Section 8.8 Reading Assignment: Improper Integrals

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for and what is really about. Please be careful with this assignment.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Feedback: #1: These are two different things though, which does Type I refer to? (1/2) #2: Same issue as #1. (1/2) #3: Good start, but reconsider how this integral is rewritten with limits. (1/2).

Section 8.8 Reading Assignment: Improper Integrals Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages ofthe textbook. Improper integrals are ones where infinity is involved in some wayr and so we must use limits to properly consider these integrals. One of the major issues is whether an improper integral converges {i.e. has a finite limit value) or diverges (Le. the limit doesn't exist or is infinite}. Exercise 1. Read the boxed definition on p. 509. Explain what it means for an improper integral to be Type |. Exercise 2. Read the boxed definition on p. 512. Explain what it means for an improper integral to be Type II. It's fairly easy to see when you have a Type I improper integral, but Type II improper integrals can be much harder to detect. It's important to watch out for this case. Integrals that seem proper may be improper, and this does cause problems sometimes. Also important for us is to use the correct notation when writing out improper integrals. The limit notation used in the textbook is the model you should follow for this. This is especially worth keeping in mind since MML can't evaluate this. There are important technical reasons for writing these out as limits, so writing these out like the book has will be a part of the grade for this material. Exercise 3. Read Example 5 (p. 513). How can you determine algebraically that the integrand has a vertical asymptote? Explain how the improper integral is broken up and how each part is written using limits. We will he skipping the subsection \"Tests for Convergence and Divergence\" {p. 514 516]. We will be considering a version of the limit comparison test later on in Chapter 10, it's probably for the best that we don't mix these two up, so we'll avoid these tests here. Chapter 8 Techniques of Integration 8.8 Improper Integrals 509 0.2 0.1 2 3 FIGURE 8.12 Are the areas under these infinite curves finite? We will see that the answer is yes for both curves, Infinite Limits of Integration Consider the infinite region (unbounded on the right) that lies under the curve y = e * in the first quadrant (Figure 8.13a). You might think this region has infinite area, but we will Area = 2 see that the value is finite. We assign a value to the area in the following way. First find the area A(b) of the portion of the region that is bounded on the right by x = b (Figure 8.13b). [a) A(b ) = exz dx = -20-1/2 " = -20-b/2 + 2 Then find the limit of A(b) as b - co lim A(b) = lim (-2e-/2 + 2) = 2. Area - -2e-W/2 + 2 The value we assign to the area under the curve from 0 to co is exit dx = lim |ex/ dx = 2.FIGURE 8.13 (a) The area in the first quadrant under the curve y = (-s- (b) The area is an improper integral of the DEFINITION Integrals with infinite limits of integration are improper inte- first type. grals of Type I. 1. If f(x) is continuous on [a, oo), then f(x) dx = lim [ sex) dx. 2. If f(x) is continuous on (-co, b ], then f(x) dx = lim [ f(x) dx. 3. If f(x) is continuous on (-co, co), then where c is any real number. In each case, if the limit exists and is finite, we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exist, the improper integral diverges.Chapter 8 Techniques of Integration 512 Chapter 8 Techniques of Integration Integrands with Vertical Asymptotes 1 Another type of improper integral arises when the integrand has a vertical asymptote-an infinite discontinuity-at a limit of integration or at some point between the limits of inte- gration. If the integrand f is positive over the interval of integration, we can again interpret the improper integral as the area under the graph of f and above the x-axis between the Arca = 2 - 2va limits of integration. Consider the region in the first quadrant that lies under the curve y = 1/ Vx from x = 0 to x = 1 (Figure 8.12b). First we find the area of the portion from a to 1 (Figure 8.16): 1. = 2Vx =2-2Va. Then we find the limit of this area as a - Q: FIGURE 8.16 The area under this curve lim ax = lim (2 - 2Va) = 2. is an example of an improper integral of the second kind. Therefore the area under the curve from 0 to I is finite and is defined to beDEFINITION Integrals of functions that become infinite at a point within the interval of integration are improper integrals of Type II. 1. If f(x) is continuous on (a, b ] and discontinuous at a, then 2. If f(x) is continuous on [ a, b) and discontinuous at b, then 3. If f(x) is discontinuous at c, where a f= (x+ 3)/((x - 1)*(r2+ 1)); Then use the integration command > int( f. x - 2..infinity); Maple returns the answer 7 " + In (5) + arctan (2).Chapter 8 Techniques of Integration 514 Chapter 8 Techniques of Integration To obtain a numerical result, use the evaluation command evalf and specify the num- ber of digits as follows: > evalf( %. 