DOCSLIB.ORG

Improper Integrals and Arise in a Variety of Practical Situations

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Improper Integrals and Arise in a Variety of Practical Situations 472 Chapter 6 Further Topics in Integration 4 This section extends the concept of the definite integral to integrals of the form ϩϱ Improper ͵ f(x) dx Integrals a in which the upper limit of integration is not a finite number. Such integrals are known as improper integrals and arise in a variety of practical situations. ϩϱ GEOMETRIC INTERPRETATION If f is nonnegative, the improper integral ͵ f(x) dx can be interpreted as the area of a the region under the graph of f to the right of x ϭ a (Figure 6.21a). Although this region has infinite extent, its area may be either finite or infinite, depending on how quickly f(x) approaches zero as x increases. yy y = f(x) y = f(x) yyy xx a aN (a) (b) yyy ϱ N FIGURE 6.21 Area ͵ f(x) dx ϭ lim ͵ f(x) dx. ϩϱ a Nfi a A reasonable strategy for finding the area of such a region is to first use a defi- nite integral to compute the area from x ϭ a to some finite number x ϭ N and then to let N approach infinity in the resulting expression. That is, N Total area ϭ lim (area from a to N) ϭ lim ͵ f(x) dx ϩ ϩ Nfi ϱ Nfi ϱ a This strategy is illustrated in Figure 6.21b and motivates the following definition of the improper integral. The Improper Integral I If f(x) is continuous for x Ն a, then ϩϱ N ͵ ϭ ͵ f(x) dx lim f(x) dx ϩ a Nfi ϱ a Chapter 6 I Section 4 Improper Integrals 473 If the limit defining the improper integral is a finite number, the integral is said to converge. Otherwise the integral is said to diverge. Here are some examples. EXAMPLE 4.1 ϩϱ ͵ 1 Evaluate 2 dx. 1 x Solution First compute the integral from 1 to N and then let N approach infinity. Arrange your work as follows: ϩϱ N N ͵ 1 ϭ ͵ 1 ϭ Ϫ1 ϭ Ϫ 1 ϩ ϭ dx lim dx lim lim 1 1 2 ϩϱ 2 ϩϱ ΂ ΃ ϩϱ ΂ ΃ 1 x Nfi 1 x Nfi x Խ1 Nfi N Refer to Example 4.1. Place EXAMPLE 4.2 1 ϩϱ f(x) ϭ into Y1 of the 1 x2 Evaluate ͵ dx. x equation editor and store the 1 numerical integral into Y2, using Solution the interval from x ϭ 1 to X. Set ϩϱ 1 N 1 N the table feature to start at x ϭ ͵ dx ϭ lim ͵ dx ϭ lim ln ԽxԽ ϭ lim ln N ϭϩϱ ϩϱ ϩϱ ΂ ΃ ϩϱ 1 x Nfi 1 x Nfi Խ1 Nfi 500 increasing in increments of 500. Explain what you observe. 1 Notice that the improper integral of the function f(x) ϭ in Example 4.1 x2 1 converged, while that of the function f(x) ϭ in Example 4.2 diverged. In geometric x 1 terms, this says that the area to the right of x ϭ 1 under the curve y ϭ is finite, x2 1 while the corresponding area under the curve y ϭ is infinite. The reason for the x 1 1 difference is that, as x increases, approaches zero more quickly than does (see x2 x Figure 6.22). 474 Chapter 6 Further Topics in Integration y 3 2 1 1 y = x Finite area 1 Infinite area y = x2 x 1 1 1 FIGURE 6.22 The areas under y ϭ and y ϭ . x x2 It can be shown that lim xm eϪkx ϭ 0 xfi ϩϱ for all m Ͼ 0 and k Ͼ 0. This kind of limit occurs frequently in practical problems involving improper integrals. Here is an example. EXAMPLE 4.3 ϩϱ Evaluate ͵ xeϪ2x dx. 0 Solution ϩϱ N ͵ Ϫ2x ϭ ͵ Ϫ2x xe dx lim xe dx ϩ 0 Nfi ϱ 0 N N 1 Ϫ 1 Ϫ ϭ lim Ϫ xe 2x ϩ ͵ e 2x dx (integration by parts) ϩϱ ΂ ΃ Nfi 2 Խ0 2 0 N 1 Ϫ2x 1 Ϫ2x ϭ lim Ϫ xe Ϫ e ϩϱ ΂ ΃ Nfi 2 4 Խ0 Chapter 6 I Section 4 Improper Integrals 475 1 1 1 ϭ lim ΂Ϫ NeϪ2N Ϫ eϪ2N ϩ 0 ϩ ΃ Nfi ϩϱ 2 4 4 1 ϭ (since NeϪ2N fi 0) 4 APPLICATIONS OF THE The following applications of the improper integral generalize applications of the def- IMPROPER INTEGRAL inite integral that you saw in Sections 2 and 3. In each, the strategy is to use the char- acterization of the definite integral as the limit of a sum to construct an appropriate definite integral and then to let the upper limit of integration increase without bound. As you read these examples, you may want to refer back to the corresponding exam- ples in Sections 2 and 3. PRESENT VALUE