Math 138 Calculus II for Honours Mathematics

1 Math 138 Calculus IIfor Honours MathematicsCourse NotesBarbara A. Forrest and Brian E. ForrestVersion Barbara A. Forrest and Brian E. rights 1, 2021 All rights, including copyright and images in the content of these course notes, areowned by the course authors Barbara Forrest and Brian Forrest. By accessing thesecourse 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 inany way the content of these course notes for any other purpose whatsoever withoutthe prior written permission of the course Contact Information:Barbara Forrest Forrest REFERENCE PAGE 1 Right Angle Trigonometrysin =o p positehy potenusecos =ad jacenthy potenusetan =o p positead jacentcsc =1sin sec =1cos cot =1tan RadiansThe angle inradians equals thelength of the directedarcBP, taken positivecounter-clockwise andnegative , radians=180 or 1rad=180.

2 Definition of Sine and CosineFor any , cos and sin aredefined to be thex andy coordinates of the pointPon theunit circle such that the radiusOPmakes an angle of radianswith the positivex axis. Thussin =AP, and cos = Unit CircleiiQUICK REFERENCE PAGE 2 Trigonometric IdentitiesPythagoreancos2 +sin2 =1 IdentityRange 1 cos 1 1 sin 1 Periodicitycos( 2 )=cos sin( 2 )=sin Symmetrycos( )=cos sin( )= sin Sum and Difference Identitiescos(A+B)=cosAcosB sinAsinBcos(A B)=cosAcosB+sinAsinBsin(A+B)=sinAcosB+co sAsinBsin(A B)=sinAcosB cosAsinBComplementary Angle Identitiescos( 2 A)=sinAsin( 2 A)=cosADouble-Anglecos 2A=cos2A sin2 AIdentitiessin 2A=2 sinAcosAHalf-Anglecos2 =1+cos 2 2 Identitiessin2 =1 cos 2 2 Other1+tan2A=sec2 AiiiQUICK REFERENCE PAGE 3 Differentiation RulesFunctionDerivativef(x)=cxa,a,0,c Rf (x)=caxa 1f(x)=sin(x)f (x)=cos(x)f(x)=cos(x)f (x)= sin(x)f(x)=tan(x)f (x)=sec2(x)f(x)=sec(x)f (x)=sec(x) tan(x)f(x)=arcsin(x)f (x)=1 1 x2f(x)=arccos(x)f (x)= 1 1 x2f(x)=arctan(x)f (x)=11+x2f(x)=exf (x)=exf(x)=axwitha>0f (x)=axln(a)f(x)=ln(x) forx>0f (x)=1xTable of Integrals xnd x=xn+1n+1+C 1xd x=ln(|x|)

3 +C exd x=ex+C sin(x)d x= cos(x)+C cos(x)d x=sin(x)+C sec2(x)d x=tan(x)+C 11+x2d x=arctan(x)+C 1 1 x2d x=arcsin(x)+C 1 1 x2d x=arccos(x)+C sec(x) tan(x)d x=sec(x)+C axd x=axln(a)+CInverse Trigonometric SubstitutionsIntegralTrig SubstitutionTrig Identity a2 b2x2d xbx=asin(u)sin2(x)+cos2(x)=1 a2+b2x2d xbx=atan(u)sec2(x) 1=tan2(x) b2x2 a2d xbx=asec(u)sec2(x) 1=tan2(x)Additional FormulasIntegration by Parts f(x)g (x)d x=f(x)g(x) f (x)g(x)d xAreas Between CurvesA= ba|g(t) f(t)|dtVolumes of Revolutions: Disk IV= ba f(x)2d xVolumes of Revolutions: Disk IIV= ba (g(x)2 f(x)2)d xVolumes of Revolutions: ShellV= ba2 x(g(x) f(x))d xArc LengthS= ba 1+(f (x))2d xTaylor Series (Maclaurin Series)11 x= n=0xn=1+x+x2+x3+ R=1ex= n=0xnn!=1+x1!+x22!+x33!+ R= cos(x)= n=0( 1)nx2n(2n)!=1 x22!+x44! x66!+ R= sin(x)= n=0( 1)nx2n+1(2n+1)!=x x33!+x55! x77!+ R= Differential EquationsSeparabley =f(x)g(y)Solveg(y)?=0, 1g(y)dy= f(x)d xFOLDEy =f(x)y+g(x)Solvey= g(x)I(x)d xI(x),I(x)=e f(x)d xivQUICK REFERENCE PAGE 4 LIST of THEOREMS:Chapter 1: IntegrationIntegrability Theoremfor Continuous FunctionsProperties of Integrals TheoremIntegrals over Subintervals TheoremAverage Value Theorem(Mean Value Theorem for Integrals)Fundamental Theorem of Calculus (Part 1)Extended Version of theFundamental Theorem of CalculusPower Rule for AntiderivativesFundamental Theorem of Calculus (Part 2)Change of Variables TheoremChapter 2: Techniques of IntegrationIntegration by Parts TheoremIntegration of Partial Fractionsp-Test for Type I Improper IntegralsProperties of Type I Improper IntegralsThe Monotone Convergence Theoremfor FunctionsComparison Testfor Type I Improper IntegralsAbsolute Convergence Theoremfor Improper Integralsp-Test for Type II Improper IntegralsChapter 3: Applications of IntegationArea Between CurvesVolumes of Revolution.

