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Unformatted text preview: M 106 Integral Calculus and Applications Contents 1 The Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.1 Antiderivatives and Indefinite Integrals 1.1.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.1.2 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Properties of the Indefinite Integral 7 1.3 Integration By Substitution 9 2 The Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.1 Summation Notation 16 2.2 Riemann Sum and Area 19 2.3 Properties of the Definite Integral 23 2.4 The Fundamental Theorem of Calculus 26 2.5 Numerical Integration 33 2.5.1 Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.5.2 Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3 Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.1 The Natural Logarithmic Function 3.1.1 Properties of the Natural Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.1.2 Differentiating and Integrating the Natural Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.2 The Natural Exponential Function 3.2.1 Properties of the Natural Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.2.2 Differentiating and Integrating the Natural Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.3 General Exponential and Logarithmic Functions 3.3.1 General Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.3.2 General Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4 Inverse Trigonometric and Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 4.1 Inverse Trigonometric Functions 62 4.2 Hyperbolic Functions 66 4.2.1 Properties of the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.2.2 Differentiating and Integrating the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.3 Inverse Hyperbolic Functions 4.3.1 Properties the Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 4.3.2 Differentiating and Integrating the Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5 Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 5.1 Integration by Parts 4 44 50 53 72 81 5.2 5.2.1 5.2.2 Trigonometric Functions 84 Integration of Powers of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Integration of Forms sin ux cos vx, sin ux sin vx and cos ux cos vx . . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.3 Trigonometric Substitutions 89 5.4 Integrals of Rational Functions 92 5.5 Integrals Involving Quadratic Forms 95 5.6 5.6.1 5.6.2 Miscellaneous Substitutions 97 Fractional Functions in sin x and cos x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Integrals of Fractional Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 5.6.3 Integrals of Form 6 Indeterminate Forms and Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 6.1 Indeterminate Forms 103 6.2 6.2.1 6.2.2 Improper Integrals Infinite Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Discontinuous Integrands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 108 109 7 Application of Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 7.1 Areas 113 7.2 Solids of Revolution 117 7.3 7.3.1 7.3.2 7.3.3 Volumes of Solids of Revolution Disk Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Washer Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Method of Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 119 122 128 7.4 7.4.1 7.4.2 Arc Length and Surfaces of Revolution 131 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Surfaces of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 8 Parametric Equations and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 8.1 8.1.1 8.1.2 Parametric Equations of Plane Curves 142 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Arc Length and Surface Area of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 8.2 8.2.1 8.2.2 8.2.3 Polar Coordinates System 150 The Relationship between Rectangular and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Tangent Line to Polar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Graphs in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 8.3 8.3.1 Area in Polar Coordinates Arc Length and Surface Area of Revolution in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . p n f (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 161 166 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Appendix (1): Basic Mathematical Concepts 173 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Appendix (1): Integration Rules and Integrals Table 187 Appendix (2): Answers to Exercises 190 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 4 Chapter 1 The Indefinite Integrals 1.1 Antiderivatives and Indefinite Integrals We begin with the definition of the antiderivatives and indefinite integrals. Then, we provide basic integration rules. 