Evaluating a Definite Integral

12 Key excerpts on "Evaluating a Definite Integral"

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  Introduction to Integral Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  If we consider the shaded region of Figure 5.3, wherein the height [of the curve y ¼ f(x)] varies as we travel across the region from left to right, then the area of the region R changes continuously. This is the type of situation where integral Calculus comes into play. More complicated situations are considered in Chapter 8a of this book. 164 PREPARATION FOR THE DEFINITE INTEGRAL: THE CONCEPT OF AREA 6a The Fundamental Theorems of Calculus 6a.1 INTRODUCTION Until now, the limiting processes of the derivative and definite integral have been considered as distinct concepts. We shall now bring these fundamental ideas together and establish the relationship that exists between them. As a result, definite integrals can be evaluated more efficiently. We have defined the definite integral Ð b a f ðxÞdx, as the limit of a sum and have had some practice in estimating the integral. Calculating definite integrals this way is always tedious, usually difficult, and sometimes impossible. (1) Since evaluation of the definite integral Ð b a f ðxÞdx has a great variety of important applications, it is highly desirable to have an easy way to compute Ð b a f ðxÞdx. The purpose of this section is to develop a general method for evaluating Ð b a f ðxÞdx that does not necessitate computing various sums. The method will allow us to evaluate many (but not all) of the definite integrals that arise in applications. It turns out that the exact value of Ð b a f ðxÞdx can be easily found if we can compute Ð f ðxÞdx [i.e., if we can find the antiderivative of f (x)]. 6a.2 DEFINITE INTEGRALS In the previous chapter, we evaluated certain definite integrals using two methods: first- as the limit of a sum (which is based on the definition of definite integral) and second- by applying the second fundamental theorem of Calculus, for which it is only necessary (as we will see shortly) that one should be able to compute Ð f ðxÞdx to evaluate Ð b a f ðxÞdx.

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  Calculus

  Single Variable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Adam H. Spiegler(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  5.3 THE FUNDAMENTAL THEOREM AND INTERPRETATIONS 297 Calculating Definite Integrals: Computational Use of the Fundamental Theorem The Fundamental Theorem of Calculus owes its name to its central role in linking rates of change (derivatives) to total change. However, the Fundamental Theorem also provides an exact way of computing certain definite integrals. Example 5 Compute  3 1 2  by two different methods. Solution Using left- and right-hand sums, we can approximate this integral as accurately as we want. With  = 100, for example, the left-sum is 7.96 and the right sum is 8.04. Using  = 500 we learn 7.992 <  3 1 2  < 8.008. The Fundamental Theorem, on the other hand, allows us to compute the integral exactly. We take  () = 2. We know that if  () =  2 , then   () = 2. So we use  () = 2 and  () =  2 and obtain  3 1 2  =  (3) −  (1) = 3 2 − 1 2 = 8. Notice that to use the Fundamental Theorem to calculate a definite integral, we need to know the antiderivative,  . Chapter 6 discusses how antiderivatives are computed. Exercises and Problems for Section 5.3 Online Resource: Additional Problems for Section 5.3 EXERCISES 1. If  () is measured in dollars per year and  is measured in years, what are the units of ∫    () ? 2. If  () is measured in meters/second 2 and  is measured in seconds, what are the units of ∫    () ? 3. If  () is measured in pounds and  is measured in feet, what are the units of ∫    () ? In Exercises 4–7, explain in words what the integral repre- sents and give units. 4. ∫ 3 1 () , where () is velocity in meters/sec and  is time in seconds. 5. ∫ 6 0 () , where () is acceleration in km/hr 2 and  is time in hours. 6. ∫ 2011 2005  () , where  () is the rate at which world pop- ulation is growing in year , in billion people per year.

