Calculus II Lecture Notes

Transcription

1 Calculus II Lecture Notes David M. McClendon Department of Mathematics Ferris State University 206 edition

2 Contents Contents 2 Review of Calculus I 5. Limits Derivatives The definite integral Homework exercises Basic integration techniques Integrals to memorize Rewriting the integrand Elementary u-substitutions More complicated u-substitutions Homework exercises Intermediate integration techniques Integration by parts Partial fractions Summary of integration techniques Mathematica commands for integration Homework exercises Improper integrals 6 4. Boundedness vs. unboundedness Horizontally unbounded regions Vertically unbounded regions Theoretical approaches Homework exercises

3 Contents 5 Applications of the integral Area Volume General principles behind all applications of integration Arc length One-dimensional moments and centers of mass Two-dimensional moments and centers of mass Probability Homework exercises Parametric equations Introduction to parametric equations Parametric equations of common graphs Calculus with parametric equations Transformations on parametric equations Homework exercises Introduction to infinite series Motivation and big-picture questions Convergence and divergence Σ-notation Elementary properties of convergence and divergence Changing indices Homework exercises Geometric series and the Ratio Test Definitions The Geometric Series Test Applications of geometric series The Ratio Test Homework exercises Convergence tests for positive series Classifying series according to sign The Integral Test; harmonic and p-series The n th -term Test The Comparison Test Homework exercises Absolute and conditional convergence Alternating series The triangle inequality Absolute and conditional convergence

4 Contents 0.4 Rearrangement of infinite series Summary of classification techniques Examples Homework exercises Taylor series 236. Uniqueness of power series Applications of Taylor series General theory of power series Homework exercises Index 263 4

5 Chapter Review of Calculus I Big picture issues. What is calculus? 2. What are some typical kinds of problems you learn how to solve / mathematical procedures you learn in Calculus I? 3. What is the difference between a math problem that is a calculus problem and a math problem that is NOT a calculus problem? 5

6 Most of Calculus I theory in one chart CALCULUS OBJECT MOTIVATING PROBLEM APPROXIMATION OF THE SOLUTION HOW THE APPROXIMATION IMPROVES THEORETICAL SOLUTION HOW THE THEORETICAL SOLUTION IS COMPUTED IN PRACTICE APPLICATIONS 6

7 .. Limits What is the difference between a math problem that is a calculus problem and a math problem that is NOT a calculus problem? A calculus problem is one that contains a limit; a non-calculus problem contains no limit. What is calculus? Calculus is the study of limits. LIMIT DERIVATIVE INTEGRAL APPLICATIONS APPLICATIONS??. Limits Recall: To say lim f(x) = L x a means that as x gets closer and closer to a, then f(x) gets closer and closer to L. This suggests that the graph of f looks like one of the following three pictures: The graph on the left is continuous at a; the other two graphs are not. More precisely, 7

8 .. Limits Definition. A function f : R R is continuous at a if lim x a f(x) = f(a). f is called continuous if it is continuous at every point in its domain. Theorem.2 Any function which is the quotient of functions made up of powers of x, sines, cosines, arcsines, arctangents, exponentials and/or logarithms is continuous everywhere except where the denominator is zero. This theorem suggests that to evaluate most limits, you should start by plugging in a for x. If you get a number, that is usually the answer. Example: Evaluate these limits:. lim (3 sin 2x) x π/3 x+3 2. lim x 4 x 2 Example: What about these: 3. lim x 5 x 2 (x 2) 2 x 4. lim 2 3x 0 x 5 x 5 When you plug in a for x and you get 0 in a denominator, you have to work a bit harder to evaluate a limit. When you get nonzero 0 : Example 3 above: lim x 5 x 2 (x 2) 2 8

9 .. Limits When you get 0 0 : x Example 4 above: lim 2 3x 0 x 5 x 5 Techniques for dealing with 0 0 : Factor and cancel (as in Example 4) Conjugate square roots Clear denominators of inside fractions L Hôpital s Rule x Example 4 revisited: lim 2 3x 0 x 5 x 5 Theorem.3 (L Hôpital s Rule) Suppose f and g are differentiable functions. Suppose also that either lim f(x) =lim g(x) = 0 or lim f(x) =lim g(x) = ±. x a x a x a x a Then: lim x a f(x) g(x) L f (x) =lim x a g (x). Limits at infinity We can also take a limit of a function as x. To say lim x f(x) = L means that as x grows larger and larger without bound, then f(x) approaches L. Graphically, this means f has a horizontal asymptote (HA) y = L: 9

10 .. Limits Although is not a number, it can be manipulated in some ways as if it is a number, so one can evaluate infinite limits by plugging in for x and applying the following arithmetic rules for (the cs represent arbitrary constants): ± c = = = c = { if c > 0 if c < 0 = ln = e = c = { if c > 0 0 if c < 0 c = 0 0 = ± c = ± so long as c 0 0 In the last two situations, you need to analyze the sign carefully to determine whether the answer is or. Warning: The following expressions are indeterminate forms, i.e. they can work out to be different things depending on the context (these are evaluated using L Hôpital s Rule): ln x Example 5: Evaluate the limit x lim. x 4x Example 6: Evaluate the limit x lim 3 +2x+. 2x 3 +x 2 +2 The last example generalizes into the following principle which has many useful applications: 0

11 .2. Derivatives Theorem.4 (Limits at infinity for rational functions) Suppose f is a rational function, i.e. has form Then: f(x) = a mx m + a m x m + a m 2 x m a 2 x 2 + a x + a 0 b n x n + b n x n + b n 2 x n b 2 x 2 + b x + b 0.. If m < n (i.e. largest power in numerator < largest power in denominator), then lim x f(x) = If m > n (i.e. largest power in numerator > largest power in denominator), then lim x f(x) = ±. 3. If m = n (i.e. largest powers in numerator and denominator are equal), then am f(x) = b n. lim x.2 Derivatives Definition.5 (Limit definition of the derivative) Let f : R R be a function and let x be in the domain of f. If the limit f(x + h) f(x) lim h 0 h exists and is finite, say that f is differentiable at x. In this case, we call the value of this limit the derivative of f and denote it by f (x) or df dy or. dx dx Differentiable functions are smooth, i.e. are continuous and do not have sharp corners, vertical tangencies or cusps. Assuming it exists, the derivative f (x) computes:. the slope of the line tangent to f at x; 2. the slope of the graph of f at x; 3. the instantaneous rate of change of y with respect to x; 4. the instantaneous velocity at time x (if f is the position of the object at time x).