1.1.1 Antiderivatives Definition 1.1 A function F is called an antiderivative of f on an interval I if 0 F (x) = f (x) for every x ∈ I. Example 1.1 (1) Let F(x) = x2 + 3x + 1 and f (x) = 2x + 3. 0 Since F (x) = f (x), then the function F(x) is an antiderivative of f (x). (2) Let G(x) = sin x + x and g(x) = cos x + 1. 0 Since G (x) = cos x + 1, then the function G(x) is an antiderivative of g(x). If F(x) is an antiderivative of f (x), then every function F(x)+c is also antiderivative of f (x), where c is a constant. The upcoming theorem states that any antiderivative G(x), which is different from F(x) can be expressed as F(x) + c where c is an arbitrary constant. Theorem 1.1 If functions F and G are antiderivatives of a function f on an interval I, there exists a constant c such that G(x) = F(x) + c. Proof. Let H be a function defined as follows: H(x) = G(x) − F(x) ∀x ∈ I where F and G are antiderivatives of the function f . Let a, b ∈ I such that a < b. Since F and G are antiderivatives of f , then H 0 (x) = G0 (x) − F 0 (x) = f (x) − f (x) = 0 for every x ∈ I. Since the function H is differentiable, it is continuous. From the mean value theorem on [a, b], there is a number c ∈ (a, b) such that H(b) − H(a) H 0 (c) = . b−a 1 Since H 0 (x) = 0 on I, then H 0 (c) = 0. This implies H(a) = H(b) and this means H is a constant function.  1 If f is continuous on [a, b] and differentiable on (a, b), there exists a number c ∈ (a, b) such that f 0 (c) = f (b)− f (a) . b−a 5 Example 1.2 Let f (x) = 2x. The functions F(x) = x2 + 2, G(x) = x2 − 21 , √ H(x) = x2 − 3 2, are antiderivatives of the function f . Therefore, F(x) = x2 + c is a general form of the antiderivatives of the function f (x) = 2x. Example 1.3 Find the general form of the antiderivatives of f (x) = 6x5 . Solution: If F(x) = x6 , then F 0 (x) = 6x5 . The function F(x) = x6 + c is the general antiderivative of f . 1.1.2 Indefinite Integrals From Theorem 1.1, if the function F(x) + c is an antiderivative of f (x), then there exist no antiderivatives in different forms for the function f (x). This leads us to define the indefinite integral. Definition 1.2 Let f be a continuous function on an interval I. The indefinite integral of f is the general antiderivative of f on I: Z f (x) dx = F(x) + c. Z The function f is called the integrand, the symbol is the integral sign, x is called the variable of the integration and c is the constant of the integration. Now, by using the previous definition, the general antiderivatives in Example 1.1 are Z 1 (2x + 3) dx = x2 + 3x + c. Z 2 (cos x + 1) dx = sin x + x + c. We can now work out how to evaluate some integrals. To do that, we should remember differentiation rules of some functions. Basic Integration Rules Rule 1: Power of x. d n+1 x = (n + 1)xn , so dx Z (n + 1)xn dx = xn+1 + c Generally, for n 6= −1, xn+1 + c. n+1 In words, to integrate the function xn , we add 1 to the power and divide the function by n + 1. If n = 1, we have a special case Z xn dx = Z 1 dx = x + c. Rule 2: Trigonometric functions. Z Therefore, sin x dx = − cos x + c. d sin x = cos x, so dx Z d cos x = − sin x, so dx Z cos x dx = sin x + c − sin x dx = cos x + c 6 The other trigonometric functions with the previous rules are listed in the following table: Derivative 1 dx = x + c n+1 d x n dx ( n+1 ) = x , d dx (sin x) = cos x xn dx = xn+1 n+1 +c cos x dx = sin x + c Z x) = sin x x) = sec2 d dx (− cot d dx (sec n 6= −1 Z Z d dx (− cos d dx (tan Indefinite Integral Z d dx (x) = 1 sec2 x dx = tan x + c Z csc2 x dx = − cot x + c x x) = csc2 x sin x dx = − cos x + c Z Z x) = sec x tan x d dx (− csc sec x tan x dx = sec x + c Z x) = csc x cot x csc x cot x dx = − csc x + c Table 1.1: The list of basic integration rules. Example 1.4 Evaluate the integral. Z x−3 dx (1) Z (2) 1 dx cos2 x Solution: Z x−3 dx = Z 1 dx = cos2 x (1) (2) x−2 −2 + c = − 2x12 + c. Z sec2 x dx = tan x + c. (sec x = 1 cos x ⇒ sec2 x = 1 ) cos2 x Note that we sometimes need to express an integrand in a form in which we can recognize its derivative like item 2 in the previous example. Exercise 1.1 1-8 Z Evaluate the integral. 