4 Disk MethodsVolumes of Revolution: Shell MethodArc LengthChapter 4: Differential EquationsTheorem for SolvingFirst-order Linear Differential EquationsExistence and Uniqueness Theoremfor FOLDEC hapter 5: Numerical SeriesGeometric Series TestDivergence TestArithmetic for Series TheoremsThe Monotone Convergence Theoremfor SequencesComparison Test for SeriesLimit Comparison TestIntegral Test for Convergencep-Series TestAlternating Series Test (AST) andthe Error in the ASTA bsolute Convergence TheoremRearrangement TheoremRatio TestPolynomial versus Factorial Growth TheoremRoot TestChapter 6: Power SeriesFundamental Convergence Theoremfor Power SeriesTest for the Radius of ConvergenceEquivalence of Radius of ConvergenceAbel s Theorem: Continuity of Power SeriesAddition of Power SeriesMultiplication of Power Series by (x a)mPower Series of Composite FunctionsTerm-by-Term Differentiation of Power SeriesUniqueness of Power Series RepresentationsTerm-by-Term Integration of Power SeriesTaylor s TheoremTaylor s Approximation Theorem IConvergence Theorem for Tayor SeriesBinomial TheoremGeneralized Binomial TheoremvTable of ContentsPage1 Areas Under Curves.

5 Estimating Areas.. Approximating Areas Under Curves.. The Relationship Between Displacement and Velocity.. Riemann Sums and the Definite Integral.. Properties of the Definite Integral.. Additional Properties of the Integral.. Geometric Interpretation of the Integral.. The Average Value of a Function.. An Alternate Approach to the Average Value of a The Fundamental Theorem of Calculus (Part 1).. The Fundamental Theorem of Calculus (Part 2).. Antiderivatives.. Evaluating Definite Integrals.. Change of Variables.. Change of Variables for the Indefinite Integral.. Change of Variables for the Definite Integral..522 Techniques of Inverse Trigonometric Substitutions.. Integration by Parts.. Partial Fractions.. Introduction to Improper Integrals.. Properties of Type I Improper Integrals.. Comparison Test for Type I Improper Integrals.

6 The Gamma Function.. Type II Improper Integrals..963 Applications of Areas Between Curves.. Volumes of Revolution: Disk Method.. Volumes of Revolution: Shell Method.. Arc Length..1184 Differential Introduction to Differential Equations.. Separable Differential Equations.. First-Order Linear Differential Equations.. Initial Value Problems.. Graphical and Numerical Solutions to Differential Equations.. Direction Fields.. Euler s Method.. Exponential Growth and Decay.. Newton s Law of Cooling.. Logistic Growth..1525 Numerical Introduction to Series.. Geometric Series.. Divergence Test.. Arithmetic of Series.. Positive Series.. Comparison Test.. Limit Comparison Test.. Integral Test for Convergence of Series.. Integral Test and Estimation of Sums and Errors.. Alternating Series.. Absolute versus Conditional Convergence.