12 .2. Derivatives We do not compute derivatives using the limit definition. We use differentiation rules, which consist of a list of functions whose derivative we memorize and a list of rules which tell us how to differentiate more complicated functions made up of pieces whose derivatives we know. Functions whose derivatives we memorize: text Differentiation rules for more complicated functions: Constant Multiple Rule: (cf) (x) = c f (x) Sum Rule: (f + g) (x) = f (x) + g (x) Difference Rule: (f g) (x) = f (x) g (x) 2

13 .2. Derivatives Example: Suppose f(x) = 3x 6 sin x. Find f (x). Example: Suppose y = 4 x 2 2 ln x + 9e 2x. Find dy dx. Example: Find the slope of the line tangent to the graph of f(x) = 3 cos x+2 sin x at x = π 4. Solution: The slope of the tangent line at π is given by f ( π ). By usual rules, 4 4 f (x) = 3 sin x + 2 cos x ) f ( π 4 = 3 sin π cos π 4 = = 2. Derivatives have many applications. The most important (for Math 230 purposes) is that given a differentiable function f, you can approximate values of f near a using the tangent line to f at a: 3

14 .3. The definite integral Definition.6 Given a differentiable function f and a number a at which f is differentiable, the tangent line to f at a is the line whose equation is y = f(a) + f (a)(x a) (a.k.a. L(x) = f(a) + f (a)(x a)). For values of x near a, f(x) L(x); approximating f(x) via this procedure is called linear approximation. You can also conclude many things about the graph of f from looking at its derivative and its higher-order derivatives, as defined below: Definition.7 Let f : R R be a function. The zeroth derivative of f, sometimes denoted f (0), is just the function f itself. The first derivative of f, sometimes denoted f () or dy dx, is just f. The second derivative of f, denoted f or f (2) or d2 y dx 2, is the derivative of f : f = (f ). More generally, the n th derivative of f, denoted f (n) or dn y dx n,is the derivative of f (n ) : f (n) = ((((f ) ) ) The first derivative of a function measures its tone (i.e. whether it is increasing or decreasing). The second derivative of a function measures its concavity (i.e. whether its graph is cupped upward or downward). In Math 230, we will see what the higher-order derivatives of f have to do with the function..3 The definite integral Definition.8 Given function f : [a, b] R, the definite integral of f from a to b is b n f(x) dx = f(c k ) x k, a lim P 0 k= where the expression inside the limit is a Riemann sum for f. 4

15 .3. The definite integral Note: In Math 230, the limit above always exists (but it doesn t always exist for crazy functions f... take Math 430 for more on that). The definite integral of a function is a number which is supposed to give the signed area of the region between the graph of f and the x-axis. Area above the x-axis is counted as positive area; area below the x-axis is counted as negative area. As with derivatives, we do not compute integrals with this definition. We use the following important circle of ideas: Definition.9 Given function f, an antiderivative of f is a function F such that F = f. Ex: F (x) = sin x is an antiderivative of f(x) = cos x. Every continuous function has an antiderivative (although you may not be able to write its formula down); any two antiderivatives of the same function must differ by a constant (so if you know one antiderivative, you know them all by adding a +C to the one you know). Definition.0 Given function f, the indefinite integral of f, denoted f(x) dx, is the set of all antiderivatives of f. Ex: cos x dx = sin x + C. Theorem. (Fundamental Theorem of Calculus Part II) Let f be continuous on [a, b]. Suppose F is any antiderivative of f. Then b a f(x) dx = F (x) b a = F (b) F (a). Despite the similar notation, f(x) dx and b a f(x) dx are very different objects. The first object is a set of functions; the second object is a number. 5

16 .3. The definite integral Exercise: Without looking anything up, evaluate each of the following indefinite integrals, if they are easy to solve. If the integral isn t easily solved, put a?. The idea here is that an integral which is easy should use facts from algebra regarding exponents, logarithms, etc. and the rules memorized in Calculus I. Easy integrals do not use sophisticated calculus techniques or trig identities.. 0 dx = 2. dx = 3. 5 dx = 4. x 4 dx = 5. x dx = 6. x dx = 7. x 3 dx = 8. x 5 dx = 9. x 2 + dx = 0. x 3 + dx =. x dx = 2. 3 x dx = 3. 3 x 2 dx = 4. x 2/5 dx = 5. x /2 dx = 6. 5 x dx = 7. sin x dx = 8. cos x dx = 9. tan x dx = 20. sec x dx = 2. csc x dx = 22. cot x dx = 23. sin 2 x dx = 24. cos 2 x dx = 25. sec 2 x dx = 26. csc 2 x dx = 27. tan 2 x dx = 28. cot 2 x dx = 29. sec x tan x dx = 30. sec x cot x dx = 3. csc x tan x dx = 32. csc x cot x dx = 33. csc x tan x dx = 34. csc x sec x dx = 35. e x dx = 36. e x2 dx = x dx = 38. ln x dx = 39. log 0 x dx = 40. arctan x dx = 4. arcsin x dx = 6

17 .3. The definite integral Theorem.2 (Integration Rules to Memorize) 0 dx = C M dx = Mx + C x n dx = xn+ + C (so long as n ) n + dx = ln x + C (I don t care so much about the ) x sin x dx = cos x + C cos x dx = sin x + C sec 2 x dx = tan x + C csc 2 x dx = cot x + C sec x tan x dx = sec x + C csc x cot x dx = csc x + C e x dx = e x + C dx = arctan x + C x 2 + (also dx = arcsin x + C) x 2 We know a little about how to integrate functions whose pieces are integrals we memorize, but we don t know as much about how to compute integrals as we know about how to compute derivatives: Theorem.3 (Linearity of Integration) Suppose f and g are integrable functions. Then: b a b a b a [f(x) + g(x)] dx = [f(x) g(x)] dx = [k f(x)] dx = k b a b a b a f(x) dx + f(x) dx b a b a g(x) dx; g(x) dx; f(x) dx for any constant k (and the same rules hold for indefinite integrals as well). 7

18 .3. The definite integral NOTE: Integration is not multiplicative nor divisive: ( f(x)g(x) dx ( ) f(x) dx g(x) ) ( f(x) dx f(x) dx g(x) dx ) g(x) dx 8