In Example 2.2 of Section 2, you saw that the present value of an investment that generates income over a finite period of time is given by a definite integral. The pres- ent value of an investment that generates income in perpetuity is given by an improper integral. Here is an example. EXAMPLE 4.4 A donor wishes to make a gift to a private college from which the college will draw $7,000 per year in perpetuity to support the operation of its computer center. Assum- ing that the prevailing annual interest rate will remain fixed at 10% compounded con- tinuously, how much should the donor give the college? That is, what is the present value of the endowment? Solution To find the present value of a gift that generates $7,000 per year for N years, divide the N-year time interval 0 Յ t Յ N into n subintervals of length ⌬nt years and let tj denote the beginning of the jth subinterval (Figure 6.23). –0.1tj ∆ ∆ 7,000e nt 7,000 nt t 0 tj N ∆ nt FIGURE 6.23 The present value of the money generated during the jth subinterval. 476 Chapter 6 Further Topics in Integration Then Amount generated during 7,000 ⌬ t jth subinterval n Present value of amount Ϫ0.1tj and generated during 7,000e ⌬nt jth subinterval n Present value of Ϫ0.1tj Hence, ϭ lim ͚ 7,000e ⌬nt ϩ N-year gift nfi ϱ jϭ1 N ϭ ͵ 7,000eϪ0.1t dt 0 To find the present value of the total gift, take the limit of this integral as N approaches infinity. That is, N Present value of ϭ ͵ Ϫ0.1t lim 7,000e dt ϩ total gift Nfi ϱ 0 N Ϫ0.1t ϭ lim Ϫ70,000e ϩϱ ΂ ΃ Nfi Խ0 Ϫ0.1N ϭ lim Ϫ70,000(e Ϫ 1) Nfi ϩϱ ϭ $70,000 NUCLEAR WASTE The next example is similar in structure to the problem involving survival and renewal functions in Example 3.2 of Section 3. EXAMPLE 4.5 It is estimated that t years from now, a certain nuclear power plant will be producing radioactive waste at the rate of f(t) ϭ 400t pounds per year. The waste decays exponentially at the rate of 2% per year. What will happen to the accumulation of radioactive waste from the plant in the long run? Solution To find the amount of radioactive waste present after N years, divide the N-year inter- val 0 Յ t Յ N into n equal subintervals of length ⌬nt years and let tj denote the begin- ning of the jth subinterval (Figure 6.24). Then, Amount of waste produced 400t ⌬ t during jth subinterval j n Chapter 6 I Section 4 Improper Integrals 477 N – tj years ∆ –0.02(N – t ) ∆ 400tj nt 400tj e j nt t 0 tj N ∆ nt FIGURE 6.24 Radioactive waste generated during the jth subinterval. Since the waste decays exponentially at the rate of 2% per year, and since there are N Ϫ tj years between times t ϭ tj and t ϭ N, it follows that Ϫ0.02(NϪtj) Amount of waste produced Ӎ 400tje ⌬nt during jth subinterval still present at t ϭ N Thus, n Amount of waste present Ϫ0.02(NϪtj) ϭ lim ͚ 400tje ⌬nt ϩ after N years nfi ϱ jϭ1 N ϭ ͵ Ϫ0.02(NϪt) 400te dt 0 N ϭ Ϫ0.02N ͵ 0.02t 400e te dt 0 The amount of radioactive waste present in the long run is the limit of this expres- sion as N approaches infinity. That is, Amount of waste N ϭ Ϫ0.02N ͵ 0.02t present in the lim 400e te dt ϩϱ long run Nfi 0 N Ϫ0.02N 0.02t 0.02t ϭ lim 400e (50te Ϫ 2,500e ) ϩϱ Nfi Խ0 Ϫ0.02N 0.02N 0.02N ϭ lim 400e (50Ne Ϫ 2,500e ϩ 2,500) Nfi ϩϱ Ϫ0.02N ϭ lim 400(50N Ϫ 2,500 ϩ 2,500e ) Nfi ϩϱ ϭϩϱ That is, in the long run, the accumulation of radioactive waste from the plant will increase without bound. 478 Chapter 6 Further Topics in Integration PROBABILITY DENSITY Improper integrals also appear in the study of probability and statistics, which is an FUNCTIONS important tool in certain areas of the social, managerial, and life sciences. We close this section with a brief introduction to this application. For example, the life span of a lightbulb selected at random from a manufac- turer’s stock is a quantity that cannot be predicted with certainty. The process of selecting a bulb is called a random experiment, and the life span X of the bulb is a continuous random variable.
