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Unformatted text preview: Math 138 Calculus II for Honours Mathematics Course Notes Barbara A. Forrest and Brian E. Forrest Version 1.4 c Barbara A. Forrest and Brian E. Forrest. Copyright All rights reserved. January 1, 2020 All rights, including copyright and images in the content of these course notes, are owned by the course authors Barbara Forrest and Brian Forrest. By accessing these course notes, you agree that you may only use the content for your own personal, non-commercial use. You are not permitted to copy, transmit, adapt, or change in any way the content of these course notes for any other purpose whatsoever without the prior written permission of the course authors. Author Contact Information: Barbara Forrest ([email protected]) Brian Forrest ([email protected]) i QUICK REFERENCE PAGE 1 Right Angle Trigonometry sin θ = opposite hypotenuse cos θ = ad jacent hypotenuse tan θ = opposite ad jacent csc θ = 1 sin θ sec θ = 1 cos θ cot θ = 1 tan θ Radians Definition of Sine and Cosine The angle θ in radians equals the length of the directed arc BP, taken positive counter-clockwise and negative clockwise. Thus, π radians = 180◦ or 1 rad = 180 . π For any θ, cos θ and sin θ are defined to be the x− and y− coordinates of the point P on the unit circle such that the radius OP makes an angle of θ radians with the positive x− axis. Thus sin θ = AP, and cos θ = OA. The Unit Circle ii QUICK REFERENCE PAGE 2 Trigonometric Identities cos2 θ + sin2 θ = 1 Pythagorean Identity −1 ≤ cos θ ≤ 1 Range −1 ≤ sin θ ≤ 1 cos(θ ± 2π) = cos θ Periodicity sin(θ ± 2π) = sin θ cos(−θ) = cos θ Symmetry sin(−θ) = − sin θ Sum and Difference Identities cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B Complementary Angle Identities cos( π2 − A) = sin A sin( π2 − A) = cos A Double-Angle cos 2A = cos2 A − sin2 A Identities sin 2A = 2 sin A cos A Half-Angle cos2 θ = 1+cos 2θ 2 Identities sin2 θ = 1−cos 2θ 2 Other 1 + tan2 A = sec2 A iii QUICK REFERENCE PAGE 3 Table of Integrals Differentiation Rules xn+1 xn dx = +C n+1 R 1 dx = ln(| x |) + C R x e x dx = e x + C R sin(x) dx = − cos(x) + C R cos(x) dx = sin(x) + C R sec2 (x) dx = tan(x) + C R 1 dx = arctan(x) + C 1 + x2 R 1 dx = arcsin(x) + C √ 1 − x2 R −1 dx = arccos(x) + C √ 1 − x2 R sec(x) tan(x) dx = sec(x) + C R x ax +C a dx = ln(a) R Function Derivative f (x) = cxa , a , 0, c ∈ R f 0 (x) = caxa−1 f (x) = sin(x) f 0 (x) = cos(x) f (x) = cos(x) f 0 (x) = − sin(x) f (x) = tan(x) f 0 (x) = sec2 (x) f (x) = sec(x) f 0 (x) = sec(x) tan(x) f (x) = arcsin(x) 1 f 0 (x) = √ 1 − x2 1 f 0 (x) = − √ 1 − x2 1 f 0 (x) = 1 + x2 f (x) = arccos(x) f (x) = arctan(x) f (x) = e x f 0 (x) = e x f (x) = a x with a > 0 f 0 (x) = a x ln(a) 1 f 0 (x) = x f (x) = ln(x) for x > 0 Inverse Trigonometric Substitutions Integral R √ a2 − b2 x2 dx R √ a2 + b2 x2 dx R √ b2 x2 − a2 dx Trig Substitution Trig Identity bx = a sin(u) sin2 (x) + cos2 (x) = 1 bx = a tan(u) sec2 (x) − 1 = tan2 (x) bx = a sec(u) sec2 (x) − 1 = tan2 (x) Additional Formulas f (x)g0 (x) dx = f (x)g(x) − Rb A = a |g(t) − f (t)| dt Rb V = a π f (x)2 dx Rb V = a π(g(x)2 − f (x)2 ) dx Rb V = a 2πx(g(x) − f (x)) dx Rb p S = a 1 + ( f 0 (x))2 dx R Integration by Parts Areas Between Curves Volumes of Revolutions: Disk I Volumes of Revolutions: Disk II Volumes of Revolutions: Shell Arc Length Taylor Series (Maclaurin Series) 1 1−x = ex = ∞ P xn = 1 + x + x2 + x3 + · · · n=0 xn n! = n=0 ∞ P ∞ P cos(x) = sin(x) = 1+ x 1! + x2 2! x2n + x3 3! (−1)n (2n)! = 1 − n=0 ∞ P x2n+1 (−1)n (2n+1)! n=0 + ··· x2 2! = x− R=1 + x3 3! x4 4! + − x5 5! x6 6! − + ··· x7 7! + ··· iv R f 0 (x)g(x) dx Differential Equations R=∞ Solve y0 = f (x)g(y) R 1 R ? g(y) = 0, g(y) dy = f (x)dx R=∞ FOLDE y0 = f (x)y + g(x) R=∞ Solve y= Separable R g(x)I(x)dx , I(x) I(x) = e− R f (x) dx QUICK REFERENCE PAGE 4 LIST of THEOREMS: Chapter 1: Integration