1 √ dx x 1 Z 2 1 x 1 √ dx 5 x tan x dx cos x Z √ x 7 dx x3 Z p 8 sin4 x csc x dx Z dx Z 1 dx sin2 x Z − csc2 x tan2 x dx 3 4 5 4 Z 5 6  7 1.2 Properties of the Indefinite Integral In this section, we list some properties of the indefinite integrals. Theorem 1.2 Assume f and g have antiderivatives on an interval I, then 1. d dx Z Z d (F(x)) dx = F(x) + c. dx 2. Z f (x) dx = f (x). Z Z  f (x) ± g(x) dx = f (x) dx ± g(x) dx. 3. Z Z k f (x) dx = k 4. f (x) dx, where k is a constant. Proof. For items 1 and 2, let F be an antiderivative of f . 1. d dx Z Z d (F(x)) dx = dx 2. f (x) dx = d dx  F(x) + c = f (x). Z f (x) dx = F(x) + c. 3. Let F and G be antiderivatives of f and g, respectively. By differentiating the left side, we have d  dx Z Z Hence,    d f (x) ± g(x) dx = F(x) ± G(x) dx = f (x) ± g(x).  f (x) ± g(x) dx = F(x) ± G(x) + c1 . From the right side, we have Z f (x) dx ± Z g(x) dx = F(x) ± G(x) + c2 For any special case, we can choose the values of the constants such that c1 = c2 and this prove item 3. 4. By differentiating the left side, we have d  dx Z Z    d  d k f (x) dx = k f (x) dx = k F(x) dx dx = k f (x). Z Hence, k f (x) dx = kF(x) + c1 . From the right side, we have Z k f (x) dx = kF(x) + c2 We can choose the values of the constants such that c1 = c2 and this prove item 4.  In the following example, we use the previous properties and the table of the basic integration rules to evaluate some indefinite integrals. Example 1.5 Evaluate the integral. Z (4x + 3) dx (1) Z (2) Z (3) (2 sin x + 3 cos x) dx √ ( x + sec2 x) dx 8 d (sin x) dx dx Z √ d x + 1 dx (5) dx Z (4) Solution: Z (4x + 3) dx = (1) Z (2) Z (3) 4x2 2 + 3x + c = 2x2 + 3x + c. (2 sin x + 3 cos x) dx = −2 cos x + 3 sin x + c. 3 3 √ x2 + tan x + c = 2x32 + tan x + c. ( x + sec2 x) dx = 3/2 d (sin x) dx = sin x + c. dx Z √ √ d (5) x + 1 dx = x + 1. dx Z (4) Z Example 1.6 If f (x) dx = x2 + c1 and Z Z g(x) dx = tan x + c2 , find  3 f (x) − 2g(x) dx. Solution: Z From the third and fourth properties, Z Z  3 f (x) − 2g(x) dx = 3 f (x) dx − 2 g(x) dx = 3x2 − 2 tan x + c, where c = 3c1 − 2c2 . Example 1.7 Solve the differential equation f 0 (x) = x3 subject to the initial condition f (0) = 1. Solution: Z f 0 (x) dx = Z x3 dx 1 ⇒ f (x) = x4 + c. 4 If x = 0, then f (0) = 14 (0)4 + c = 1 and this implies c = 1. Hence, the solution of the differential equation is f (x) = 41 x4 + 1. Example 1.8 Solve the differential equation f 0 (x) = 6x2 + x − 5 subject to the initial condition f (0) = 2. Solution: Z f 0 (x) dx = Z (6x2 + x − 5) dx 1 ⇒ f (x) = 2x3 + x2 − 5x + c. 2 Use the condition f (0) = 2 by substituting x = 0 into the function f (x). We have f (0) = 0 + 0 − 0 + c = 2 ⇒ c = 2. Therefore, the solution of the differential equation is f (x) = 2x3 + 21 x2 − 5x + 2. Example 1.9 Solve the differential equation f 00 (x) = 5 cos x + 2 sin x subject to the initial conditions f (0) = 3 and f 0 (0) = 4. Solution: Z f 00 (x) dx = Z (5 cos x + 2 sin x) dx ⇒ f 0 (x) = 5 sin x − 2 cos x + c Using the condition f 0 (0) = 4 gives f 0 (0) = 0 − 2 + c = 4 ⇒ c = 6. (use values of the trigonometric functions given on page 180) 9 Hence, f 0 (x) = 5 sin x − 2 cos x + 6. Now, again Z f 0 (x) dx = Z (5 sin x − 2 cos x + 6) dx ⇒ f (x) = −5 cos x − 2 sin x + 6x + c. Use the condition f (0) = 3 by substituting x = 0 into f (x). We obtain f (0) = −5 − 0 + 0 + c = 3 ⇒ c = 8. Hence, the solution of the differential equation is f (x) = −5 cos x − 2 sin x + 6x + 8. Notes: We can always check our answers by differentiating the results. In the previous examples, we use x as a variable of the integration. However, for this role, we can use any variable such as y, z, t, etc . That is, instead of f (x) dx, we can integrate f (y) dy or f (t) dt. The properties of the indefinite integral and the table of the basic integrals are elementary for simple functions. Meaning that, for more complex functions, we need some techniques to simplify the integrals. Section 1.3, we shall provide one of these techniques. Exercise 1.2 1 - 10 1 Evaluate the integral. Z p Z 2 Z 3 Z 4 Z 5 3 7 x(x3 + 2x + 1) dx 8 (x2 + sec2 x) dx 9 √ (csc2 x − x) dx 10 13 - 17 3 sin2 x + 4 dx sin2 x Z x2 − 1 dx x4 Z  2 2 4x 5 − 2x 3 + x dx Z  3 2 + + 1 dx x3 x4 Z x2 + x + 1 √ dx 3 x Evaluate. Z p d dx ( Z 12 Z 6 (x 4 + x2 + 1) dx 11 - 12 11 x5 dx cos3 x + 1 dx) d p 3 ( cos x + 1) dx dx Solve the differential equation subject to the given conditions. 