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  Calculus: Single and Multivariable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, David Mumford, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Ad(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  5.3 THE FUNDAMENTAL THEOREM AND INTERPRETATIONS 297 Calculating Definite Integrals: Computational Use of the Fundamental Theorem The Fundamental Theorem of Calculus owes its name to its central role in linking rates of change (derivatives) to total change. However, the Fundamental Theorem also provides an exact way of computing certain definite integrals. Example 5 Compute  3 1 2  by two different methods. Solution Using left- and right-hand sums, we can approximate this integral as accurately as we want. With  = 100, for example, the left-sum is 7.96 and the right sum is 8.04. Using  = 500 we learn 7.992 <  3 1 2  < 8.008. The Fundamental Theorem, on the other hand, allows us to compute the integral exactly. We take  () = 2. We know that if  () =  2 , then   () = 2. So we use  () = 2 and  () =  2 and obtain  3 1 2  =  (3) −  (1) = 3 2 − 1 2 = 8. Notice that to use the Fundamental Theorem to calculate a definite integral, we need to know the antiderivative,  . Chapter 6 discusses how antiderivatives are computed. Exercises and Problems for Section 5.3 Online Resource: Additional Problems for Section 5.3 EXERCISES 1. If  () is measured in dollars per year and  is measured in years, what are the units of ∫    () ? 2. If  () is measured in meters/second 2 and  is measured in seconds, what are the units of ∫    () ? 3. If  () is measured in pounds and  is measured in feet, what are the units of ∫    () ? In Exercises 4–7, explain in words what the integral repre- sents and give units. 4. ∫ 3 1 () , where () is velocity in meters/sec and  is time in seconds. 5. ∫ 6 0 () , where () is acceleration in km/hr 2 and  is time in hours. 6. ∫ 2011 2005  () , where  () is the rate at which world pop- ulation is growing in year , in billion people per year.

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  Calculus

  Single Variable

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

    (Publisher)

  273 CHAPTER 5 Applications of the Definite Integral in Geometry, Science, and Engineering In the last chapter we introduced the definite integral as the limit of Riemann sums in the context of finding areas. However, Riemann sums and definite integrals have applications that extend far beyond the area problem. In this chapter we will show how Riemann sums and definite integrals arise in such problems as finding the volume and surface area of a solid, finding the length of a plane curve, calculating the work done by a force, finding the center of gravity of a planar region, finding the pressure and force exerted by a fluid on a submerged object. Although these problems are diverse, the required calculations can all be approached by the same procedure that we used to find areas—breaking the required calculation into “small parts,” making an approximation for each part, adding the approximations from the parts to produce a Riemann sum that approximates the entire quantity to be calculated, and then taking the limit of the Riemann sums to produce an exact result. Alexander Tolstykh/Shutterstock.com 3D printers generate objects by stacking horizontal layers of the object. A calculus analogue computes the volume of a solid from the areas of its cross sections. 5.1 Area Between Two Curves In the last chapter we showed how to find the area between a curve y = f (x) and an interval on the x-axis. Here we will show how to find the area between two curves. A Review of Riemann Sums Before we consider the problem of finding the area between two curves it will be helpful to review the basic principle that underlies the calculation of area as a definite integral.

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  Mathematical Applications for the Management, Life, and Social Sciences

  • Ronald Harshbarger, James J. Reynolds(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Definite Integral Note that for some intervals, values of f may be negative. In this case, the product f (x i *)∆x i will be negative and can be thought of geometrically as a “signed area.” (Remember that area is a positive number.) Thus a definite integral can be thought of geometrically as the sum of signed areas, just as a derivative can be thought of geometrically as the slope of a tangent line. In the case where f (x) is positive for all x from a to b, the definite integral equals the area between the graph of y 5 f (x) and the x-axis. The obvious question is how this definite integral is related to the indefinite integral (the antiderivative) discussed in Chapter 12. The connection between these two concepts is the most important result in calculus because it connects derivatives, indefinite integrals, and definite integrals. To help see the connection, consider the marginal revenue function R9(x) 5 300 2 0.2x and the revenue function which is the indefinite integral of the marginal function: R(x) 5 2 (300 2 0.2x) dx 5 300x 2 0.1x 2 Figure 13.9 (a) on the next page shows the graph of R(x). Fundamental Theorem of Calculus max Δx i S 0 1 n S ∞ 2 max Δx i S 0 1 n S ∞ 2 Copyright 2019 Cengage Learning. 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.