Integrability Theorem for Continuous Functions Properties of Integrals Theorem Integrals over Subintervals Theorem Average Value Theorem (Mean Value Theorem for Integrals) Fundamental Theorem of Calculus (Part 1) Extended Version of the Fundamental Theorem of Calculus Power Rule for Antiderivatives Fundamental Theorem of Calculus (Part 2) Change of Variables Theorem Chapter 2: Techniques of Integration Integration by Parts Theorem Integration of Partial Fractions p-Test for Type I Improper Integrals Properties of Type I Improper Integrals The Monotone Convergence Theorem for Functions Comparison Test for Type I Improper Integrals Absolute Convergence Theorem for Improper Integrals p-Test for Type II Improper Integrals Chapter 3: Applications of Integation Area Between Curves Volumes of Revolution: Disk Methods Volumes of Revolution: Shell Method Arc Length Chapter 5: Numerical Series Geometric Series Test Divergence Test Arithmetic for Series Theorems The Monotone Convergence Theorem for Sequences Comparison Test for Series Limit Comparison Test Integral Test for Convergence p-Series Test Alternating Series Test (AST) and the Error in the AST Absolute Convergence Theorem Rearrangement Theorem Ratio Test Polynomial versus Factorial Growth Theorem Root Test Chapter 6: Power Series Fundamental Convergence Theorem for Power Series Test for the Radius of Convergence Equivalence of Radius of Convergence Abel’s Theorem: Continuity of Power Series Addition of Power Series Multiplication of Power Series by (x − a)m Power Series of Composite Functions Term-by-Term Differentiation of Power Series Uniqueness of Power Series Representations Term-by-Term Integration of Power Series Taylor’s Theorem Taylor’s Approximation Theorem I Convergence Theorem for Tayor Series Binomial Theorem Generalized Binomial Theorem Chapter 4: Differential Equations Theorem for Solving First-order Linear Differential Equations Existence and Uniqueness Theorem for FOLDE v Table of Contents Page 1 Integration 1.1 Areas Under Curves . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Estimating Areas . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Approximating Areas Under Curves . . . . . . . . . . . . 1.1.3 The Relationship Between Displacement and Velocity . . 1.2 Riemann Sums and the Definite Integral . . . . . . . . . . . . . . 1.3 Properties of the Definite Integral . . . . . . . . . . . . . . . . . . 1.3.1 Additional Properties of the Integral . . . . . . . . . . . . 1.3.2 Geometric Interpretation of the Integral . . . . . . . . . . 1.4 The Average Value of a Function . . . . . . . . . . . . . . . . . . 1.4.1 An Alternate Approach to the Average Value of a Function 1.5 The Fundamental Theorem of Calculus (Part 1) . . . . . . . . . . 1.6 The Fundamental Theorem of Calculus (Part 2) . . . . . . . . . . 1.6.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 Evaluating Definite Integrals . . . . . . . . . . . . . . . . 1.7 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.1 Change of Variables for the Indefinite Integral . . . . . . . 1.7.2 Change of Variables for the Definite Integral . . . . . . . . 1 1 1 2 8 13 18 19 22 27 28 30 40 41 43 47 48 52 2 Techniques of Integration 2.1 Inverse Trigonometric Substitutions . . . . . . . . . . . 2.2 Integration by Parts . . . . . . . . . . . . . . . . . . . 2.3 Partial Fractions . . . . . . . . . . . . . . . . . . . . . 2.4 Introduction to Improper Integrals . . . . . . . . . . . . 2.4.1 Properties of Type I Improper Integrals . . . . 2.4.2 Comparison Test for Type I Improper Integrals 2.4.3 The Gamma Function . . . . . . . . . . . . . . 2.4.4 Type II Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 56 62 71 80 86 88 94 96 3 Applications of Integration 3.1 Areas Between Curves . . . . . . . . 3.2 Volumes of Revolution: Disk Method . 3.3 Volumes of Revolution: Shell Method . 