19 .4. Homework exercises.4 Homework exercises In problems -22, evaluate the following limits. If they are equal to or, say so; if the limits fail to exist and are not infinite, say so lim x 3 lim x x x 2 + lim 7 x 4 x 2 9 x 2 + x 2 x + 2 lim x 2 + x 2 4 lim x e x lim x 3 x + 3 x+2 + lim x 2x lim x 3x x lim x x 4x 2 3x + lim x 3 2x 2 lim x 4x 2 e x Find dy dx if y = x3 3x Find f (x) if f(x) = x e 2x lim x 0 x sin x lim x 0 x lim ln x x lim ln x x 0 + lim x 0 ex lim x 0 e/x lim sin x x lim arctan x x lim x 0 e /x lim f(x) where f(x) = x + x lim x 0 x { x 2 x 5x x >. 9

20 .4. Homework exercises 25. Find the derivative of f(x) = 5x If f(x) = 2x 4 3x 2 + 5x + 7, find f ( ). 27. Find f (8) if f(x) = 4 3 x. 28. If the position of an object at time t is given by f(t) = (3 ln t) 2, find the velocity and acceleration of the object at time t = e. 29. If f(x) = 3 ln x + ex 5, find f (x). 30. If f(x) = x2, find f (x). x 3. Let y = 3 x x 3/7 + 3 dy x. Find. 2 dx 32. Let f(x) = 7e x sin x df. Find. 4 dx 33. Find d (2 tan x 3 cos x). dx 34. Find the derivative of f(x) = 4 cot x + 2 sec x. 35. Find f (x) if f(x) = e 4x. 36. Find h (x) if h(x) = x tan x. 37. Find the derivative of f(x) = sin x. 38. Find dy dx if y = 2x4 sin x. 39. Find the second derivative of f(x) = cos(3x 2 ). 40. Find the ninth derivative of f(x) = 4 sin x. 4. Find f (x) if f(x) = x2 3x+4 2x 3 x Find the derivative of f(x) = x ln(3x 2 + ). 43. Find the derivative of f(x) = 4 arctan x Find the zeroth derivative of f(x) = 8x cot 4x. 45. Find the slope of the line tangent to y = 2x 3 3x when x = Find the equation of the line tangent to y = 4x(x 2 3) 4 when x = Find dy dx if y = xex. 48. Find f (0) if f(x) = x. 49. Let y = 4x 2. Find dy. 20

21 .4. Homework exercises 50. Let y = 2 sin 3x. Find dy. In problems 5-60, evaluate the given integral x3 dx 52. (3x 2 2x 7 + 5) dx 53. ( x x 4 ) dx x dx 55. (4e x + ) dx 56. (x ) 2 x dx dx 58. π/2 0 3 cos x dx sec 2 x dx x 2 + dx 6. Compute (sec x tan x + csc x cot x) dx. 62. Compute x2 dx. Hint: Think about the shape of the graph of the integrand. 63. Compute 4 3 x dx. Hint: Think about the shape of the graph of the integrand. 64. Find the derivative of f(x) = x 2 sin x. What does your answer tell you about the value of π ( 2x sin x + x 2 cos x ) dx? Find the derivative of the function F (x) = x 2 3t2 dt. Hint: Use the part of the Fundamental Theorem of Calculus not mentioned in these notes. 66. Find the area under the graph of the function f(x) = 8 x from x = to x = Suppose that an object s velocity at time t is given by v(t) = 4 sin t. Find the displacement of the object between times 0 and 2π Suppose that the acceleration of an object is given by a(t) = 48t. If the velocity of the object at time 0 is 2 and the position of the object at time 0 is 0, what is the position of the object at time 3? 2

22 .4. Homework exercises Answers DNE 9. π DNE 23. 3x 2 2x x v(e) = 8 e ; a(e) = x 2 + ex x 2 + x x 3 7 x 4/7 2 3 x 5/ e x cos x sec 2 x + 3 sin x csc 2 x + 2 sec x tan x 35. 4e 4x 36. tan x x sec 2 x tan 2 x sin x cos x 38. 8x 3 sin x + 2x 4 cos x x 2 cos(3x 2 ) 6 sin(3x 2 ) cos x 4. (2x 3)(2x 3 x+) (6x 2 )(x 2 3x+4) (2x 3 x+) ln(3x 2 + ) + 6x2 3x (x/2) x cot 4x y = (x 2) 47. e x + xe x 48. DNE 49. dy = 8x dx 50. dy = 6 cos 3x dx 22

23 .4. Homework exercises x 3 4 x8 + 5x + C x3/2 5 3 x 3 + C ln x + C 55. 4e x + x + C x2 2x + ln x + C 57. C tan x + C arctan x + C 6. sec x csc x + C 62. 9π f (x) = 2x sin x + x 2 cos x; it tells you that the value of the integral is [x 2 sin x] π 0 = x

24 Chapter 2 Basic integration techniques Recall: we know from Calculus I that to evaluate an integral like b a f(x) dx we use the Fundamental Theorem of Calculus, which says that b a f(x) dx = [F (x)] b a = F (b) F (a) where F is any antiderivative of f, i.e. F = f. This means that we need to learn various methods to compute antiderivatives (so that we can evaluate integrals). Recall also that the indefinite integral of a function f, denoted f(x) dx, is the set of all antiderivatives of f. From Calculus I we know that if F is any single antiderivative of f, then all antiderivatives of f are given by F (x) + C where C is an arbitrary constant. In the next two chapters we will discuss various methods to compute antiderivatives/integrals. The general idea is that there are a variety of techniques which work in various circumstances; one needs to learn which technique to use in which situation. 24

25 2.. Integrals to memorize 2. Integrals to memorize In Calculus I you learn many derivative rules, such as d dx (x2 ) = 2x and d (sin x) = cos x. dx Every derivative rule you learn is also an integration rule; for example the above two rules translate as and We also learn the following rules, called the linearity rules for integration, which allow us to split integrals into pieces: Theorem 2. (Linearity of Integration) Suppose f and g are integrable functions. Then: b a b a b a [f(x) + g(x)] dx = [f(x) g(x)] dx = [k f(x)] dx = k b a b a b a f(x) dx + f(x) dx b a b a g(x) dx; g(x) dx; f(x) dx for any constant k; [f(x) + g(x)] dx = f(x) dx + g(x) dx; [f(x) g(x)] dx = f(x) dx g(x) dx; [k f(x)] dx = k f(x) dx for any constant k. WARNING: Integration is not multiplicative nor divisive: ( f(x)g(x) dx ( ) f(x) dx g(x) ) ( f(x) dx f(x) dx g(x) dx ) g(x) dx 25