Load more

Recommended publications
  • MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79
    MATH 305 Complex Analysis, Spring 2016 Using Residues to Evaluate Improper Integrals Worksheet for Sections 78 and 79 One of the interesting applications of Cauchy's Residue Theorem is to find exact values of real improper integrals. The idea is to integrate a complex rational function around a closed contour C that can be arbitrarily large. As the size of the contour becomes infinite, the piece in the complex plane (typically an arc of a circle) contributes 0 to the integral, while the part remaining covers the entire real axis (e.g., an improper integral from −∞ to 1). An Example Let us use residues to derive the formula p Z 1 x2 2 π 4 dx = : (1) 0 x + 1 4 Note the somewhat surprising appearance of π for the value of this integral. z2 First, let f(z) = and let C = L + C be the contour that consists of the line segment L z4 + 1 R R R on the real axis from −R to R, followed by the semi-circle CR of radius R traversed CCW (see figure below). Note that C is a positively oriented, simple, closed contour. We will assume that R > 1. Next, notice that f(z) has two singular points (simple poles) inside C. Call them z0 and z1, as shown in the figure. By Cauchy's Residue Theorem. we have I f(z) dz = 2πi Res f(z) + Res f(z) C z=z0 z=z1 On the other hand, we can parametrize the line segment LR by z = x; −R ≤ x ≤ R, so that I Z R x2 Z z2 f(z) dz = 4 dx + 4 dz; C −R x + 1 CR z + 1 since C = LR + CR.
    [Show full text]
  • Section 8.8: Improper Integrals
    Section 8.8: Improper Integrals One of the main applications of integrals is to compute the areas under curves, as you know. A geometric question. But there are some geometric questions which we do not yet know how to do by calculus, even though they appear to have the same form. Consider the curve y = 1=x2. We can ask, what is the area of the region under the curve and right of the line x = 1? We have no reason to believe this area is finite, but let's ask. Now no integral will compute this{we have to integrate over a bounded interval. Nonetheless, we don't want to throw up our hands. So note that b 2 b Z (1=x )dx = ( 1=x) 1 = 1 1=b: 1 − j − In other words, as b gets larger and larger, the area under the curve and above [1; b] gets larger and larger; but note that it gets closer and closer to 1. Thus, our intuition tells us that the area of the region we're interested in is exactly 1. More formally: lim 1 1=b = 1: b − !1 We can rewrite that as b 2 lim Z (1=x )dx: b !1 1 Indeed, in general, if we want to compute the area under y = f(x) and right of the line x = a, we are computing b lim Z f(x)dx: b !1 a ASK: Does this limit always exist? Give some situations where it does not exist. They'll give something that blows up.
    [Show full text]
  • Math 1B, Lecture 11: Improper Integrals I
    Math 1B, lecture 11: Improper integrals I Nathan Pflueger 30 September 2011 1 Introduction An improper integral is an integral that is unbounded in one of two ways: either the domain of integration is infinite, or the function approaches infinity in the domain (or both). Conceptually, they are no different from ordinary integrals (and indeed, for many students including myself, it is not at all clear at first why they have a special name and are treated as a separate topic). Like any other integral, an improper integral computes the area underneath a curve. The difference is foundational: if integrals are defined in terms of Riemann sums which divide the domain of integration into n equal pieces, what are we to make of an integral with an infinite domain of integration? The purpose of these next two lecture will be to settle on the appropriate foundation for speaking of improper integrals. Today we consider integrals with infinite domains of integration; next time we will consider functions which approach infinity in the domain of integration. An important observation is that not all improper integrals can be considered to have meaningful values. These will be said to diverge, while the improper integrals with coherent values will be said to converge. Much of our work will consist in understanding which improper integrals converge or diverge. Indeed, in many cases for this course, we will only ask the question: \does this improper integral converge or diverge?" The reading for today is the first part of Gottlieb x29:4 (up to the top of page 907).