6): The symbol % instructs the computer to evaluate the last expression on the screen, in this case (-1/2)# + In (5) + arctan (2). Maple returns 1.14579. Using Mathematica, entering In/ 1):= Integrate [(x + 3)/((x - D)(x*2 + 1)), {x, 2, Infinity } ] returns Out! !)= -5 + ArcTan [2] + Log [5]. To obtain a numerical result with six digits, use the command "N [ %, 6]"; it also yields 1.14579. Tests for Convergence and Divergence When we cannot evaluate an improper integral directly, we try to determine whether it converges or diverges. If the integral diverges, that's the end of the story. If it converges, we can use numerical methods to approximate its value. The principal tests for conver- gence or divergence are the Direct Comparison Test and the Limit Comparison Test. EXAMPLE 6 Does the integral Je s dx converge? Solution By definition, edx = lim er dx. b-+03./We cannot evaluate this integral directly because it is nonelementary. But we can show that its limit as b - co is finite. We know that ], e * dx is an increasing function of b because the area under the curve increases as b increases. Therefore either it becomes infi- nite as b -co or it has a finite limit as b -co. For our function it does not become infi- nite: For every value of x 2 1, we have e" s e" (Figure 8.19) so that eds ed=-ette 1 something positive and less than 0.37. Here we are relying on the completeness property (Example 6). of the real numbers, discussed in Appendix 6. The comparison of e" and er in Example 6 is a special case of the following test. THEOREM 2-Direct Comparison Test Let f and g be continuous on [ a, co) with 0 s ((x) S g(x) for all > > a. Then 1. If g(x) dx converges, then f(x) dx also converges. 2. If f(x)dx diverges, then g(x) dx also diverges.Chapter 8 Techniques of Integration 8.8 Improper Integrals 515 Proof The reasoning behind the argument establishing Theorem 2 is similar to that in Example 6. If 0 a, then from Rule 7 in Theorem 2 of Section 5.3 we have From this it can be argued, as in Example 6, that [ fix)dx converges if 8(x) dx converges, Turning this around to its contrapositive form, this says that 8(x) dx diverges if diverges. Although the theorem is stated for Type I improper integrals, a similar result is true for integrals of Type II as well. EXAMPLE 7 These examples illustrate how we use Theorem 2. HISTORICAL BIOGRAPHY (a) converges because Karl Weiersiras (1815-1897) www. goo . g1/3RH2ro O s sin'x S on [1, co) and dx converges. Example 3 (b) VI - 0.1 = dx diverges because 1 By on [1, 00) and diverges. Example 3 VP- - 0.1 (c) cos I dx Vx converges because0. 0 = cox = Ion Vx and dx = lim dx Vx = lim V4x 2VI = VAI 0-0 = lim ( V24 - V40) = V2% converges. THEOREM 3-Limit Comparison Test If the positive functions f and g are continuous on [ a. co), and if f(x) lim then and either both converge or both diverge.Chapter 8 Techniques of Integration 516 Chapter 8 Techniques of Integration We omit the proof of Theorem 3, which is similar to that of Theorem 2. Although the improper integrals of two functions from a to co may both converge, this does not mean that their integrals necessarily have the same value, as the next example shows. EXAMPLE 8 Show that converges by comparison with /,"(1/x3) 2x. Find and compare the two integral values. Solution The functions f(x) = 1/x and g(x) - 1/(1 + 27) are positive and continu- ous on [ 1. co). Also, 1/x2 lim lim x 50 1/(1 + x3) = lim *-100 = lim 12+ 1) = 0+1 =1, which is a positive finite limit (Figure 8.20). Therefore, 1 + x2 a converges because converges. The integrals converge to different values, however: 1 F = 2 - 1 1 = 1 Example 3 FIGURE 8.20 The functions in Exumple 8. andlim = lim [tan b - tan ' 1 ] = 1+x 100 TABLE 8.5 EXAMPLE 9 Investigate the convergence of dr. dx Solution The integrand suggests a comparison of f(x) = (1 - e )/x with g(x) = 1/x. However, we cannot use the Direct Comparison Test because f(x) = g(x) and the integral 0.5226637569 of g(x) diverges. On the other hand, using the Limit Comparison Test we find that 1.3912002736 f(x) 10 2.0832053156 lim = lim = lim (1 - 2 )) = 1, X-160 100 4.3857862516 1000 6.6883713446 which is a positive finite limit. Therefore, -dr diverges because / I 10000 8.9909564376 diverges. Approximations to the improper integral are given in Table 8.5. Note that the 100000 11.2935415306 values do not appear to approach any fixed limiting value as b - co