26 2.. Integrals to memorize Example : Evaluate the following integral: 4 csc2 x dx Example 2: Evaluate the following integral: 0 (2 x ) 3 + x9 dx Example 3: Evaluate the following integral: ( ) dx x x Example 4: Evaluate the following integral: (4 sec 2 x 2x 4 + 7e x) dx 26

27 2.. Integrals to memorize Here is a list of integration rules (discussed in Chapter ) which, together with linearity, allows you to do most easy integrals: Theorem 2.2 (Integration Rules to Memorize) 0 dx = C M dx = Mx + C x n dx = xn+ + C (so long as n ) n + dx = ln x + C (I don t care so much about the ) x sin x dx = cos x + C cos x dx = sin x + C sec 2 x dx = tan x + C csc 2 x dx = cot x + C sec x tan x dx = sec x + C csc x cot x dx = csc x + C e x dx = e x + C dx = arctan x + C x 2 + dx = arcsin x + C x 2 27

28 2.2. Rewriting the integrand 2.2 Rewriting the integrand It is often useful to rewrite the integrand using algebra, a log rule or a trig identity. Example 5: Evaluate the following integral: tan 2 x dx Example 6: Evaluate the following integral: (x 2 ) 2 dx x Example 7: Evaluate the following integral: ln(2 x ) dx 28

29 2.3. Elementary u-substitutions Useful log/exp rules and trig identities for rewriting integrands ln(ac) = ln A + ln C tan 2 x = sec 2 x ( ) A ln = ln A ln C C cot 2 x = csc 2 x ln(a C ) = C ln A log b (A) = ln A ln b A B = e B ln A tan x = sin x cos x cot x = cos x sin x cos 2 + cos 2x x = 2 sin 2 cos 2x x = Elementary u-substitutions In Calculus I you learn how to compute integrals using a u substitution: Theorem 2.3 (Integration by u substitution - Indefinite Integrals) f(g(x)) g (x) dx = f(u) du by setting u = g(x). Theorem 2.4 (Integration by u substitution - Definite Integrals) by setting u = g(x). b a f(g(x)) g (x) dx = g(b) g(a) f(u) du Notice that for this formula to work, the integrand must have the form f (g(x)) g (x). That means that for an elementary u substitution to be a valid integration technique, the integrand must consist of two terms multiplied together, where one 29

30 2.3. Elementary u-substitutions term is the derivative of part of the other term. Example 8: Evaluate the following integral: xe x2 dx Example 9: Evaluate the following integral: x (x 2 + ) 2 dx 30

31 2.3. Elementary u-substitutions Example 0: Evaluate the following integral: 7x 5 sec 2 x 6 dx Example : Evaluate the following integral: π/4 0 e cos x sin x dx 3

32 2.3. Elementary u-substitutions Example 2: We know that cos x dx = sin x + C. Evaluate the following integrals: cos 2x dx ( ) 5 cos 3 x dx cos (7x 4) dx cos (3 x) dx 32

33 2.3. Elementary u-substitutions The technique in Example 2 is completely general, and it is a good idea (to save time) to remember the pattern. I call this kind of reasoning the Linear Replacement Principle: Theorem 2.5 (Linear Replacement Principle) Suppose you know Then for any constants m and b, f(x) dx = F (x) + C. f(mx + b) dx = F (mx + b) + C. m Example 3: Evaluate the following integral: e 6x dx Example 4: Evaluate the following integral: ( ) x sin dx 2 Example 5: Evaluate the following integral: (3x 2) 3 dx Example 6: Evaluate the following integral: csc 2 4x dx 33

34 2.3. Elementary u-substitutions Sometimes you need to rewrite an integral before using a u substitution: Example 7: Evaluate the following integral: tan x dx Example 8: Evaluate the following integral: sin 2 x dx Note: The following power reducing identities are helpful to evaluate sin 2 x dx and cos 2 x dx: sin 2 cos 2x x = cos 2 + cos 2x x =

35 2.4 More complicated u-substitutions 2.4. More complicated u-substitutions We have seen how to perform u substitution when the integral consists of two terms multiplied together. In some cases, one can also perform u substitution when the integral consists of three terms multiplied together, so long as one term is the derivative of part of one of the other terms. Example 9: Evaluate the following integral: x 3 x 2 + dx To execute a more complicated u substitution, we think of the integrand as being three terms multiplied together as follows: 35

36 2.4. More complicated u-substitutions Example 20: Evaluate the following integral: x 2 x + 3 dx Examples 9 and 20 above illustrate a general principle. To evaluate integrals of the form x n (mx ± b) q dx where n is an integer but q is not a positive integer, use the u substitution u = mx ± b. 36

37 2.4. More complicated u-substitutions Example 2: Evaluate the following integral: x + 5 2x + dx Example 22: Evaluate the following integral: x x + 2 dx 37

38 2.4. More complicated u-substitutions Sometimes you can use trig identities to rewrite an integrand before using a complicated u substitution: Example 23: Evaluate the following integral: sin 4 x cos 3 x dx Integrals which contain powers of tangent and secant, or contain powers of cosecant and cotangent, are handled similarly. 38

39 2.4. More complicated u-substitutions Here are two last examples which illustrates how if you are just willing to try something, sometimes things work out: Example 24: Evaluate the following integral: x x dx 39

40 2.4. More complicated u-substitutions Example 25: Evaluate the following integral: 9 x dx 40

41 2.5. Homework exercises 2.5 Homework exercises In Problems -33, compute the following integrals by hand: ( t 2 t 2 ) dt t (3 x) x dx ( + 2x 2 ) 2 dx ( x + x) 2 dx π/3 π/4 3 cot 2 x dx (x 7) 2 dx x x 2 x 3 dx x 2 + x 3 + 3x 2 dx 3 0. e 2x e 2x + dx e 2x + e 2x cot x dx dx sin x cos x dx e 7 3x dx 7x + dx sin 4x dx ( ) 2 cos 3 x dx 2 sec 2 3x dx e x/2 dx csc πx cot πx dx 3x + 2 5x dx (4x + 5) 3 dx x 2 x dx 2x x 4 dx x x + dx x 2 (x + 3) 8 dx cos 2 3x dx sin 2 4x dx sin 5 x dx cos 3 x sin x dx sin 4 2x cos 2x dx + sin x cos x dx cos z dz 33. ln x 2 x dx 4

