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CHAPTER 3 Topics in Differential Calculus 3.1 Tangent Lines Everyone knows that the Earth is not flat, but locally, e.g. in your immediate vicinity, isn’t the Earth effectively flat? In other words, “flat” is a fairly good approximation of the Earth’s surface “near” you, and it simplifies matters enough for you to do some ... | ElementaryCalculus_Page_66_Chunk3601 |
Tangent Lines • Section 3.1 57 Example 3.1 x y 1 1 −1 0 (1,1) y = x2 y = 2x−1 Figure 3.1.2 Find the tangent line to the curve y = x2 at x = 1. Solution: By formula (3.1), the equation of the tangent line is y −f (a) = f ′(a)·(x−a) with a = 1 and f (x) = x2. So f (a) = f (1) = 12 = 1. Both the curve y = x2 and the tange... | ElementaryCalculus_Page_67_Chunk3602 |
58 Chapter 3 • Topics in Differential Calculus §3.1 There are several important things to note about tangent lines: • The slope of a curve’s tangent line is the slope of the curve. Since the slope of a tangent line equals the derivative of the curve at the point of tangency, the slope of a curve at a particular point c... | ElementaryCalculus_Page_68_Chunk3603 |
Tangent Lines • Section 3.1 59 • Smooth curves have tangent lines, nonsmooth curves do not. For example, think of the absolute value function f (x) = |x|. Its graph has a sharp edge at the point (0,0), making it nonsmooth there, as shown in Figure 3.1.6(a) below. There is no real way to define a tangent line at (0,0), b... | ElementaryCalculus_Page_69_Chunk3604 |
60 Chapter 3 • Topics in Differential Calculus §3.1 The tangent line to a curve y = f (x) makes an angle φ(x) with the positive x-axis, given by φ(x) = tan−1 f ′(x) . (3.2) Example 3.4 y x y = e2x ¡ −1 2, 1 e ¢ L φ Find the angle φ that the tangent line to the curve y = e2x at x = −1 2 makes with the positive x-axis, s... | ElementaryCalculus_Page_70_Chunk3605 |
Tangent Lines • Section 3.1 61 Exercises A For Exercises 1-12, find the equation of the tangent line to the curve y = f (x) at x = a. 1. f (x) = x2 + 1; at x = 2 2. f (x) = x2 −1; at x = 2 3. f (x) = −x2 + 1; at x = 3 4. f (x) = 1; at x = −1 5. f (x) = 4x; at x = 1 6. f (x) = ex; at x = 0 7. f (x) = x2 −3x + 7; at x = 2... | ElementaryCalculus_Page_71_Chunk3606 |
62 Chapter 3 • Topics in Differential Calculus §3.2 3.2 Limits: Formal Definition So far only the intuitive notion of a limit has been used, namely: A real number L is the limit of f (x) as x approaches a if the values of f (x) can be made arbitrarily close to L by picking values of x sufficiently close to a. That notion... | ElementaryCalculus_Page_72_Chunk3607 |
Limits: Formal Definition • Section 3.2 63 Example 3.6 Show that lim x→a x = a for any real number a. Solution: Though the limit is obvious, the following “epsilon-delta” proof shows how to use the formal definition. The idea is to let ǫ > 0 be given, then “work backward” from the inequality | f (x)−L| < ǫ to get an ineq... | ElementaryCalculus_Page_73_Chunk3608 |
64 Chapter 3 • Topics in Differential Calculus §3.2 The proofs of the other limit rules are similar.1 In general, using the formal definition will not be necessary for evaluating limits of specific functions—in many cases a simple analysis of the function is all that is needed, often from its graph. Example 3.8 x y 1 1 2... | ElementaryCalculus_Page_74_Chunk3609 |
Limits: Formal Definition • Section 3.2 65 Call L the right limit of a function f (x) as x approaches a, written as lim x→a+ f (x) = L , if f (x) approaches L as x approaches a for values of x larger than a. Call L the left limit of a function f (x) as x approaches a, written as lim x→a−f (x) = L , if f (x) approaches L... | ElementaryCalculus_Page_75_Chunk3610 |
66 Chapter 3 • Topics in Differential Calculus §3.2 So far only finite limits have been considered, that is, L = lim x→a f (x) where L is a real (i.e. finite) number. Define an infinite limit, with L = ∞or −∞, as follows: For a real number a, the limit of a function f (x) equals infinity as x approaches a, written as lim x→... | ElementaryCalculus_Page_76_Chunk3611 |
Limits: Formal Definition • Section 3.2 67 Example 3.12 x y 0 y = 1 x2 Figure 3.2.7 Evaluate lim x→0 1 x2 . Solution: For x ̸= 0 the function f (x) = 1 x2 is defined, and its graph is shown in Figure 3.2.7. As x approaches 0 from either the right or the left, 1/x2 approaches ∞, that is, lim x→0+ 1 x2 = ∞= lim x→0− 1 x2 .... | ElementaryCalculus_Page_77_Chunk3612 |