3.4 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 100 108 114 118 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Differential Equations 122 4.1 Introduction to Differential Equations . . . . . . . . . . . . . . . . 122 4.2 Separable Differential Equations . . . . . . . . . . . . . . . . . . 124 4.3 First-Order Linear Differential Equations . . . . . . . . . . . . . . 131 vi 4.4 Initial Value Problems . . . . . . . . . . . . . . . . . . . . . 4.5 Graphical and Numerical Solutions to Differential Equations 4.5.1 Direction Fields . . . . . . . . . . . . . . . . . . . . 4.5.2 Euler’s Method . . . . . . . . . . . . . . . . . . . . . 4.6 Exponential Growth and Decay . . . . . . . . . . . . . . . . 4.7 Newton’s Law of Cooling . . . . . . . . . . . . . . . . . . . 4.8 Logistic Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 140 140 142 145 149 152 5 Numerical Series 5.1 Introduction to Series . . . . . . . . . . . . . . . . . . . 5.2 Geometric Series . . . . . . . . . . . . . . . . . . . . . 5.3 Divergence Test . . . . . . . . . . . . . . . . . . . . . . 5.4 Arithmetic of Series . . . . . . . . . . . . . . . . . . . . 5.5 Positive Series . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Comparison Test . . . . . . . . . . . . . . . . . . 5.5.2 Limit Comparison Test . . . . . . . . . . . . . . 5.6 Integral Test for Convergence of Series . . . . . . . . . 5.6.1 Integral Test and Estimation of Sums and Errors 5.7 Alternating Series . . . . . . . . . . . . . . . . . . . . . 5.8 Absolute versus Conditional Convergence . . . . . . . . 5.9 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 162 167 169 173 177 180 185 189 199 202 213 218 227 6 Power Series 6.1 Introduction to Power Series . . . . . . . . . . . . . . 6.1.1 Finding the Radius of Convergence . . . . . . 6.2 Functions Represented by Power Series . . . . . . . . 6.2.1 Building Power Series Representations . . . . 6.3 Differentiation of Power Series . . . . . . . . . . . . . 6.4 Integration of Power Series . . . . . . . . . . . . . . . 6.5 Review of Taylor Polynomials . . . . . . . . . . . . . . 6.6 Taylor’s Theorem and Errors in Approximations . . . . 6.7 Introduction to Taylor Series . . . . . . . . . . . . . . . 6.8 Convergence of Taylor Series . . . . . . . . . . . . . . 6.9 Binomial Series . . . . . . . . . . . . . . . . . . . . . 6.10 Additional Examples and Applications of Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 229 236 240 241 244 253 257 269 277 281 285 290 vii . . . . . . . . . . . . Chapter 1 Integration Many operations in mathematics have an inverse operation: addition and subtraction; multiplication and division; raising a number to the nth power and finding its nth root; taking a derivative and finding its antiderivative. In each case, one operation “undoes” the other. In this chapter, we begin the study of the integral and integration. Soon you will understand that integration is the inverse operation of differentiation. 1.1 Areas Under Curves The two most important ideas in calculus - differentiation and integration - are both motivated from geometry. The problem of finding the tangent line led to the definition of the derivative. The problem of finding area will lead us to the definition of the definite integral. 1.1.1 Estimating Areas Our objective is to find the area under the curve of some function. What do we mean by the area under a curve? The question about how to calculate areas is actually thousands of years old and it is one with a very rich history. To motivate this topic, let’s first consider what we know about finding the area of some familiar shapes. We can easily determine the area of a rectangle or a right-angled triangle, but how could we explain to someone why the area of a circle with radius r is πr2 ? The problem of calculating the area of a circle was studied by the ancient Greeks. In particular, both Archimedes and Eudoxus of Cnidus