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Math 113 Lecture #16 §7.8: Improper Integrals
    Math 113 Lecture #16 x7.8: Improper Integrals Improper Integrals over Infinite Intervals. How do we make sense of and evaluate an improper integral like Z 1 1 2 dx? 1 x Can we apply the Fundamental Theorem of Calculus to evaluate this? No, because the FTC only applies to integrals on finite intervals! So instead we convert the integral over an infinite interval into a limit of integrals over finite intervals: Z A 1 lim 2 dx: A!1 1 x Now we can apply the FTC to get Z A 1 1 A 1 lim dx = lim − = lim − + 1 = 1: A!1 2 A!1 A!1 1 x x 1 A Since this limit exists, we say that the improper integral converges, and the value of this limit we take as the value of the improper integral: Z 1 1 2 dx = 1: 1 x Z t Definition. If f(x) dx exists for every t ≥ a, then the improper integral a Z 1 Z A f(x) dx = lim f(x) dx a A!1 a converges provided the limit exists (as a finite number); otherwise it diverges. Z b If f(x) dx exists for every t ≤ b, then the improper integral t Z b Z b f(x) dx = lim f(x) dx 1 B!1 B converges provided the limit exists (as a finite number); otherwise it diverges. Z 1 Z a If both improper integrals f(x) dx and f(x) dx are convergent, then we define a −∞ Z 1 Z 1 Z a f(x) dx = f(x) dx + f(x) dx: −∞ a −∞ The choice of a is arbitrary.
    [Show full text]
  • Lecture 18 : Improper Integrals
    1 Lecture 18 : Improper integrals R b We de¯ned a f(t)dt under the conditions that f is de¯ned and bounded on the bounded interval [a; b]. In this lecture, we will extend the theory of integration to bounded functions de¯ned on unbounded intervals and also to unbounded functions de¯ned on bounded or unbounded intervals. ImproperR integral of the ¯rst kind: SupposeR f is (Riemann) integrable on [a; x] for all x > a, i.e., x f(t)dt exists for all x > a. If lim x f(t)dt = L for some L 2 R; then we say that a x!1 a R 1 R 1 the improper integral (of the ¯rst kind) a f(t)dt converges to L and we write a f(t)dt = L. R 1 Otherwise, we say that the improper integral a f(t)dt diverges. Observe that the de¯nition of convergence of improper integrals is similar to the one given for R x series. For example, a f(t)dt; x > a is analogous to the partial sum of a series. R 1 1 R x 1 1 Examples : 1. The improper integral 1 t2 dt converges, because, 1 t2 dt = 1 ¡ x ! 1 as x ! 1: R 1 1 R x 1 On the other hand, 1 t dt diverges because limx!1 1 t dt = limx!1 log x. In fact, one can show R 1 1 1 that 1 tp dt converges to p¡1 for p > 1 and diverges for p · 1.
    [Show full text]
  • Improper Integrals
    330 Chapter 3 | Techniques of Integration 3.7 | Improper Integrals Learning Objectives 3.7.1 Evaluate an integral over an infinite interval. 3.7.2 Evaluate an integral over a closed interval with an infinite discontinuity within the interval. 3.7.3 Use the comparison theorem to determine whether a definite integral is convergent. 1 Is the area between the graph of f (x) = x and the x-axis over the interval [1, +∞) finite or infinite? If this same region is revolved about the x-axis, is the volume finite or infinite? Surprisingly, the area of the region described is infinite, but the volume of the solid obtained by revolving this region about the x-axis is finite. In this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity on the interval. Integrals of these types are called improper integrals. We examine several techniques for evaluating improper integrals, all of which involve taking limits. Integrating over an Infinite Interval +∞ t How should we go about defining an integral of the type f (x)dx? We can integrate f (x)dx for any value of ∫a ∫a t, so it is reasonable to look at the behavior of this integral as we substitute larger values of t. Figure 3.17 shows that t f (x)dx may be interpreted as area for various values of t. In other words, we may define an improper integral as a ∫a limit, taken as one of the limits of integration increases or decreases without bound. Figure 3.17 To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.