68 Chapter 3 • Topics in Differential Calculus §3.2 Example 3.13 From Figures 3.2.6 and 3.2.7, it is clear that lim x→∞ 1 x = 0 = lim x→−∞ 1 x and lim x→∞ 1 x2 = 0 = lim x→−∞ 1 x2 Note that the x-axis (i.e. the line y = 0) is a horizontal asymptote for f (x) = 1 x and f (x) = 1 x2 . Some limits are obvious, and you can... | ElementaryCalculus_Page_78_Chunk3613 |
Limits: Formal Definition • Section 3.2 69 Some limits need algebraic manipulation before they can be evaluated. Example 3.15 Evaluate lim x→∞ ³p x+1 −px ´ . Solution: Note that both p x+1 and px approach ∞as x goes to ∞, resulting in a limit of the form ∞−∞. This is an example of an indeterminate form, which can equal ... | ElementaryCalculus_Page_79_Chunk3614 |
70 Chapter 3 • Topics in Differential Calculus §3.2 Example 3.17 Evaluate lim x→∞ x ln x . Solution: This limit is of the form ∞/∞: lim x→∞ x ln x →∞ ∞ = lim x→∞ 1 1 x by L’Hôpital’s Rule = lim x→∞x = ∞ Note that one way of interpreting the limit being ∞is that x grows much faster than ln x. In fact, using L’Hôpital’s ... | ElementaryCalculus_Page_80_Chunk3615 |
Limits: Formal Definition • Section 3.2 71 Example 3.20 Evaluate lim x→0 1 −cos x x . Solution: This limit is of the form 0/0: lim x→0 1 −cos x x →0 0 = lim x→0 sin x 1 by L’Hôpital’s Rule = sin0 1 = 0 There is an intuitive justification for L’Hôpital’s Rule: since the limit limx→a f (x) g(x) uses a ratio to compare how ... | ElementaryCalculus_Page_81_Chunk3616 |
72 Chapter 3 • Topics in Differential Calculus §3.2 Exercises A For Exercises 1-18 evaluate the given limit. 1. lim x→2 x2 +3x−10 x2 −x−2 2. lim x→∞ x2 +3x−10 2x2 −x−2 3. lim x→∞ x2 +3x−10 2x3 −x−2 4. lim x→∞ x3 +3x−10 2x2 −x−2 5. lim x→π/2 cos x x−π/2 6. lim x→∞ x2 ex 7. lim x→−∞x2ex 8. lim x→0+ ln x e1/x 9. lim x→0 t... | ElementaryCalculus_Page_82_Chunk3617 |
Continuity • Section 3.3 73 3.3 Continuity Recall from the previous section that a limit limx→a f (x) can exist without being equal to f (a), or with f (a) not even being defined. Many functions encountered in applications, however, will meet those conditions, and they have a special name: A function f is continuous at ... | ElementaryCalculus_Page_83_Chunk3618 |
74 Chapter 3 • Topics in Differential Calculus §3.3 Example 3.22 The floor function ⌊x⌋is defined as ⌊x⌋= the largest integer less than or equal to x . In other words, ⌊x⌋rounds a non-integer down to the previous integer, and integers stay the same. For example, ⌊0.1⌋= 0, ⌊0.9⌋= 0, ⌊0⌋= 0, and ⌊−1.3⌋= −2. The graph of ⌊x... | ElementaryCalculus_Page_84_Chunk3619 |
Continuity • Section 3.3 75 numbers. This is a property of R. So the value of f will keep jumping between 0 and 1 no matter how close you get to x. In other words, for any number a in R, f (a) exists but it will never equal limx→a f (x) because that limit will not exist. By the various rules for limits, it is straightf... | ElementaryCalculus_Page_85_Chunk3620 |
76 Chapter 3 • Topics in Differential Calculus §3.3 There is an important relationship between differentiability and continuity: Every differentiable function is continuous. Proof: If a function f is differentiable at x = a then f ′(a) = lim x→a f (x)−f (a) x−a exists, so lim x→a (f (x)−f (a)) = lim x→a (f (x)−f (a))· ... | ElementaryCalculus_Page_86_Chunk3621 |
Continuity • Section 3.3 77 Example 3.25 Show that there is a solution to the equation cos x = x. Solution: Let f (x) = cos x −x. Since f is continuous for all x, in particular it is continuous on [0,1]. So since f (0) = 1 > 0 and f (1) = −0.459698 < 0, then by the Intermediate Value Theorem there is a number c in the ... | ElementaryCalculus_Page_87_Chunk3622 |
78 Chapter 3 • Topics in Differential Calculus §3.3 Exercises A For Exercises 1-18, indicate whether the given function f (x) is continuous or discontinuous at the given value x = a by comparing f (a) with limx→a f (x). 1. f (x) = |x|; at x = 0 2. f (x) = |x−1|; at x = 0 3. f (x) = ⌊x⌋; at x = 0 4. f (x) = ⌊x⌋; at x = ... | ElementaryCalculus_Page_88_Chunk3623 |
Implicit Differentiation • Section 3.4 79 3.4 Implicit Differentiation A function y = f (x) is usually given by an explicit formula, such as y = x2. It is then straight- forward to find dy dx using the differentiation rules you have learned so far. But suppose instead that you were given merely an equation involving x a... | ElementaryCalculus_Page_89_Chunk3624 |
80 Chapter 3 • Topics in Differential Calculus §3.4 An algebraic curve is defined as the set of all points (x, y) satisfying a polynomial equation in the variables x and y, such as x2 −3xy4 + 1 = x5 −y2. An elliptic curve is a special case of an algebraic curve, where the polynomial has the specific form x3 + ax + b = y2... | ElementaryCalculus_Page_90_Chunk3625 |