42 2.5. Homework exercises 34. a) Evaluate the integral sin x cos x dx by performing the u substitution u = sin x. Answers b) Evaluate the integral of part (a) by performing the u substitution u = cos x. c) Reconcile the (apparently) different answers you get to parts (a) and (b) of this question. Note: with indefinite integrals, answers can sometimes vary a little bit.. 4 t4 t 2 + C x x3 + x + C 4. 2 x2 + 2x + ln x + C π x2 4x + 49 ln x + C (x3 ) 3/2 + C 8. 3 ln(x3 + 3x 2) + C 9. 2 ln(e2x + ) + C e 6. ln(sin x) + C 2. 2 cos x + C 3. 3 e7 3x + C 4. ln(7x + ) + C cos 4x + C sin ( 2 3 x) + C 2 7. tan 3x + C e x/2 + C 9. π csc πx + C x + 3 ln(5x ) + C (4x + 5) 2 + C ln x + 8 ln(x 4) + C (x + 3) 3 5 (x + 3)0 + (x + 3) 9 + C 26. x + sin 6x + C x sin 8x + C cos5 x cos3 x cos x + C cos4 x + C sin5 2x + C 3. ln( sin x) + C Hint: First, rewrite the integrand by multiplying through the numerator and denominator by ( sin x). 32. csc z + cot z + C 42

43 2.5. Homework exercises Hint: First, rewrite the integrand by multiplying through the numerator and denominator by (cos z + ). 33. ln 2 x + C Hint: First, rewrite ln x 2 as 2 ln x. 34. (a) 2 sin2 x + C; (b) 2 cos2 x + D; (c) Set the two answers equal to one another; rewriting this equation gives = D C so since C and D are arbitary 2 constants, as long as they are related by = D C, the answers reconcile. 2 43

44 Chapter 3 Intermediate integration techniques 3. Integration by parts Motivation: d ( x 7 sec x ) = dx More generally, the Product Rule tells us (fg) (x) = Rewrite the Product Rule as follows: 44

45 3.. Integration by parts Theorem 3. (Integration by Parts (IBP) Formula) u dv = uv v du. The IBP formula is used (usually) to find integrals of the form u dv (you choose u and dv) where v du is easier than u dv. Example : x sin x dx Thought Process: Solution: Example 2: x 2 e x dx Solution: 45

46 3.. Integration by parts More generally, to handle integrals like x n f(x) dx where f(x) is exponential, sine or cosine, you would (at least theoretically) use integration by parts n times. Example 3: 0 x2 e x dx Solution: Example 4: ln x dx Solution: (The integrals arctan x dx and arcsin x dx are similar.) Parts vs. u-substitutions x sin x dx vs. sin x cos x dx 46

47 3.. Integration by parts How to choose u and dv Suppose you know you want to use integration by parts. How do you decide which part of the integral should be u and which should be dv? General principle: The integral v du should be easier than u dv. So think ahead! Specific guidelines: HIGH PRIORITY u (choose these to be u if possible): MEDIUM PRIORITY u: LOW PRIORITY u (avoid choosing these as u if possible): Example: For each of the following integrals, decide if integration by parts is an appropriate method to try. If so, write what you would choose for u and dv: (a) 3x 2 arctan x dx (b) x ln x dx (c) 2x sin 4x dx (d) e 3x cos 2x dx (e) ln x x (f) ln x x 2 dx dx (g) 4x 3 e 2x dx (h) x 2 +3x 3 2x+3 dx (i) 2x+3 x 2 +3x 3 dx 47

48 3.2. Partial fractions 3.2 Partial fractions Motivation: Consider the following expression: This is equal to 2(x ) (x 3)(x ) + (x 3) (x )(x 3) 2 x 3 + x = [2x 2] + [x 3] (x )(x 3) = 3x 5 x 2 4x + 3. This means ( 2 x 3 + ) dx = x 3x 5 x 2 4x + 3 dx Goal: Rewrite expressions like 3x 5 x 2 4x+3 as 2 x 3 + x. Definition 3.2 A polynomial in x is an expression of the form a 0 + a x + a 2 x a n x n where a 0, a,..., a n are constants. The degree of a polynomial is the highest power of x that appears. A polynomial is called irreducible if it cannot be factored into two polynomials, both of lower degree. Examples: x 4 3x + is a degree 4 polynomial x 2 3x 9 7x + 5x 6 is a degree 9 polynomial x7 +2, x,, sin x, x x 4xπ,... are not polynomials x 2 + A is an irreducible polynomial if A > 0 x 2 4 is not irreducible Ax ± B is irreducible Theorem 3.3 (Fundamental Theorem of Algebra) No polynomial of degree 3 is irreducible. In the same way that every whole number factors into a product of primes, we have: 48

49 3.2. Partial fractions Theorem 3.4 Every polynomial factors into a product of irreducibles. A precise statement of the problem: Given a rational function p(x) (a rational q(x) function is the quotient of any two polynomials) with degree(p) < degree(q), write p(x) as q(x) p(x) q(x) = p (x) q (x) + p 2(x) q 2 (x) p n(x) q n (x) where: degree(p j ) < degree(q j ) for all j; and every q j (x) is a power of an irreducible polynomial, i.e. q j (x) = ax ± b or q j (x) = (ax ± b) r or q j (x) = (ax 2 ± bx ± c) r where ax 2 ± bx ± c doesn t factor This procedure is called splitting p(x) into partial fractions or finding the partial fraction decomposition of p(x). q(x) q(x) Example: Find the partial fraction decomposition of x 2. STEP : Factor the denominator completely STEP 2: Guess the form of the decomposition STEP 3: Write the equation from step 2 so that it has common denominator STEP 4: Clear the denominator from the equation obtained in step 3 49

50 3.2. Partial fractions STEP 5: Find the unknowns A, B, C,... (two methods) STEP 6: Write answer Remark: There is no calculus in the partial fraction decomposition itself (you do no limit, derivative or integral in steps -6 above). This is really an algebraic technique, whose most common use is to rewrite integrands. 50

51 3.2. Partial fractions Example: Evaluate the following integral: 3x 2 + 0x 24 x 3 + 2x 2 8x dx 5

52 3.2. Partial fractions Example: Find the partial fraction decomposition of 2x 3 x 2 x + x 2 (x ) 2 52

