbitrary constant.

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