used the Method of Exhaustion to calculate areas. This method used various regular inscribed polygons of known area to approximate the area of an enclosed region. Chapter 1: Integration 2 In the case of a circle, as the number of sides of the inscribed polygon increased, the error in using the area of the polygon to approximate the area of the circle decreased. As a result, the Greeks had effectively used the concept of a limit as a key technique in their calculation of the area. 1.1.2 Approximating Areas Under Curves Let’s use the ideas from the Method of Exhaustion and try to find the area underneath a parabola by using rectangles as a basis for the approximation. y f (x) = x2 2 1.5 Suppose that we have the function f (x) = x2 . Consider the region R bounded by the graph of f , by the x-axis, and by the lines x = 0 and x = 1. (1, 1) 1 0.5 R −1 −0.5 0 0.5 x 1 How could we determine the area of this irregular region? y For our first estimate, we can approximate the area of R by constructing a rectangle R1 of length 1 (from x = 0 to x = 1) and height 1 (y = f (1) = 12 = 1). This rectangle (in this case a square) has area length × height = 1 × 1 = 1. f (x) = x2 2 1.5 1 R1 (1, 1) height 0.5 R −1 −0.5 0 x 0.5 1 length y f (x) = x2 2 1.5 The diagram shows that the area of rectangle R1 is larger than the area of region R. Moreover, the error is actually quite large. 1 0.5 R1 (1, 1) error R −1 −0.5 0 0.5 1 x We can find a better estimate if we split the interval [0, 1] into 2 equal subintervals, [0, 12 ] and [ 21 , 1]. Calculus 2 (B. Forrest)2 Section 1.1: Areas Under Curves 3 y Using these intervals, two rectangles are constructed. The first rectangle R1 has its length from x = 0 to x = 21 with height equal to f ( 12 ) = 212 = 41 . f (x) = x2 2 1.5 R2 1 0.5 0.25 The second rectangle R2 has its length from x = 21 to x = 1 with height f (1) = 12 = 1. −1 −0.5 0 R1 (1, 1) ( 12 , 14 ) 0.5 1 x The area of rectangle R1 is equal to R1 = length × height = 1 1 1 × 2 = 2 2 8 while the area of rectangle R2 is equal to R2 = length × height = 1 1 ×1= 2 2 Our second estimate for the area of the original region R is obtained by adding the areas of these two rectangles to get R1 + R2 = Observe from the diagram that our new estimate using two rectangles for the area under f (x) = x2 on the interval [0, 1] is much better than our first estimate since the error is smaller. The region containing the dashed lines indicates the improvement in our estimate (this is the amount by which we have reduced the error from our first estimate). 1 1 5 + = = 0.625 8 2 8 y f (x) = x2 2 1.5 1 R2 error (1, 1) 0.5 R1 0.25 error ( 12 , 41 ) −1 −0.5 0 0.5 1 x To improve our estimate even further, divide the interval [0, 1] into five equal subintervals of the form i−1 i [ , ] 5 5 where i ranges from 1 to 5. This produces the subintervals 1 1 2 2 3 3 4 4 5 [0, ], [ , ], [ , ], [ , ], [ , ] 5 5 5 5 5 5 5 5 5 each having equal lengths of 15 . Calculus 2 (B. Forrest)2 Chapter 1: Integration 4 Next we construct five new rectangles where the ith rectangle forms its length from i−1 to 5i and has height equal to the value of the function at the right-hand endpoint 5 of the interval. That is, the height of a rectangle is f (x) = x2 where x = 5i or i i f ( ) = ( )2 5 5 i2 = 2 5 The area of the ith rectangle is given by length × height = 1 i2 i2 × 2 = 3 5 5 5 y f (x) = x2 error area under curve 2 1.5 R5 1 (1,1) R4 0.5 R3 R2 R1 0 1 5 2 5 3 5 4 5 5 5 =1 x 1 5 Our new estimate is the sum of the areas of these rectangles which is 2 1 3 1 4 1 5 1 1 1 R1 + R2 + R3 + R4 + R5 = [( )2 ( )] + [( )2 ( )] + [( )2 ( )] + [( )2 ( )] + [( )2 ( )] 5 5 5 5 5 5 5 5 5 5 12 22 32 42 52 = 3+ 3+ 3+ 3+ 3 5 5 5 5 5 = 1 2 (1 + 22 + 32 + 42 + 52 ) 53 = 5 1 X 2 i 53 i=1 Note: It can be shown that for any n n X i=1 Calculus 2 i2 = (n)(n + 1)(2n + 1) 6 (B. Forrest)2 Section 1.1: Areas Under Curves 5 This means that the sum of the areas of the rectangles is 5 1 (5)(5 + 1)(2(5) + 1) 1 X 2 i = 3× 3 5 i=1 5 6 = 1 (5)(6)(11) × 53 6 = 11 25 = 0.44 So far the estimates for the area under the curve of f (x) = x2 on the interval [0, 1] are: Number of Subintervals (Rectangles) Length of Subinterval (Width of Rectangle) Estimate for Area under Curve 1 1 1 2 1 2 0.625 5 1 5 0.44 Observe from the diagram that the estimate for the area is getting better while the error in the estimate is getting smaller. y f (x) = x2 2 1.5 Let’s repeat this process again by using 10 equal subintervals. 