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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)

  (See Figure 1.) ■ 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. If we didn’t have the benefit of the Evaluation Theorem, we would have to compute a difficult limit of sums using obscure trigonometric identities (or a computer algebra system as in Exercise 5.1.23). It was even more difficult for Roberval because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s, when the Evaluation Theorem was discovered by Newton and Leibniz, such problems became very easy, as you can see from Example 2. ■ Indefinite Integrals We need a convenient notation for antiderivatives that makes them easy to work with. Because of the relation given by the Evaluation Theorem between antiderivatives and integrals, the notation y f s x d dx is traditionally used for an antiderivative of f and is called an indefinite integral . Thus y f s x d dx -F s x d means F 9 s x d -f s x d You should distinguish carefully between definite and indefinite integrals. A definite integral y b a f s x d dx is a number, whereas an indefinite integral y f s x d dx is a function (or family of functions). The connection between them is given by the Evaluation Theorem: If f is continuous on f a , b g , then y b a f s x d dx -y f s x d dx g a b Recall from Section 4.6 that if F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is F s x d 1 C , where C is an arbitrary constant. For instance, the formula y 1 x dx -ln | x | 1 C is valid (on any interval that doesn’t contain 0) because s d y dx d ln | x | -1 y x . So an indefinite integral y f s x d dx can represent either a particular antiderivative of f or an entire family of antiderivatives (one for each value of the constant C ). The effectiveness of the Evaluation Theorem depends on having a supply of anti-derivatives of functions.

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  Anton's Calculus

  Early Transcendentals

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

    (Publisher)

  However, if we ignore this and mistakenly apply Formula (2) over the interval [−1, 1], we might incorrectly compute ∫ 1 −1 (1∕x 2 ) dx by evaluating an antiderivative, −1∕x, at the endpoints, arriving at the answer − 1 x ] 1 −1 = −[1 − (−1)] = −2 But f (x) = 1∕x 2 is a nonnegative function, so clearly a negative value for the definite integral is impossible. The Fundamental Theorem of Calculus can be applied without modification to definite integrals 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 5.5.3(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 5.5.5. Example 6 Evaluate ∫ 3 0 f (x) dx if f (x) = { x 2 , x < 2 3x − 2, x ≥ 2 Solution. See Figure 5.6.5. From Theorem 5.5.5 we can integrate from 0 to 2 and from 2 1 1 2 3 4 2 3 4 5 6 7 x y Figure 5.6.5 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) 298 Chapter 5 / Integration (Figure 5.6.6). To compute total area using Formula (7), begin by dividing the interval of (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 5.6.6 integration into subintervals on which f (x) does not change sign.

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  Calculus for The Life Sciences

  • Sebastian J. Schreiber, Karl J. Smith, Wayne M. Getz(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  (For more information about this institute, visit http://www.claymath.org.) So you can become a millionaire doing mathematics! Write a paper on Georg Riemann; in particular, discuss this million dollar prize. 5.3 The Definite Integral Previously we defined area under a nonnegative function as the limit of a Riemann sum. In this section we define this limit for any continuous function (positive or neg- ative) and develop its geometrical meaning as well as its properties. For a nonnegative continuous function f (x) from x = a to x = b, we defined the area under the curve as area = lim n→∞ [ f (x 1 )x + f (x 2 )x + · · · + f (x n )x] = lim n→∞ n  i =1 f (x i )x where x = b − a n and x i is a point from the interval [a + (i − 1) x, a + i x]. Theorem 5.1 from the previous section implies that lim n→∞ n  i =1 f (x i )x exists and is independent of the sample points x i whenever f is continuous. When f takes on negative values, the integral no longer corresponds to the area under the curve, but the signed area as we soon shall see. The existence of the limit is so important that Leibniz (see Historical Quest , page 538) developed the following special notation for it. Definite Integral Let f be continuous on [a, b]. Then the definite integral of f from a to b is defined to be  b a f (x) dx = lim n→∞ n  i =1 f (x i )x In the definition of the definite integral, the function f that is being integrated is called the integrand; the interval [a, b] is the interval of integration; and the end- points a and b are called, respectively, the lower and the upper limits of integration. The variable x is called the variable of integration. Notice that in taking the limit the Greek letters are supplanted by the Roman letters:  becomes a d and  becomes an elongated S. 370 Chapter 5 Integration Example 1 From sums to integrals Write the sum lim n→∞ n  i =1 sin  2π i n  2π n as a definite integral.