7 Ratio Test.. Root Test..2276 Power Introduction to Power Series.. Finding the Radius of Convergence.. Functions Represented by Power Series.. Building Power Series Representations.. Differentiation of Power Series.. Integration of Power Series.. Review of Taylor Polynomials.. Taylor s Theorem and Errors in Approximations.. Introduction to Taylor Series.. Convergence of Taylor Series.. Binomial Series.. Additional Examples and Applications of Taylor Series..290viiChapter 1 IntegrationMany operations in Mathematics have an inverse operation: addition and subtraction;multiplication and division; raising a number to the nth power and finding its nthroot; 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 you will understand thatintegration is the inverse operation of Areas Under CurvesThe two most important ideas in Calculus - differentiation and integration - are bothmotivated from geometry.

8 The problem of finding the tangent line led to the definitionof thederivative. The problem of finding area will lead us to the definition of thedefinite Estimating AreasOur objective is to find the area under the curve of some do we mean by thearea under a curve?The question about how to calculate areas is actually thousands of years old and it isone with a very rich history. To motivate this topic, let s first consider what we knowabout finding the area of some familiar shapes. We can easily determine the area ofa rectangle or a right-angled triangle, but how could we explain to someone why thearea of a circle with radiusris r2?The problem of calculating the area of a circle was studied by the ancient Greeks. Inparticular, both Archimedes and Eudoxus of Cnidus used theMethod of Exhaustionto calculate areas. This method used various regular inscribed polygons of knownarea to approximate the area of an enclosed 1: Integration2In the case of a circle, as the number of sides of the inscribed polygon increased, theerror in using the area of the polygon to approximate the area of the circle a result, the Greeks had effectively used the concept of alimitas a key techniquein their calculation of the Approximating Areas Under CurvesLet s use the ideas from the Method of Exhaustion and try to find the area underneatha parabola by using rectangles as a basis for the that we have thefunctionf(x)=x2.

9 Consider theregionRbounded by the graph off, by thex-axis, and by the linesx=0 andx=1.(1,1)f(x)= 1 could we determine the area of this irregular region?For our first estimate, we canapproximate the area ofRbyconstructing a rectangleR1oflength 1 (fromx=0 tox=1)and height 1 (y=f(1)=12=1).This rectangle (in this case asquare) has arealength height=1 1=1.(1,1)f(x)= 1 diagram shows that the areaof rectangleR1is larger than thearea of regionR. Moreover, theerroris actually quite large.(1,1)f(x)= 1 can find a better estimate if we split the interval [0,1] into 2 equal subintervals,[0,12] and [12,1]. Calculus 2(B. Forrest)2 Section : Areas Under Curves3 Using these intervals, tworectangles are constructed. Thefirst rectangleR1has its lengthfromx=0 tox=12with heightequal tof(12)=122= second rectangleR2has itslength fromx=12tox=1 withheightf(1)=12=1.(1,1)f(x)= 1 (12,14)The area of rectangleR1is equal toR1=length height=12 122=18while the area of rectangleR2is equal toR2=length height=12 1=12 Our second estimate for the area of the original regionRis obtained by adding theareas of these two rectangles to getR1+R2=18+12=58= from the diagram thatour new estimate using tworectangles for the area underf(x)=x2on the interval [0,1] ismuch better than our firstestimate since the error issmaller.

10 The region containingthe dashed lines indicates theimprovement in our estimate (thisis the amount by which we havereduced the error from our firstestimate).(1,1)f(x)= 1 (12,14)errorerrorTo improve our estimate even further, divide the interval [0,1] into five equalsubintervals of the form[i 15,i5]whereiranges from 1 to produces the subintervals[0,15],[15,25],[25,35],[35,4 5],[45,55]each having equal lengths 2(B. Forrest)2 Chapter 1: Integration4 Next we construct five new rectangles where theithrectangle forms its length fromi 15toi5and has height equal to the value of the function at theright-hand endpointof the interval. That is, the height of a rectangle isf(x)=x2wherex=i5orf(i5)=(i5)2=i252 The area of theithrectangle is given bylength height=15 i252=i253(1,1)R1f(x)=x2errorarea under new estimate is the sum of the areas of these rectangles which isR1+R2+R3+R4+R5=[(15)2(15)]+[(25)2(15)] +[(35)2(15)]+[(45)2(15)]+[(55)2(15)]=125 3+2253+3253+4253+5253=153(12+22+32+42+52 )=1535 i=1i2 Note:It can be shown that for anynn i=1i2=(n)(n+1)(2n+1)6 Calculus 2(B.)