1 (1,1) Ri 0.5 0 Calculus 2 i−1 10 i 10 x (B. Forrest)2 Chapter 1: Integration 6 In this case, the sum of the areas of the 10 rectangles will be 10 X Ri = R1 + R2 + R3 + R4 + R5 + R6 + R7 + R8 + R9 + R10 i=1 = 10 1 X 2 i 103 i=1 = 1 (10)(10 + 1)(2(10) + 1) 103 6 = 77 200 = 0.385 If we were to use 1000 subintervals, the estimate for the area would be 1000 X Ri = R1 + R2 + R3 + . . . + R1000 i=1 = 1000 1 X 2 i 10003 i=1 = 1 (1000)(1000 + 1)(2(1000) + 1) 10003 6 = 0.3338335 You should begin to notice that as we increase the number of rectangles (number of subintervals), the total area of these rectangles seems to be getting closer and closer to the actual area of the original region R. In particular, if we were to produce an accurate diagram that represents 1000 rectangles, we would see no noticeable difference between the estimated area and the true area. For this reason we would expect that our latest estimate of 0.3338335 is actually very close to the true value of the area of region R. We could continue to divide the interval [0, 1] into even more subintervals. In fact, we can repeat this process with n subintervals for any n ∈ N. In this generic case, the estimated area Rn would be n 1 X 2 Rn = 3 i n i=1 = = 1 (n)(n + 1)(2(n) + 1) n3 6 1 (2n3 n3 + 3n2 + n) 6 2 + n3 + = 6 Calculus 2 1 n2 (B. Forrest)2 Section 1.1: Areas Under Curves 7 Note that if we let the number of subintervals n approach ∞, then 2 + n3 + n→∞ 6 lim Rn = lim n→∞ = 2 6 = 1 3 1 n2 By calculating the area under the graph of f (x) = x2 using an increasing number of rectangles, we have constructed a sequence of estimates where each estimate is larger than the actual area. Though it appears that the limiting value 31 is a plausible guess for the actual value of the area, at this point the best that we can say is that the area should be less than or equal to 13 . Number of Subintervals (Rectangles) Length of Subinterval (Width of Rectangle) Estimate for Area under Curve 1 1 1 2 1 2 0.625 5 1 5 0.44 10 1 10 0.385 1000 1 1000 0.3338335 approaches ∞ approaches 0 approaches 1 3 y f (x) = x2 Alternately, we can use a similar process that would produce an estimate for the area that will be less than the actual value. To do so we again divide the interval [0, 1] into n subintervals of length n1 with the i-th interval [ i−1 , i ]. This interval again n n forms the length of a rectangle Li , but this time we will use the left-hand endpoint of the interval so that the value f ( i−1 ) is the n height of the rectangle. Calculus 2 2 1.5 1 (1,1) Li 0.5 First rectangle has height 0 0 i−1 n i n x (B. Forrest)2 Chapter 1: Integration 8 In this case, notice that since f (0) = 0 the first rectangle is really just a horizontal line with area 0. Then the estimated area Ln for this generic case would be Ln = n 1 X (i − 1)2 n3 i=1 = 1 (n − 1)(n + 1 − 1)(2(n − 1) + 1) n3 6 = 1 (n − 1)(n)(2n − 1) n3 6 = 1 2n3 − 3n2 + n n3 6 2 − n3 + = 6 Finally, observe that 1 n2 2 − 3n + n→∞ 6 lim Ln = lim n→∞ 1 n2 1 = . 3 In summary, we have now shown that if R is the area of the region under the graph of f (x) = x2 , above the x-axis, and between the lines x = 0 and x = 1, then for each n ∈ N, Ln ≤ R ≤ Rn . y It would be reasonable to f (x) = x2 conclude that the area under the 2 graph of f (...
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