    [Show full text]
  • Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose Fx( ) Is Continuous Anti-Derivative : an Anti-Derivative of Fx( ) on [Ab, ]
    Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose fx( ) is continuous Anti-Derivative : An anti-derivative of fx( ) on [ab, ] . Divide [ab, ] into n subintervals of is a function, Fx( ) , such that Fx′( ) = fx( ) . width ∆ x and choose x* from each interval. = + i Indefinite Integral : ∫ f( x) dx F( x) c b n Then f( x) dx= lim f x* ∆ x . where Fx( ) is an anti-derivative of fx( ) . ∫a n→∞ ∑ ( i ) i=1 Fundamental Theorem of Calculus Part I : If fx( ) is continuous on [ab, ] then Variants of Part I : d ux( ) x f( t) dt= u′( x) f u( x) g( x) = f( t) dt is also continuous on [ab, ] ∫a ∫a dx d x d b and g′( x) = f( t) dt= f( x) . ∫ f( t) dt= − v′( x) f v( x) dx ∫a dx vx( ) d ux( ) Part II : fx( ) is continuous on[ab, ] , Fx( ) is ftdtuxf( ) = ′′( ) [ux()]− vxf( ) [vx()] dx ∫vx( ) an anti-derivative of fx( ) (i.e. F( x) = ∫ f( x) dx ) b then f( x) dx= F( b) − F( a) . ∫a Properties ∫f( x) ±= g( x) dx ∫∫ f( x) dx ± g( x) dx ∫∫cf( x) dx= c f( x) dx , c is a constant b bb bb f( x) ±= g( x) dx f( x) dx ± g( x) dx cf( x) dx= c f( x) dx , c is a constant ∫a ∫∫ aa ∫∫aa a b f( x) dx = 0 c dx= c( b − a) ∫a ∫a ba bb f( x) dx= − f( x) dx f( x) dx≤ f( x) dx ∫∫ab ∫∫aa b cb f( x) dx= f( x) dx + f( x) dx for any value of c.
    [Show full text]
  • 7. Integration Techniques
    7. Integration Techniques Integration by Substitution Each basic rule of integration that you have studied so far was derived from a corresponding differentiation rule. Even though you have learned all the necessary tools for differentiating exponential, logarithmic, trigonometric, and algebraic functions, your set of tools for integrating these functions is not yet complete. In this chapter we will explore different ways of integrating functions and develop several integration techniques that will greatly expand the set of integrals to which the basic integration formulas can be applied. Before we do that, let us review the basic integration formulas that you are already familiar with from previous chapters. 1. The Power Rule : 2. The General Power Rule : 3. The Simple Exponential Rule: 4. The General Exponential Rule: 5. The Simple Log Rule: 6. The General Log Rule: It is important that you remember the above rules because we will be using them extensively to solve more complicated integration problems. The skill that you need to develop is to determine which of these basic 325 rules is needed to solve an integration problem. Learning Objectives A student will be able to: • Compute by hand the integrals of a wide variety of functions by using the technique of -substitution. • Apply the -substitution technique to definite integrals. • Apply the -substitution technique to trig functions. Probably one of the most powerful techniques of integration is integration by substitution. In this technique, you choose part of the integrand to be equal to a variable we will call and then write the entire integrand in terms of The difficulty of the technique is deciding which term in the integrand will be best for substitution by However, with practice, you will develop a skill for choosing the right term.