Implicit Differentiation • Section 3.4 81 Example 3.28 x y 0 1 x2 + y2 = 1 (4/5,3/5) Find the tangent line to the curve x2 + y2 = 1 at the point (4/5,3/5). Solution: This curve is the unit circle, shown in the picture on the right. First use implicit differentiation to find dy dx : d dx (x2 + y2) = d dx (1) ⇒ 2x + 2y· d... | ElementaryCalculus_Page_91_Chunk3626 |
82 Chapter 3 • Topics in Differential Calculus §3.5 3.5 Related Rates If several quantities are related by an equation, then differentiating both sides of that equation with respect to a variable (usually t, representing time) produces a relation between the rates of change of those quantities. The known rates of chang... | ElementaryCalculus_Page_92_Chunk3627 |
Related Rates • Section 3.5 83 Example 3.31 The radius of a right circular cylinder is decreasing at the rate of 3 cm/min, while the height is increas- ing at the rate of 2 cm/min. Find the rate of change of the volume of the cylinder when the radius is 8 cm and the height is 6 cm. Solution: Let r, h, and V be the radi... | ElementaryCalculus_Page_93_Chunk3628 |
84 Chapter 3 • Topics in Differential Calculus §3.6 3.6 Differentials An ideal gas satisfies the equation PV = RT, where R is a constant and P, V, and T are the pressure, volume per mole, and temperature, respectively, of the gas. It will be proved that dP P + dV V = dT T . (3.6) Recall that dP, dV, and dT represent infi... | ElementaryCalculus_Page_94_Chunk3629 |
Differentials • Section 3.6 85 Let f and g be differentiable functions, and let c be a constant. Then: (a) d(c) = 0 (b) d(cf ) = c df (Constant Multiple Rule) (c) d(f + g) = df + dg (Sum Rule) (d) d(f −g) = df −dg (Difference Rule) (e) d(f g) = f dg + g df (Product Rule) (f) d µ f g ¶ = g df −f dg g2 (Quotient Rule) (g... | ElementaryCalculus_Page_95_Chunk3630 |
86 Chapter 3 • Topics in Differential Calculus §3.6 Example 3.33 Suppose that M is the total mass of a rocket and its unburnt fuel at any time t (so M is a function of t). Over an infinitesimal time dt a mass dm of fuel is burnt and the gas byproducts are expelled out the rear of the rocket at a velocity vE relative to ... | ElementaryCalculus_Page_96_Chunk3631 |
Differentials • Section 3.6 87 Example 3.34 Prove the relation dP P + dV V = dT T using logarithmic differentiation. Solution: Take the natural logarithm and then the differential of both sides of the equation PV = RT: ln(PV) = ln(RT) ⇒ ln P + ln V = ln R + ln T ⇒ d(ln P + ln V) = d(ln R + ln T) ⇒ dP P + dV V = 0 + dT ... | ElementaryCalculus_Page_97_Chunk3632 |
88 Chapter 3 • Topics in Differential Calculus §3.6 There are many benefits to using differentials—i.e. infinitesimals—in calculus.10 For exam- ple, recall Example 3.31 in Section 3.5 on related rates, where the volume V of a right circular cylinder with radius r and height h changes with time t as dV dt = µ 2πr · dr dt ... | ElementaryCalculus_Page_98_Chunk3633 |
Differentials • Section 3.6 89 7. Let y = cun, where c and n are constants. Show that dy y = n du u . 8. Obviously the derivative of the constant π2 is not 2π. But is d(π2) = 2πd(π) true? Explain. B 9. The continuity relation for an ideal gas is PM p T = constant where P and T are the pressure and temperature, respecti... | ElementaryCalculus_Page_99_Chunk3634 |
CHAPTER 4 Applications of Derivatives 4.1 Optimization Many physical problems involve optimization: finding either a maximum or minimum value of some quantity. Optimization problems often have a constraint involving two variables which allows you to rewrite the objective function—the function to optimize—as a function o... | ElementaryCalculus_Page_100_Chunk3635 |
Optimization • Section 4.1 91 Every global maximum [minimum] is a local maximum [minimum], but not vice versa. In physical applications global maxima or minima1 are the primary interest. The Extreme Value Theorem in Section 3.3 guarantees the existence of at least one global maximum and at least one global minimum for ... | ElementaryCalculus_Page_101_Chunk3636 |
92 Chapter 4 • Applications of Derivatives §4.1 There is a simple visual mnemonic device for remembering the Second Derivative Test, due to a generic minimum or maximum resembling a smile or frown, respectively: + + f ′′ > 0 local min. − − f ′′ < 0 local max. o o f ′′ = 0 test fails The “eyes” in the faces represent th... | ElementaryCalculus_Page_102_Chunk3637 |
Optimization • Section 4.1 93 and hence A = x µ L 2 −x ¶ = Lx 2 −x2 is now a function of x alone, on the open interval (0,L/2) (since the length x is positive). Now find the critical points of A: A′(x) = 0 ⇒ L 2 −2x = 0 ⇒ x = L 4 is the only critical point This problem is thus the case of a function defined on an open in... | ElementaryCalculus_Page_103_Chunk3638 |