13 f 0 (x) = 4x3 + 2x + 1; f (0) = 1. 14 f 00 (x) = sin x + 2 cos x; f (0) = 1 and f 0 (0) = 3. √ 15 f 0 (x) = x; f (0) = 0. 16 f 0 (x) = cos x; f (π) = 1. 17 f 0 (x) = sec2 x; f ( π4 ) = 0.  1.3 Integration By Substitution The integration by substitution (known as u-substitution) is a technique for solving some composite functions. The method is based on changing the variable of the integration to obtain a simple indefinite integral. The following theorem shows how the substitution technique works. 10 Theorem 1.3 Let g be a differentiable function on an interval I where the derivative is continuous. Let f be continuous on the interval J contains the range of the function g. If F is an antiderivative of the function f on J, then Z Proof. Since F is an antiderivative of f , then 0 f (g(x))g (x) dx = F(g(x)) + c, x ∈ I. 0 0 d dx F(g(x)) = F (g(x))g (x) = Z Z 0 f (g(x))g (x) dx = 0 f (g(x))g (x). Hence, d F(g(x)) dx = F(g(x)) + c.  dx 0 The task here is to recognize whether an integrand has the form f (g(x))g (x). The following two examples explain this task. Z Example 1.10 Evaluate the integral 2x (x2 + 1)3 dx. Solution: We can use the previous theorem as follows:  0 let f (x) = x3 and g(x) = x2 + 1, then f g(x) = (x2 + 1)3 . Since g (x) = 2x, then from Theorem 1.3, we have Z 2x(x2 + 1)3 dx = (x2 + 1)4 + c. 4 We can end with the same solution by using the five steps of the substitution method given below. Steps of the integration by substitution: Step 1: Choose a new variable u. Step 2: Determine the value of du. Step 3: Make the substitution i.e., eliminate all occurrences of x in the integral by making the entire integral is in terms of u. Step 4: Evaluate the new integral. Step 5: Return the evaluation to the initial variable x. In Example 1.10, let u = x2 + 1, then du = 2x dx. By substituting that into the original integral, we have Z u3 du = Z Now, by returning the evaluation to the initial variable x, we have Z Example 1.11 Evaluate the integral Solution: Z We use Theorem 1.3 for the integral 2 then we have u4 + c. 4 2x(x2 + 1)3 dx = (x2 +1)4 4 + c. √ sec2 x √ dx. x √  √ √ √ sec2 x 0 √ dx. Let f (x) = sec2 x and g(x) = x, then f g(x) = sec2 x. Since g (x) = 1/(2 x), 2 x √ Z √ sec2 x √ dx = 2 tan x + c. x By using the steps of the substitution method, let u = Z 2 √ x, then du = 1 √ 2 x dx. By substitution, we obtain sec2 u du = 2 tan u + c = 2 tan √ x + c. 11 x2 − 1 Z Example 1.12 Evaluate the integral (x3 − 3x + 1)6 dx. Solution: Let u = x3 − 3x + 1, then du = 3(x2 − 1) dx. By substitution, we have 1 3 Z u−6 du = 1 1 −1 +c = + c. 3 3 −5u5 15(x − 3x + 1)5 The upcoming corollary simplifies the process of the substitution method for some functions. Z Corollary 1.1 If f (x) dx = F(x) + c, then for any a 6= 0, Z 1 f (ax ± b) dx = F(ax ± b) + c. a Proof. To verify the previous result, it is sufficient to choose the variable u = ax ± b, then du = a dx. This implies dx = substitution, we have Z Z Z du 1 1 1 f (ax ± b) dx = f (u) = f (u) du = F(u) = F(ax ± b) + c.  a a a a Example 1.13 Evaluate the integral. Z √ du. By Z 2x − 5 dx (1) 1 a (2) cos (3x + 4) dx Solution: From Corollary 1.1, we have Z √ (2x−5)3/2 (1) 2x − 5 dx = 12 3/2 + c = Z (2) (2x−5)3/2 3 + c. 1 3 cos (3x + 4) dx = sin (3x + 4) + c. Notes: The substitution method turns the integral into a simpler integral involving the variable u. The new integral can be evaluated by using either the table of the basic integrals or other techniques of the integration. When using the substitution method, we need to return to the original variable. All examples above expressed in terms of the original variable x. Students should distinguish between integrals that can be evaluated by the substitution method. We must choose u so that du is already R sitting in the integrand, regardless of a constant k. For example, the integral cos x2 dx cannot be evaluated by the substitution method. To see this, let u = x2 , this implies du = 2x dx. However, the term x is not in the integrand. Therefore, the integral cannot be evaluated by the substitution method. The substitution method may be used as a first step in simplifying an integral. It might be followed by other techniques given in Chapter 5. Exercise 1.3 1 - 16 Evaluate the integral. 12 Z 1 x Z 1 + x2 dx ...
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