    [Show full text]
  • Sufficient Conditions for a Summability Method for Improper Lebesgue
    Iowa State University Capstones, Theses and Retrospective Theses and Dissertations Dissertations 1961 Sufficient conditions for a summability method for improper Lebesgue-Stieltjes integrals involving an integral-to-function transformation to be regular Marvin Glen Mundt Iowa State University Follow this and additional works at: https://lib.dr.iastate.edu/rtd Part of the Mathematics Commons Recommended Citation Mundt, Marvin Glen, "Sufficient conditions for a summability method for improper Lebesgue-Stieltjes integrals involving an integral- to-function transformation to be regular " (1961). Retrospective Theses and Dissertations. 1979. https://lib.dr.iastate.edu/rtd/1979 This Dissertation is brought to you for free and open access by the Iowa State University Capstones, Theses and Dissertations at Iowa State University Digital Repository. It has been accepted for inclusion in Retrospective Theses and Dissertations by an authorized administrator of Iowa State University Digital Repository. For more information, please contact [email protected]. f This dissertation has been 62—1363 microfilmed exactly as received MUNDT, Marvin Glen, 1933- SUFFICEENT CONDITIONS FOR A SUMMABILITY METHOD FOR IMPROPER LEBESGUE-STIELTJES INTEGRALS INVOLVING AN INTEGRAL-TO- FUNCTION TRANSFORMATION TO BE REGULAR. Iowa State University of Science and Technology Ph.D., 1961 ^iîniversity^icrofilms, Inc., Ann Arbor, Michigan SUFFICIENT CONDITIONS FOR A SUMMABILITY METHOD FOR IMPROPER LEBESGUE-STIELTJES INTEGRALS INVOLVING AN INTEGRAL-TO-FUNCTION TRANSFORMATION TO BE REGULAR Marvin Glen Mundt A Dissertation Submitted to the Graduate Faculty in Partial Fulfillment of The Requirements for the Degree of DOCTOR OF PHILOSOPHY Major Subject: Mathematics Approved: Signature was redacted for privacy. In Charge of Major'Work Signature was redacted for privacy.
    [Show full text]
  • 4.10 Improper Integrals
    4.10. IMPROPER INTEGRALS 237 4.10 Improper Integrals 4.10.1 Introduction b In Calculus II, students defined the integral a f (x) dx over a finite interval [a, b]. The function f was assumed to be continuous, or at least bounded, otherwise the integral was not guaranteed to exist. UnderR these conditions, assuming an b antiderivative of f could be found, a f (x) dx always existed, and was a number. In this section, we investigate what happens when these conditions are not met. R Definition 337 (Improper Integral) An integral is an improper integral if either the interval of integration is not finite (improper integral of type 1) or if the function to integrate is not continuous (not bounded that is becomes infinite) in the interval of integration (improper integral of type 2). 1 x Example 338 e dx is an improper integral of type 1 since the upper limit 0 of integration isZ infinite. 1 dx 1 Example 339 is an improper integral of type 2 because is not con- 0 x x tinuous at 0. Z dx Example 340 1 is an improper integral of type 1 since the upper limit 0 x 1 1 of integration isR infinite. It is also an improper integral of type 2 because x 1 is not continuous at 1 and 1 is in the interval of integration. 2 dx 1 Example 341 2 is an improper integral of type 2 because 2 is 2 x 1 x 1 not continuous atZ 1 and 1. Example 342 tan xdx is an improper integral of type 2 because tan x is Z0 not continuous at .
    [Show full text]
  • 08A-Improper Integral Practice Solutions
    Improper Integral Practice Problems These problems are taken from old quizzes I have given on improper integrals. Solutions will be posted on the course webpage later, so you can use these to gauge your preparedness for the quiz. 1) Evaluate each improper integral below using antiderivatives. Make sure your work is correct! e ∞ ⌠ dx ⌠ dx a) ⎮ Improper at x = 1 b) ⎮ Improper at infinity x ln(x) x ln(x) ⌡1 ⌡e So this integral converges to 2/. which diverges. ∞ ⌠ dx π 2 c) ⎮ Improper at 0 and infinity d) sec (x)dx Improper at π/2 x ln(x) ∫0 ⌡1 We must split this integral into two We must split this one up into two pieces, parts, for it is improper at x = 0 and at also. So: infinity. We could split it into the integral from parts a) and b), and since the integral from part b) diverges, this whole thing diverges. Now as a gets close to π/2, tan(a) blows up, so the first piece, hence the entire integral diverges. 1) (Evaluate using antiderivatives—continued) 2 ∞ ⌠ dx ⌠ dx e) Improper at x = 1 f) Improper at infinity ⎮ x ln2 (x) ⎮ x ln2 (x) ⌡1 ⌡2 But as c goes to one, ln(c) goes to zero so this integral diverges. 2 ⌠ dz g) Improper at both endpoints ⎮ 2 ⌡−2 4 − z The antiderivative can be found by the trig substitution z = 2sin(θ); the antiderivative turns out to be sin–1(z/2). So breaking the integral into two pieces (zero seems convenient) and using appropriate limits on each part gives: 2) Test each integral for convergence or divergence.
    [Show full text]