94 Chapter 4 • Applications of Derivatives §4.1 Since both r and h are lengths and have to be positive, then 0 < r < ∞. So this is another case of a function defined on an open interval having only one critical point. Use the Second Derivative Test to verify that this critical point r = 3q V 2π is a local minimum for S:... | ElementaryCalculus_Page_104_Chunk3639 |
Optimization • Section 4.1 95 Then L is the value of x > 0 that makes y = 0: 0 = − gL2 2v2 0 cos2 θ + Ltan θ ⇒ L = 2v2 0 sin θ cos θ g = v2 0 sin 2θ g So L is now a function of θ, with 0 < θ < π/2 (why?). So if there is a single local maximum then it must be the global maximum. Now get the critical points of L: L′(θ) =... | ElementaryCalculus_Page_105_Chunk3640 |
96 Chapter 4 • Applications of Derivatives §4.1 Since time is distance divided by speed, and since the speed of light is constant, then minimizing the total time elapsed is equivalent to minimizing the total distance traveled, namely D = d1+d2. The basic idea here is that Fermat’s Principle implies that for the light t... | ElementaryCalculus_Page_106_Chunk3641 |
Optimization • Section 4.1 97 Example 4.5 A man is in a boat 4 miles off a straight coast. He wants to reach a point 10 miles down the coast in the minimum possible time. If he can row 4 mi/hr and run 5 mi/hr, where should he land the boat? p x2 +16 4 x 10−x 10 Y Solution: Let T be the total time traveled. The goal is ... | ElementaryCalculus_Page_107_Chunk3642 |
98 Chapter 4 • Applications of Derivatives §4.1 Example 4.7 Find the width and height of the rectangle with the largest possible perimeter inscribed in a semicircle of radius r. h r w 2 w Solution: Let w be the width of the rectangle and let h be the height, as in the picture. Then the perimeter is P = 2w +2h. By symme... | ElementaryCalculus_Page_108_Chunk3643 |
Optimization • Section 4.1 99 6. A man is in a boat 5 miles off a straight coast. He wants to reach a point 15 miles down the coast in the minimum possible time. If he can row 6 mi/hr and run 10 mi/hr, where should he land the boat? 7. The current I in a voltaic cell is I = E R + r , where E is the electromotive force ... | ElementaryCalculus_Page_109_Chunk3644 |
100 Chapter 4 • Applications of Derivatives §4.1 16. The stress τ in the xy-plane along a varying angle φ is given by τ = τ(φ) = σx −σy 2 sin 2φ + τxy cos 2φ , where σx, σy, and τxy are stress components that can be treated as constants. Show that the maxi- mum stress is τ = q¡ σx −σy ¢2 + 4τ2xy 2 . (Hint: Draw a right... | ElementaryCalculus_Page_110_Chunk3645 |
Optimization • Section 4.1 101 28. Find the equation of the tangent line to the ellipse x2 a2 + y2 b2 = 1 in the first quadrant that forms with the coordinate axes the right triangle with minimal area. 29. A “cold” star that has exhausted its nuclear fuel—called a white dwarf—has total energy E, given by E = ħ2 (3π2Nq)5... | ElementaryCalculus_Page_111_Chunk3646 |
102 Chapter 4 • Applications of Derivatives §4.2 4.2 Curve Sketching A function can increase between two points in different ways, as shown in Figure 4.2.1. y x y = f (x) (a) f ′′ = 0: straight y x y = f (x) (b) f ′′ > 0: concave up y x y = f (x) (c) f ′′ < 0: concave down Figure 4.2.1 Increasing function f : f ′ > 0, ... | ElementaryCalculus_Page_112_Chunk3647 |
Curve Sketching • Section 4.2 103 Concavity Theorem: Suppose that f is a twice-differentiable function on [a,b]. Then: (a) If f ′′(x) > 0 on (a,b) then f (x) is below the line l(x) joining the points (a, f (a)) and (b, f (b)) for all x in (a,b). (b) If f ′′(x) < 0 on (a,b) then f (x) is above the line l(x) joining the ... | ElementaryCalculus_Page_113_Chunk3648 |
104 Chapter 4 • Applications of Derivatives §4.2 y x 0 (a) f (x) = x3: inflection point at x = 0 y x 0 (b) f (x) = x4: non-inflection point at x = 0 Figure 4.2.3 Inflection vs non-inflection point at x = 0 with f ′′(0) = 0 Figure 4.2.3(b) shows that a point where the second derivative is 0 is a possible inflection point, bu... | ElementaryCalculus_Page_114_Chunk3649 |
Curve Sketching • Section 4.2 105 Example 4.9 Sketch the graph of f (x) = −x 1 + x2 . Find all local maxima and minima, inflection points, where the func- tion is increasing or decreasing, and where the function is concave up or concave down. Also indicate any asymptotes. Solution: Since f ′(x) = x2−1 (1+x2)2 then x = 1... | ElementaryCalculus_Page_115_Chunk3650 |