53 3.2. Partial fractions 2? Question: In a partial fraction problem, what should your guess be in Step It is hard to write down a one-size fits all statement; here are some examples: p(x) q(x) GUESSED FORM OF DECOMPOSITION 6x (x 5)(x+2) 4 x(x 3)(x+4) 2 (x 4) 2 (x+6) 7 x 4 (x ) 3 (x+2) x (x 2 +5)(x 2) x 3 3x+4 (x 2 +) 3 x 2 (x 7)(x 2 +6) General guidelines:. Make sure the denominator is factored completely first. 2. For every linear term (x ± a) r in the denominator, the guessed form needs r terms of the form A x ± a + B (x ± a) + C const (x ± a) 3 (x ± a) r 3. For every unfactorable quadratic (x 2 +ax+b) r in the denominator, the guessed form needs r terms of the form Ax + B x 2 + ax + b + Cx + D (x 2 + ax + b) + Ex + F const x + const (x 2 + ax + b) 3 (x 2 + ax + b) r 53

54 3.2. Partial fractions Remark: Partial fractions is a useful technique to evaluate integrals under the following circumstances:. The integrand must be a rational function (i.e. contain only nonnegative powers of x and no trig/exp/log functions). 2. The degree of the numerator must be less than the degree of the denominator. 3. The denominator must be easily factored. 4. The numerator must be sufficiently complicated. Example: Solution # : Use partial fractions: x (x 2) 3 dx x (x 2) = A 3 x 2 + B (x 2) + C 2 (x 2) 3 x A(x 2)2 = + (x 2) 3 (x 2) 3 B(x 2) (x 2) 3 + C (x 2) 3 x = A(x 2) 2 + B(x 2) + C x = A(x 2 4x + 4) + Bx 2B + C x = Ax 2 + (B 4A)x + (4A 2B + C) Thus A = 0, B 4A = and 4A 2B + C = 0. That means B = and C = 2. So ( ) x (x 2) dx = 3 (x 2) + 2 dx = 2 (x 2) 3 Solution # 2: x 2 + (x 2) 2 + C. 54

55 3.2. Partial fractions Example: Evaluate the following integral: 5x x + 6 x 3 + 2x 2 + x dx Elementary integral formulas useful in conjunction with partial fractions Each term in a partial fraction decomposition usually works out to be something like one of these terms below: dx = ln x ± a + C x ± a ax ± b dx = ln ax ± b + C a ( ) (x a) dx = + C (if n ) n n (x a) n dx = arctan x + C x 2 + x 2 + a dx = 2 a arctan x a + C 55

56 3.3 Summary of integration techniques 3.3. Summary of integration techniques. Check to see if the integral is one you can just write the answer to. 2. If the integrand contains terms which are added or subtracted, consider using linearity rules to split the integrand into pieces you can just write the answer to. 3. See if you can rewrite the integrand using algebra, log rules or a trig identity. 4. If the integrand is a function you can integrate by hand with a mx ± b instead of an x, use the Linear Replacement Principle. 5. If the integrand contains terms which are multiplied together, where one part is the derivative of something in the other part (up to a constant), use a u substitution. 6. If the integrand has terms multiplied together which are unrelated, try integration by parts. 7. If the integrand is a rational function where the denominator factors (and the degree of the denominator is greater than the degree of the numerator), use partial fractions to decompose the integrand. 8. If all else fails, use Mathematica (see below). 3.4 Mathematica commands for integration To find the indefinite integral of a function: Integrate[f unction, x] To find the exact value of a definite integral of a function from a to b: Integrate[f unction, {x,a,b}] To find a decimal approximation to a definite integral of a function from a to b: NIntegrate[f unction, {x,a,b}] To find a partial fraction decomposition (without integrating): Apart[expression] 56

57 3.5. Homework exercises 3.5 Homework exercises In Problems -0, evaluate the following integrals by hand.. xe 3x dx 6. x cos x dx 2. 4x e dx x 7. π/2 x sin 2x dx 0 3. x 3 e x4 dx 8. 2xe x/2 dx 0 4. x ln(x + 5) dx 9. arctan x dx 5. x 2 ln x dx 0. x arctan x dx In Problems -9, determine whether it is better to use parts or a u substitution to evaluate the integral. If parts are better, write PARTS and give your choices of u and dv. Otherwise, write u SUB and give your choice of u. You do not have to evaluate these integrals x 2 + 3x dx x 2 cos x dx x ln 2 x dx arcsin x dx arctan x x 2 + dx x ln x dx 3x 2 e 2x dx 5. sin x cos x dx 9. e x cos(3x) dx In Problems 20-23, write the guessed form of the partial fraction decomposition of the given expression. You do not need to solve for the constants. Example: x 2 Solution: x 2 = A x + B x + 57

58 3.5. Homework exercises x 2 3x 22. 3x x 2 +4x x+2 x 3 5x 2 24x x 2 +9x 54 In Problems 24-27, evaluate the integrals by hand. 24. x 2 6 dx x 2x 2 + x dx x 2 + x 2 dx 27. 5x 2 2x 2 dx x 3 4x In Problems 28-33, write the guessed form of the partial fraction decomposition of the given expression. You do not need to solve for the constants x+3 x(x 2 +) (x 3)(x+2) 2 3. (x 2) 2 (x 2 +4) x+ x 3 (x 2 +x+)(x 7) x 2 x 3 (x ) (x+4) 3 In Problems 34-37, evaluate the integrals by hand x 2 + 2x dx x 3 + x x 2 x + 9 (x 2 + 9) 2 dx 35. x 2 x 4 2x 2 8 dx x x 2 (x + ) dx 38. a) Evaluate the integral 2x 3 (x ) 3 dx by performing a u substitution. b) Evaluate the integral from part (a) via partial fractions. c) Verify that the answers you get to parts (a) and (b) are equal. d) Evaluate the integral using Mathematica. Which method do you think Mathematica used to compute the integral? 39. In the Elementary integral formulas used with partial fractions subsection on p. 54, we asserted that x 2 + a 2 dx = a arctan x a + C. Show that this formula is correct. Hint: Rewrite the integrand by factoring out a 2 from the denominator and then use the Linear Replacement Principle. 58

59 3.5. Homework exercises Answers. 3 xe 3x 9 e 3x + C 2. 4xe x 4e x + C 3. 4 ex4 + C 4. 4 x x + 2 x2 ln(x + 5) x x ln x + C 6. x sin x + cos x + C 7. π e /2 9. x arctan x 2 ln(x2 + ) 0. 2 x2 arctan x + 2 x + 2 arctan x + C. u-sub; u = 2 + 3x 2. PARTS; u = x 2 ; dv = cos x dx 3. PARTS; u = ln 2 x; dv = x dx 4. PARTS; u = arcsin x; dv = dx 5. u-sub; u = sin x or u = cos x 6. u-sub; u = arctan x 7. u-sub; u = ln x 8. PARTS; u = 3x 2 ; dv = e 2x dx ln(x + 5) + C 9. PARTS; u = e x ; dv = cos 3x dx (or vice versa) x 2 3x = A x + B x x+2 x 3 5x 2 24x = A x + B x 8 + C x x x 2 +4x 5 = x 2 +9x 54 = A + B x+5 x A + B x+6 3x ln(x 4) ln(x + 4) + C