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  The Fundamentals of Mathematical Analysis

  • G. M. Fikhtengol'ts, I. N. Sneddon(Authors)
  • 2014(Publication Date)
  • Pergamon

    (Publisher)

  t This important proposition was first strictly proved by Cauchy (in 1823) for a function continuous in the whole interval. If we remember a geometrical interpretation of the definite integral as an area [Sec. 175], then theorem (12) will be identified with the so-called Newton-Leibniz theorem [Sec. 156]. § 2. PROPERTIES OF DEFINITE INTEGRALS 363 and hence f(x) -εζζμ /(l — A 2 sin 2 Ç?) respectively; they vanish for ψ = 0. Remark. The statements proved in this section can easily be extended to the case of an integral with a variable lower limit, since by (1), b x j/co* = -j/(o*. * b It is evident that the derivative of this integral with respect to x is equal to —f(x) if x is a point of continuity. 364 11. DEFINITE INTEGRAL § 3. Evaluation and transformation of definite integrals 184. Evaluation using integral sums. We now give some examples of the evaluation of definite integrals by the direct consideration of the limits of the integral sums.

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  Applied Calculus for the Managerial, Life, and Social Sciences

  A Brief Approach

  • Soo Tan(Author)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2015 Cengage Learning. 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. 6.5 EVALUATING DEFINITE INTEGRALS 459 When you use Method 2, make sure you adjust the limits of integration to reflect integrating with respect to the new variable u. Refer to Example 1. You can confirm the results obtained there by using a graph- ing utility as follows: 1. Use the numerical integration operation of the graphing utility to evaluate 3 4 0 x"9 1 x 2 dx 2. Evaluate 1 2 3 25 9 !u du. 3. Conclude that 3 4 0 x"9 1 x 2 dx 5 1 2 3 25 9 !u du. Exploring with TECHNOLOGY EXAMPLE 2 Evaluate 3 2 0 xe 2x 2 dx. Solution Let u 5 2x 2 , so that du 5 4x dx, or x dx 5 1 4 du. When x 5 0, u 5 0; and when x 5 2, u 5 8. These give the lower and upper limits of integration with respect to u. Making the indicated substitutions, we find 3 2 0 xe 2x 2 dx 5 3 8 0 1 4 e u du 5 1 4 e u ` 0 8 5 1 4 1 e 8 2 1 2 EXAMPLE 3 Evaluate 3 1 0 x 2 x 3 1 1 dx. Solution Let u 5 x 3 1 1, so that du 5 3x 2 dx, or x 2 dx 5 1 3 du. When x 5 0, u 5 1; and when x 5 1, u 5 2. These give the lower and upper limits of integration with respect to u. Making the indicated substitutions, we find 3 1 0 x 2 x 3 1 1 dx 5 1 3 3 2 1 du u 5 1 3 ln 0 u 0 ` 1 2 5 1 3 1 ln 2 2 ln 1 2 5 1 3 ln 2 Finding the Area Under a Curve EXAMPLE 4 Find the area of the region R under the graph of f 1 x 2 5 e (1/2)x from x 5 21 to x 5 1.

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  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(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 represented by the definite integral A = ∫ b a f(x) dx (1) (Figure 5.6.1) y = f (x) b a x y A ▴ 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. f(x) = { 2 if x ≥ 0 1 if x < 0 f(x) = { 1 x is rational −1 x is irrational 5.6 The Fundamental Theorem of Calculus 323 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 assumption is not essential.

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  Calculus

  Concepts and Contexts, Enhanced Edition

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

    (Publisher)

  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 Fundamental Theorem of Calculus gives the precise inverse relationship between the derivative and the integral. It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method. The first part of the Fundamental Theorem deals with functions defined by an equation of the form where is a continuous function on and varies between and . Observe that depends only on , which appears as the variable upper limit in the integral. If is a fixed number, then the integral is a definite number. If we then let vary, the number also varies and defines a function of denoted by . If happens to be a positive function, then can be interpreted as the area under the graph of from to , where can vary from to . (Think of as the “area so far” func-tion; see Figure 1.) 0 y t a b x area =© y=f(t) FIGURE 1 t b a x x a f t x f t x x x x a f t dt x x x a f t dt x x t b a x a , b f t x y x a f t dt 1 5.4 The Fundamental Theorem of Calculus then is the area under the graph of from 0 to [until becomes negative, at which point becomes a difference of areas]. Use part (a) to determine the value of at which starts to decrease. [Unlike the integral in Problem 2, it is impossible to evaluate the integral defining to obtain an explicit expression for .] (c) Use the integration command on your calculator or computer to estimate , , , . . . , , . Then use these values to sketch a graph of . (d) Use your graph of from part (c) to sketch the graph of using the interpretation of as the slope of a tangent line. How does the graph of compare with the graph of ? 4.

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