106 Chapter 4 • Applications of Derivatives §4.2 Example 4.10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 −2−1.5−1−0.5 0 0.5 1 1.5 2 y x Sketch the graph of f (x) = x2/3. Solution: Clearly f (x) is continuous for all x, including x = 0 (since f (0) = 0), but f ′(x) = 2 3 3px is not defined at x = 0. Since f ′(x) changes from negative t... | ElementaryCalculus_Page_116_Chunk3651 |
Curve Sketching • Section 4.2 107 A common practice in many fields of science and engineering is to combine multiple named constants (e.g. π) or variables in a function into one variable and then sketch a graph of that function. The example below illustrates the technique. Example 4.12 A hydrogen atom has one electron, ... | ElementaryCalculus_Page_117_Chunk3652 |
108 Chapter 4 • Applications of Derivatives §4.2 Example 4.13 For a single particle with two states—energy 0 and energy ǫ—in thermal contact with a reservoir at temperature τ, the average energy U and heat capacity CV are given by U = ǫ e−ǫ/τ 1+ e−ǫ/τ and CV = kB ³ ǫ τ ´2 eǫ/τ ¡ 1+ eǫ/τ¢2 where kB ≈1.38065×10−23 J/K is... | ElementaryCalculus_Page_118_Chunk3653 |
Numerical Approximation of Roots of Functions • Section 4.3 109 4.3 Numerical Approximation of Roots of Functions When finding critical points of a function f , you encounter the problem of solving the equation f ′(x) = 0. The examples and exercises so far were set up carefully so that solutions to that equation could b... | ElementaryCalculus_Page_119_Chunk3654 |
110 Chapter 4 • Applications of Derivatives §4.3 The general formula for the number xn obtained after n ≥1 iterations in Newton’s method can be determined by considering the formula for x1. First, the tangent line to y = f (x) at the point (x0, f (x0)) has slope f ′(x0), so the equation of the line is y −f (x0) = f ′(x... | ElementaryCalculus_Page_120_Chunk3655 |
Numerical Approximation of Roots of Functions • Section 4.3 111 Example 4.14 Use Newton’s method to find the root of f (x) = cos x−x. Solution: Since the root is already known to be in the interval [0,1], choose x0 = 1 as the initial guess. The numbers xn for n ≥1 can be computed with a hand-held scientific calculator, b... | ElementaryCalculus_Page_121_Chunk3656 |
112 Chapter 4 • Applications of Derivatives §4.3 n=0: 1.0 n=1: 0.7503638678402439 n=2: 0.7391128909113617 n=3: 0.739085133385284 n=4: 0.7390851332151607 n=5: 0.7390851332151607 n=6: 0.7390851332151607 n=7: 0.7390851332151607 n=8: 0.7390851332151607 n=9: 0.7390851332151607 n=10: 0.7390851332151607 Note that the solution... | ElementaryCalculus_Page_122_Chunk3657 |
Numerical Approximation of Roots of Functions • Section 4.3 113 Since the secant line through (x0, f (x0)) and (x1, f (x1)) has slope f (x1)−f (x0) x1−x0 , the equation of that secant line is: y −f (x1) = f (x1)−f (x0) x1 −x0 (x −x1) The point (x2,0) is on that line, so that 0 −f (x1) = f (x1)−f (x0) x1 −x0 (x2 −x1) ⇒ ... | ElementaryCalculus_Page_123_Chunk3658 |
114 Chapter 4 • Applications of Derivatives §4.3 One difference you might have noticed between the secant method and Newton’s method is that the secant method does not use derivatives. The secant method replaces the derivative in Newton’s method with the slope of a secant line which approximates the derivative (recall ... | ElementaryCalculus_Page_124_Chunk3659 |
Numerical Approximation of Roots of Functions • Section 4.3 115 n=2: 0.6850733573260451 n=3: 0.736298997613654 n=4: 0.7391193619116293 n=5: 0.7390851121274639 n=6: 0.7390851332150012 n=7: 0.7390851332151607 n=8: 0.7390851332151607 n=9: NaN n=10: NaN Notice that the root was found after 6 iterations (n = 7). The undefine... | ElementaryCalculus_Page_125_Chunk3660 |
116 Chapter 4 • Applications of Derivatives §4.3 To see this, notice that Newton’s method always requires that both f (xn−1) and f ′(xn−1) be computed for the nth term xn in the sequence. The secant method needs f (xn−1) and f (xn−2) for the nth term, but a good programmer would save the value of f (xn−1) so that it co... | ElementaryCalculus_Page_126_Chunk3661 |
Numerical Approximation of Roots of Functions • Section 4.3 117 There are conditions under which Newton’s method is guaranteed to work, and convergence is fast. Newton’s method has a quadratic rate of convergence, meaning roughly that the error terms—the differences between approximate roots and the actual root—are bei... | ElementaryCalculus_Page_127_Chunk3662 |
118 Chapter 4 • Applications of Derivatives §4.4 4.4 The Mean Value Theorem The difference between instantaneous and average rates of change has been discussed in ear- lier sections. Recall that there is no difference between the two for linear functions. For nonlinear functions the average rate of change over an inter... | ElementaryCalculus_Page_128_Chunk3663 |