60 3.5. Homework exercises 25. ln(x ) ln(x + 2) + C ln(2x ) 2 ln(x + ) + C ln x + ln(x 2) + ln(x + 2) + C A Bx+C x x A x 3 + B x+2 + C (x+2) A x + B x 2 + C x 3 + D x + E (x ) 2 A 3. + B + Cx+D + Ex+F x 2 (x 2) 2 x 2 +4 (x 2 +4) A x + B x 2 + C x 3 + Dx+E x 2 +x+ + F x A x+4 + B (x+4) x C (x+4) ln x + ln(x + ) + C arctan x ln(x 2) ln(x + 2) + C (x2 + 9) + 3 arctan x 3 + C ln a) 2(x ) + 2 (x ) 2 + C b) 2 (x ) 2 2(x ) + C c) They are clearly equal. d) It isn t clear, because Mathematica simplifies the answer so much. 39. Follow the hint: x 2 + a dx = 2 a 2 x 2 + dx a 2 = a 2 ( ) 2 dx x a + = [ a 2 arctan x a a ] = a 2 a arctan x a + C = a arctan x a + C. + C (by the Lin. Repl. Principle) 60

61 Chapter 4 Improper integrals 4. Boundedness vs. unboundedness Recall Let f be continuous and let a, b R. Then b a f(x) dx gives the area under f from x = a to x = b. Since f is continuous on [a, b], it must be that the graph of f encloses a bounded region (in the sense that one can draw a box around it): This guarantees that b a f(x) dx <. New question: Can you compute the area of an unbounded region? A typical unbounded region has infinite area: 6

62 4.. Boundedness vs. unboundedness But some unbounded regions have finite area! Here is an example: Notice that in the previous picture, the region obtained was unbounded, but in the direction it was unbounded, the width/thickness of the region decreases to zero. Another situation where the width/thickness of an unbounded region decreases to zero is when the region is described by a graph with an asymptote: Horizontally unbounded region Vertically unbounded region Better question: How are such unbounded regions described? 62

63 4.2. Horizontally unbounded regions 4.2 Horizontally unbounded regions Definition 4. Improper integrals for horizontally unbounded regions] Let f(x) be continuous on [a, ). Define the improper integral a f(x) dx to be a b f(x) dx = lim f(x) dx. b a If this limit exists and is equal to a finite number L, we write a f(x) dx = L nd say a f(x) dx converges (to L). If this limit does not exist or is equal to ±, we say f(x) dx diverges. a A picture to explain the definition: Remark: If a all x), it must be the case that f(x) dx converges for a nonnegative function f (i.e. f(x) 0 for lim f(x) = 0. x The converse of this statement is false, as is seen in the following example: Example: Compute the following improper integral: x dx. 2 b 63

64 4.2. Horizontally unbounded regions Example: Compute the following improper integral: x 2 dx. 2 b Now we generalize the results of the preceding two examples: Example: Let p > 0 be a constant. Compute the following improper integral: x p dx. From the previous two examples we know: When p =, the integral diverges (i.e. the area under the curve is infinite). When p = 2, the integral converges to (i.e. the area under the curve is ). 64

65 4.2. Horizontally unbounded regions Solution for p, p > 0: The results of the previous problem are summarized in the following theorem, which should be memorized: Theorem 4.2 (Convergence/divergence of p-integrals) The improper integral x p dx converges to if p >, but diverges if p. p 65

66 4.3. Vertically unbounded regions Example: Determine whether or not the following improper integral converges or diverges: e x ln 3 x dx 4.3 Vertically unbounded regions If the region under consideration is unbounded in a vertical direction, one needs to be more careful in rewriting the improper integral as a limit. There are three situations; the first situation is when the vertical asymptote is on the righthand edge of the region whose area you want to compute: 66

67 4.3. Vertically unbounded regions Definition 4.3 (Improper integrals for vertically unbounded regions I) (VA on right-hand edge of region) Let f(x) be continuous on [a, c) and suppose lim x c f(x) = ±. Define the improper integral c a f(x) dx to be c a b f(x) dx = lim f(x) dx. b c a If this limit exists and is equal to a finite number L, we write c a f(x) dx = L and say c a f(x) dx converges (to L). If this limit does not exist or is equal to ±, we say f(x) dx diverges. c a A picture to explain the definition: a b c The next situation is when the vertical asymptote is on the left-hand edge of the region whose are you want to compute: Definition 4.4 (Improper integrals for vertically unbounded regions II) (VA on the left-hand edge of the region) Let f(x) be continuous on (a, c] and suppose lim x a +f(x) = ±. Define the improper integral c a f(x) dx to be c a c f(x) dx = lim f(x) dx. b a + b If this limit exists and is equal to a finite number L, we write c a f(x) dx = L and say c a f(x) dx converges (to L). If this limit does not exist or is equal to ±, we say f(x) dx diverges. c a A picture to explain the definition: 67

68 4.3. Vertically unbounded regions a b c The last situation is more complicated. If a region is bounded by a graph with a vertical asymptote in the middle of the region, then the improper integral splits into two parts which must be studied separately: Definition 4.5 (Improper integrals for vertically unbounded regions III) (VA in middle of region) Let f(x) be continuous on [a, c] except at a single point b (a, c). Suppose lim x b +f(x) = lim x b f(x) = or lim x b +f(x) = lim x b f(x) =. Then the improper integral c a f(x) dx is said to converge only if b a f(x) dx and f(x) dx converge (in the sense of the previous definitions). In this case we set c b c a f(x) dx = b a f(x) dx + c b f(x) dx where the two integrals on the right side are defined as in the preceding definitions. a b c Key concept: Must consider the two regions above separately. 68

69 4.3. Vertically unbounded regions Example: Determine whether or not the following improper integral converges or diverges: π/2 tan x dx 0 69

70 4.3. Vertically unbounded regions Example: Determine whether or not the following improper integral converges or diverges: dx x(x + ) 0 70