The Mean Value Theorem • Section 4.4 119 Rolle’s Theorem: Let a and b be real numbers such that a < b, and suppose that f is a function such that (a) f is continuous on [a,b], (b) f is differentiable on (a,b), and (c) f (a) = f (b) = 0. Then there is at least one number c in the interval (a,b) such that f ′(c) = 0. y x... | ElementaryCalculus_Page_129_Chunk3664 |
120 Chapter 4 • Applications of Derivatives §4.4 If f is a differentiable function on an interval I such that f ′(x) = 0 for all x in I, then f is a constant function on I. Note that I can be any interval, even the entire real line (−∞,∞). It is already known that f = constant ⇒f ′ = 0; the above result says that the c... | ElementaryCalculus_Page_130_Chunk3665 |
The Mean Value Theorem • Section 4.4 121 The Mean Value Theorem can help in proving inequalities, often used in the sciences for establishing upper or lower bounds on a quantity (e.g. worst-case scenario). Example 4.16 Show that sin x ≤x for all x ≥0. Solution: The inequality holds trivially for x = 0, since sin 0 = 0 ... | ElementaryCalculus_Page_131_Chunk3666 |
122 Chapter 4 • Applications of Derivatives §4.4 The Mean Value Theorem says that the derivative of a differentiable function will always attain one particular value on a closed interval: the function’s average rate of change over the interval. It turns out that the derivative will take on every value between its value... | ElementaryCalculus_Page_132_Chunk3667 |
The Mean Value Theorem • Section 4.4 123 9. Suppose that f and g are continuous on [a,b] and differentiable on (a,b), and that f ′(x) > g′(x) for all a < x < b. Show that f (b)−g(b) > f (a)−g(a). 10. Prove the Extended Mean Value Theorem, by applying Rolle’s Theorem to the function F(x) = f (x) −f (a) −f (b)−f (a) g(b)... | ElementaryCalculus_Page_133_Chunk3668 |
CHAPTER 5 The Integral 5.1 The Indefinite Integral Derivatives appear in many physical phenomena, such as the motion of objects. Recall, for example, that given the position function s(t) of an object moving along a straight line at time t, you could find the velocity v(t) = s′(t) and the acceleration a(t) = v′(t) of the... | ElementaryCalculus_Page_134_Chunk3669 |
The Indefinite Integral • Section 5.1 125 To see some of the issues involved, consider a simple function like f (x) = 2x. Of course you know that d dx(x2) = 2x, so it seems that F(x) = x2 is the antiderivative of f (x) = 2x. But is it the only antiderivative of f (x)? No. For example, if F(x) = x2 +1 then F′(x) = 2x = f... | ElementaryCalculus_Page_135_Chunk3670 |
126 Chapter 5 • The Integral §5.1 Though the indefinite integral R f (x) dx represents all antiderivatives of f (x), the integral can be thought of as a single object or function in its own right, whose derivative is f (x): d dx µZ f (x) dx ¶ = f (x) You might be wondering what the integral sign in the indefinite integra... | ElementaryCalculus_Page_136_Chunk3671 |
The Indefinite Integral • Section 5.1 127 Power Formula: Z xn dx = xn+1 n+1 + C if n ̸= −1 ln|x| + C if n = −1 The following rules for indefinite integrals are immediate consequences of the rules for derivatives: Let f and g be functions and let k be a constant. Then: 1. Z k f (x) dx = k Z f (x) dx 2. Z (f ... | ElementaryCalculus_Page_137_Chunk3672 |
128 Chapter 5 • The Integral §5.1 Example 5.5 Evaluate Z px dx. Solution: Use the Power Formula: Z px dx = Z x1/2 dx = x3/2 3/2 + C = 2x3/2 3 + C Example 5.6 Evaluate Z µ 1 x2 + 1 x ¶ dx. Solution: Use the Power Formula and integrate term by term: Z µ 1 x2 + 1 x ¶ dx = Z µ x−2 + 1 x ¶ dx = x−1 −1 + ln|x| + C = −1 x + l... | ElementaryCalculus_Page_138_Chunk3673 |
The Indefinite Integral • Section 5.1 129 Example 5.7 Evaluate Z (3sin x + 4cos x −5ex) dx. Solution: Integrate term by term: Z (3sin x + 4cos x −5ex) dx = 3 Z sin x dx + 4 Z cos x dx −5 Z ex dx = −3cos x + 4sin x −5ex + C Example 5.8 Recall from Section 1.1 the example of an object dropped from a height of 100 ft. Show... | ElementaryCalculus_Page_139_Chunk3674 |
130 Chapter 5 • The Integral §5.1 Free fall motion: At time t ≥0: acceleration: a(t) = −g velocity: v(t) = −gt + v0 position: s(t) = −1 2 gt2 + v0t + s0 initial conditions: s0 = s(0), v0 = v(0) Note that the units are not specified—they just need to be consistent. In metric units, g = 9.8 m/s2, while g = 32 ft/s2 in Eng... | ElementaryCalculus_Page_140_Chunk3675 |
The Indefinite Integral • Section 5.1 131 Solution: Integrating both sides of the equation yields Z dP P + Z dV V = Z dT T ln P + ln V = ln T + C (C is a constant) ln(PV) = ln T + C PV = eln T+C = eln T · eC = T eC = RT where R = eC is a constant. ✓ The integration formulas in this section depended on already knowing th... | ElementaryCalculus_Page_141_Chunk3676 |
132 Chapter 5 • The Integral §5.2 5.2 The Definite Integral Recall from the last section that the integral sign in the indefinite integral Z f (x) dx represents a summation of the infinitesimals f (x)dx = dF for an antiderivative F(x) of f (x). Why is the term “indefinite” used? Because the summation is indefinite: the x in... | ElementaryCalculus_Page_142_Chunk3677 |
The Definite Integral • Section 5.2 133 f (x) df = f (x+ dx)−f (x) y = f (x) x x+ dx dx A B C Figure 5.2.2 Area under the curve y = f (x) over [x,x+ dx] By the Microstraightness Property, the curve y = f (x) shown in Figure 5.2.1 is a straight line over the infinitesimal interval [x, x+ dx], as shown in Figure 5.2.2.1 Th... | ElementaryCalculus_Page_143_Chunk3678 |
134 Chapter 5 • The Integral §5.2 y = f (x) y x a b R Figure 5.2.3 The area A of the region R equals Rb a f (x)dx In Figure 5.2.3 the area under the curve y = f (x) between x = a and x = b is the area A of the shaded region R, namely A = Rb a f (x)dx. To calculate that area for a specific function, rect- angles can agai... | ElementaryCalculus_Page_144_Chunk3679 |
The Definite Integral • Section 5.2 135 The limit in Formula (5.2) should be taken over all partitions whose norm—the length of the largest subinterval—approaches 0. In practice, however, the partitions are usually chosen so that the subintervals are of equal length, and then simply make those equal lengths smaller and ... | ElementaryCalculus_Page_145_Chunk3680 |
136 Chapter 5 • The Integral §5.2 The following summation formulas can be helpful when calculating Riemann sums: Let n ≥1 be a positive integer. Then: (1) nX k=1 1 = n (2) nX k=1 k = 1 + 2 + ··· + n = n(n+1) 2 (3) nX k=1 k2 = 12 + 22 + ··· + n2 = n(n+1)(2n+1) 6 (4) nX k=1 k3 = 13 + 23 + ··· + n3 = n2(n+1)2 4 (5) nX k=1... | ElementaryCalculus_Page_146_Chunk3681 |
The Definite Integral • Section 5.2 137 Formulas (3)-(5) can be proved similarly by induction (see the exercises). The example below shows how Formulas (2) and (3) are used in finding the limit of a Riemann sum. Example 5.11 Use Riemann sums to calculate Z2 1 x2 dx. Solution: The definite integral is the area under the cu... | ElementaryCalculus_Page_147_Chunk3682 |
138 Chapter 5 • The Integral §5.2 It it often simpler to use a computer to calculate approximations of a definite integral, by taking the Riemann sum of a sufficiently large number of rectangles in order to achieve the desired accuracy. Choosing subintervals of equal length, as in Example 5.11, makes it easier to use an ... | ElementaryCalculus_Page_148_Chunk3683 |
The Definite Integral • Section 5.2 139 Note: In the definite integral Zb a f (x) dx the numbers a and b are called the limits of integra- tion, with a being the lower limit of integration and b the upper limit of integration. The function f (x) being integrated is called the integrand, in both definite and indefinite inte... | ElementaryCalculus_Page_149_Chunk3684 |
140 Chapter 5 • The Integral §5.3 5.3 The Fundamental Theorem of Calculus Using Riemann sums to calculate definite integrals can be tedious, as was seen in the previous section. In fact the technique shown in that section depended on the function being a low- degree polynomial, which obviously will not always be the cas... | ElementaryCalculus_Page_150_Chunk3685 |
The Fundamental Theorem of Calculus • Section 5.3 141 By the Microstraightness Property the curve y = f (x) is a straight line over the infinitesimal interval [x, x + dx], so f must be either increasing, constant, or decreasing over that interval. The three possibilities are shown in Figure 5.3.3: f (x) f (x) df y = f (... | ElementaryCalculus_Page_151_Chunk3686 |
142 Chapter 5 • The Integral §5.3 By definition A(a) = 0, since it is the area under the curve over the interval [a,a] of zero length. Thus, 0 = A(a) = F(a) + C ⇒ C = −F(a) ⇒ A(x) = F(x) −F(a) for all x in [a,b] and so Zb a f (x) dx = A(b) = F(b) −F(a) which proves Part II of the theorem.3 ✓ Note: In some textbooks Part... | ElementaryCalculus_Page_152_Chunk3687 |
The Fundamental Theorem of Calculus • Section 5.3 143 Example 5.14 Calculate Z1 −1 x3 dx. Solution: Since F(x) = x4 4 is an antiderivative of f (x) = x3, then Z1 −1 x3 dx = x4 4 ¯¯¯¯ 1 −1 = 14 4 −(−1)4 4 = 1 4 −1 4 = 0 . Example 5.14 is a special case of the following result for odd functions: If f is an odd function, ... | ElementaryCalculus_Page_153_Chunk3688 |