71 4.4. Theoretical approaches 4.4 Theoretical approaches Frequently in mathematics we encounter quantities (such as improper integrals) where it is just as important (if not moreso) to determine whether or not the quantity converges than it is to determine what number the quantity converges to. It is useful to know some tricks based on theoretical concepts which help tell us whether or not an improper integral converges. The first trick is that when integating over a horizontally unbounded region, the starting index of the integral doesn t affect whether or not the integral converges. Theorem 4.6 (Starting index is irrelevant to convergence/divergence) Suppose a < b and f is continuous on [a, b]. Then:. if a 2. if a f(x) dx converges, so does b f(x) dx; f(x) dx diverges, so does b f(x) dx. A picture to explain: Note: This theorem does not say that f(x) dx = a b f(x) dx. If f is positive and a < b, then (assuming these integrals converge) it is clear that a f(x) dx > b f(x) dx because you are integrating a positive function over a larger region. 7

72 4.4. Theoretical approaches The next set of results describe how one can split up improper integrals whose integrands are composed of terms added/subtracted together: Theorem 4.7 (Linearity of improper integrals I) Suppose a f(x) dx = L and g(x) dx = M. Then: a. a [f(x) + g(x)] dx = L + M; 2. a [f(x) g(x)] dx = L M; 3. a kf(x) dx = kl for any constant k. Theorem 4.8 (Linearity of improper integrals II) Suppose a f(x) dx = L but g(x) dx diverges. Then: a. a [f(x) + g(x)] dx diverges; 2. a [f(x) g(x)] dx diverges; 3. a kg(x) dx diverges, for any constant k 0. Theorem 4.9 (Linearity of improper integrals III) Suppose that both a f(x) dx and a g(x) dx diverge. Then you know nothing about whether or not the improper integrals a [f(x) + g(x)] dx and a [f(x) g(x)] dx converge or diverge. The preceding three theorems can be summarized in the following three morals, which we will see again and which hold throughout mathematics:

73 4.4. Theoretical approaches Example: Determine whether or not the following improper integrals converge or diverge: (a) ( ) x x dx 3 (b) ( ) 6 3 5x 2 x dx 3/2 Our last theoretical result is based on a very important and very general mathematical idea of reasoning by way of inequalities: Theorem 4.0 (Comparison Test for Improper Integrals) Suppose that for all x a. Then:. if a 2. if a 0 f(x) g(x) g(x) dx converges, so does a f(x) dx; f(x) dx diverges, so does a g(x) dx. A picture to explain: 73

74 4.4. Theoretical approaches Example: Determine whether or not the following improper integral converges or diverges: 3 x 5 + 2x + 2 dx Example: Determine whether or not the following improper integral converges or diverges: 2 dx x

75 4.4. Theoretical approaches Example: Determine whether or not the following improper integral converges or diverges: sin x dx 2 x 8 General principles to help construct inequalities. Addition in the denominator. Suppose the integrand is of the form + where, and are all positive quantities. Then, one can start with one of these two inequalities and try to apply the Comparison Test. + or + (The reason these inequalities are valid is that removing positive terms from the denominator makes the denominator smaller, which makes the entire fraction bigger.) 75

76 4.4. Theoretical approaches 2. Subtraction in the denominator. Suppose the integrand is of the form where, and are all positive quantities. Then, one can start with this inequality: and try to apply the Comparison Test. (The reason this inequality is valid is that removing negative terms from the denominator makes the denominator bigger, which makes the entire fraction smaller.) 3. Terms containing sines or cosines. Suppose the integrand contains some expression of the form cos or sin where is some expression. Then, one can start with the inequality and try to apply the Comparison Test. cos ( or sin ) 76

77 4.5. Homework exercises 4.5 Homework exercises In Problems -6, evaluate the improper integral (if the integral diverges, say so).. x 4 dx 4. 2 x 2 + dx x dx 5. ln x x dx 3. xe x dx dx (x + 3) 5/3 7. In class we showed that dx diverges. Have Mathematica evaluate this x integral. Does Mathematica recognize that this integral diverges? 8. Use Mathematica to find the values of x n e x dx 0 for n =, 2, 3, 4 and 5. Find a pattern in these answers to make a guess as to what the value of 0 x n e x dx is for an arbitrary whole number n (your answer should be in terms of n). 9. For each of the following improper integrals, write them as a limit or a sum of limits: Example: 4 0 Solution: lim b 2 dx x b 0 Example: 2 e x dx Solution: lim b b 2 e x dx 2 dx + lim 4 x B + B 2 dx x Then, sketch a picture (similar to those on pages 66-67) which reflects how the limit is being used to evaluate the improper integral. a) 2 b) 0 c) 0 d) 5 3 e) 5 3 dx x ln x x dx x 2 dx xe x dx (x 5) 2 dx (x 3) 2 77

78 4.5. Homework exercises In Problems 0-5, evaluate the improper integral (if the integral diverges, say so) x 2 dx x dx. x ln x dx 0 4. x ln x dx 2. e ln x 2 dx x 2 dx In Problems 6-25, determine whether the improper integral converges or diverges (you do not necessarily need to evaluate the integral if it converges). Note: the intent here is for you to use theory, not to perform lots of computations x 8 dx 2 x 4 + dx x dx 9. 4x 3/2 dx e x + x dx 2. cos(πx) + 3 x 4 dx 22. ( 3 x 2 ) dx x ( 2 x + 5 ) dx 3 x sin(2x + ) + 4 x 8 2x dx dx 78

79 4.5. Homework exercises Answers diverges π/2 5. diverges 6. 3 /3 2 2/3 7. Yes, Mathematica recognizes that this integral diverges. 8. ; 2; 6; 24; 20. In general, 0 x n e x dx = n!. 9. a) 2 dx = lim 2 x ln x b + b dx x ln x (Picture looks like the one on page 7, with the asymptote at x = ) b) 0 x dx = lim b x x 2 b 0 dx x 2 (Picture looks like the one on page 6, with the asymptote at x = ) c) 0 d) 5 3 e) diverges. 4 b B dx = lim dx + lim dx xe x b 0 + xe x B xe x (Picture looks like the example on page 0, with the asymptotes at x = 0 and y = 0) b 3 dx = lim dx (x 5) 2 b 5 (x 5) 2 (Picture looks like the one on page 6, with the asymptote at x = 5) 5 b dx = lim dx (x 3) 2 b 3 + (x 3) 2 (Picture looks like the one on page 7, with the asymptote at x = 3) diverges 5. diverges 6. converges 7. converges 8. diverges 9. converges 20. converges 2. converges 22. diverges 23. converges 24. diverges 25. diverges 79