144 Chapter 5 • The Integral §5.3 The following rules for definite integrals are a consequence of the corresponding rules for indefinite integrals: Let f and g be continuous functions on [a,b] and let k be a constant. Then: 1. Zb a k f (x) dx = k Zb a f (x) dx 2. Zb a (f (x)+ g(x)) dx = Zb a f (x) dx + Zb a g(x) dx 3. Zb... | ElementaryCalculus_Page_154_Chunk3689 |
The Fundamental Theorem of Calculus • Section 5.3 145 Example 5.15 Let F(x) = Zx2 0 e−t2 dt for all x > 0. Find F′(x). Solution: By the Chain Rule for integrals, with f (t) = et2 and g(x) = x2: F′(x) = f (g(x))· g′(x) = e−(x2)2 ·(2x) = 2x e−x4 Exercises A For Exercises 1-12, evaluate the given definite integral. 1. Z1 0... | ElementaryCalculus_Page_155_Chunk3690 |
146 Chapter 5 • The Integral §5.4 5.4 Integration by Substitution The integrals encountered so far—whether indefinite or definite—have been the simplest kind, since the antiderivatives had known formulas. For example, R cos x dx = sin x+C. What if the integral were R cos 2x dx instead? No formula has been discussed yet f... | ElementaryCalculus_Page_156_Chunk3691 |
Integration by Substitution • Section 5.4 147 Example 5.17 Evaluate Z e−3x dx. Solution: The −3x in the exponential function is what makes this integral unknown, so make the substitution u = −3x, which means that du = −3dx, and so dx = −1 3du. Thus: Z e−3x dx = Z eu µ −1 3 du ¶ = −1 3 Z eu du = −1 3 eu + C = −1 3 e−3x ... | ElementaryCalculus_Page_157_Chunk3692 |
148 Chapter 5 • The Integral §5.4 Example 5.21 Evaluate Z x2 dx p x3 +9 . Solution: Let u = x3 +9, so that du = 3x2 dx ⇒x2 dx = 1 3 du. Thus: Z x2 dx p x3 +9 = Z 1 3 du pu = 1 3 Z u−1/2 du = 1 3 u1/2 1/2 + C = 2 3 p x3 +9 + C Example 5.22 Evaluate Z 2x dx x2 −1. Solution: Notice that the numerator 2x in the function is... | ElementaryCalculus_Page_158_Chunk3693 |
Integration by Substitution • Section 5.4 149 For any constant a > 0: Z dx p a2 −x2 = sin−1³ x a ´ + C (if |x| < a) (5.3) Z dx a2 + x2 = 1 a tan−1³ x a ´ + C (5.4) Z dx |x| p x2 −a2 = 1 a sec−1³ x a ´ + C (if |x| > a) (5.5) For example, to prove the second formula, recall that d dx ¡ tan−1 x ¢ = 1 1+x2 . Make the subst... | ElementaryCalculus_Page_159_Chunk3694 |
150 Chapter 5 • The Integral §5.4 The following property of definite integrals comes in handy for evaluating certain types of definite integrals: For any constant a, Za 0 f (x) dx = Za 0 f (a−x) dx . (5.6) This is simple to prove, using the substitution u = a−x, so x = a−u and dx = −du, while x = 0 becomes u = a and x = ... | ElementaryCalculus_Page_160_Chunk3695 |
Integration by Substitution • Section 5.4 151 13. Z1 0 (2x+1)3 dx 14. Z1 0 (2x−1)3 dx 15. Z8 0 x p 1+ x dx 16. Zπ/2 0 4 sin(x/2) dx 17. Zπ/4 0 4 sin x cos x dx 18. Zpπ 0 5x cos(x2) dx 19. Z−1 −2 x (x2 +2)3 dx 20. Zln 3 −ln 3 ex ex +4 dx 21. Z3 1 dx px (x+1) 22. Z1 −1 x2 dx p x3 +9 23. Z2 1 dx x2 −6x+9 24. Z3 −3 x5 dx e... | ElementaryCalculus_Page_161_Chunk3696 |
152 Chapter 5 • The Integral §5.5 5.5 Improper Integrals Definite integrals so far have been defined only for continuous functions over finite closed intervals. There are times when you will need to perform integration despite those conditions not being met. For example, in quantum mechanics the Dirac delta function4 δ is... | ElementaryCalculus_Page_162_Chunk3697 |
Improper Integrals • Section 5.5 153 The limits in the above definitions are always taken after evaluating the integral inside the limit. Just as for “proper” definite integrals, improper integrals can be interpreted as representing the area under a curve. Example 5.28 Evaluate Z∞ 1 dx x . y x 0 1 y = 1 x Solution: For a... | ElementaryCalculus_Page_163_Chunk3698 |
154 Chapter 5 • The Integral §5.5 Example 5.31 Evaluate Z∞ 0 sin x dx . y x 0 2π 1 −1 y = sin x Solution: Since Z∞ 0 sin x dx = lim b→∞ Zb 0 sin x dx = lim b→∞ à −cos x ¯¯¯¯ b 0 ! = lim b→∞(−cos b + 1) then the integral is divergent, since limb→∞cos b does not exist (cosb oscillates between 1 and -1). This means that t... | ElementaryCalculus_Page_164_Chunk3699 |
Improper Integrals • Section 5.5 155 The second type of improper integral is of a function not continuous or not bounded over its interval of integration. For example, the integral in property (3) of the Dirac delta function is of that type, since δ is discontinuous at x = 0. Define this type of improper integral as fol... | ElementaryCalculus_Page_165_Chunk3700 |
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