CALCULUS II MATH 2414 Calculus I Review Worksheet-Answers 1 1 2 p 558 – Basic Integration Formulas (Thomas' CALCULUS Media Upgrade 11th edition)
REVIEW WORKSHEET FOR TEST #3 1 Find the general term of the following sequence, determine if it converges, and if so to what limit 2
Worksheet for Calculus 2 Tutor, Section 8: Arc Length 1 Calculate the length of the following lines using the arc length calculation formula
This booklet contains our notes for courses Math 152 - Calculus II at Simon Fraser University Students are expected to bring this booklet to each lecture and
Calculus II Practice Problems 1: Answers 1 Solve for x: a) 6x 362¡ x Answer Since 36 62, the equation becomes 6x 62 ¢ 2¡ x£ , so we must have x
Math 251 Worksheet: Calculus II Review Exercises This worksheet serves as a review of only the truly essential material from Math 152 that you
2 Write an equation in polar coordinates for the circle of radius ?2 centered at (x, y) = (1,1)
10 déc 2015 · (In Calculus I and II, sometimes called single variable calculus, we study functions of one variable, so the word argument is singular
For this worksheet (and on homework), we choose functions where the integrals are possible to do by hand or by using an integration table
40130_6Calculus2.pdf
Calculus II
MATH 152 Course Notes
Department. of Mathematics, SFU
Spring 2023
Copyright © 2023 Veselin Jungic and Jamie Mulholland, SFU
SELFPUBLISHED
http://www.math.sfu.caLicensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License (the
"License"). You may not use this document except in compliance with the License. You may obtain a copy of the License athttp://creativecommons.org/licenses/by-nc-sa/4.0/. Unless required by applicable law or agreed to in writing, software distributed under the License is dis- tributed on an"AS IS"BASIS,WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License.
First printing, August 2006
Contents
Preface.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Greek Alphabet.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 IPart One: Introduction to the Integral
1Integrals.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.1 Areas and Distances
14
1.2 The Definite Integral
19
1.3 The Fundamental Theorem of Calculus
28
1.4 The Net Change Theorem
34
1.5 The Substitution Rule
39
2Applications of Integration.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
2.1 Areas Between Curves
48
2.2 Areas in Polar Coordinates
52
2.3 Volumes55
2.4 Volumes by Cylindrical Shells
62 IIPart Two: Integration Techniques and Applications
3Techniques of Integration and Applications.. . . . . . . . . . . . . . . . . . . 69
3.1 Integration By Parts
70
3.2 Trigonometric Integrals
76
3.3 Trigonometric Substitutions
80
3.4 Integration of Rational Functions by Partial Fractions82
3.5 Strategy for Integration
89
3.6 Approximate Integration
95
3.7 Improper Integrals
103
4Further Applications of Integration.. . . . . . . . . . . . . . . . . . . . . . . . . . . 109
4.1 Arc Length
110
4.2 Area of a Surface of Revolution
114
4.3 Calculus with Parametric Curves
118 IIIPart Three: Sequences and Series
5Infinite Sequences and Series.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
5.1 Sequences
126
5.2 Series133
5.3 The Integral Test and Estimates of Sums
139
5.4 The Comparison Test
143
5.5 Alternating Series
147
5.6 Absolute Convergence and the Ratio and Root Test
151
5.7 Strategy for Testing Series
156
5.8 Power Series
160
5.9 Representation of Functions as Power Series
164
5.10 Taylor and Maclaurin Series
167
5.11 Applications of Taylor Polynomials
175 IVPart Four: Differential Equations
6A First Look at Differential Equations.. . . . . . . . . . . . . . . . . . . . . . . . . . 181
6.1 Modeling with Differential Equations, Directions Fields
182
6.2 Separable Equations
187
6.3 Models for Population Growth
194 VExam Preparation
7Review Materials for Exam Preparation.. . . . . . . . . . . . . . . . . . . . . . . 199
7.1 Midterm 1 Review Package
200
7.2 Midterm 2 Review Package
207
7.3 Final Exam Practice Questions
216 VIAppendix
Bibliography.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
Articles223
Books223
Web Sites223
Index.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
PrefaceThis booklet contains the note templates for coursesMath 150/151 - Calculus Iat Simon Fraser University. Students are expected to use this booklet during each lecture by following along with the instructor, filling in the details in the blanks provided. Definitions and theorems appear in highlighted boxes.
Next to some examples you"ll see [
link to applet ]. The link will take you to an online interactive
applet to accompany the example - just like the ones used by your instructor in the lecture. The link
above will take you to the following url [Mul22] containing all the applets: http://www.sfu.ca/~jtmulhol/calculus-applets/html/appletsforcalculus.html
Try it now.
Next to some section headings you"ll notice a QR code. They look like the image on the right. Each one provides a link to a webpage (could be a youtube video, or access to online Sage code). For example this one takes you to the Wikipedia page which explains what a QR code is. Use a QR code scanner on your phone or tablet and it will quickly take you off to the webpage. The app "Red Laser" is a good QR code scanner which is
available for free (iphone, android, windows phone).We offer a special thank you to Keshav Mukunda for his many contributions to these notes.
No project such as this can be free from errors and incompleteness. We will be grateful to everyone who points out any typos, incorrect statements, or sends any other suggestion on how to improve this manuscript.
Veselin Jungic Jamie Mulholland
vjungic@sfu.ca j_mulholland@sfu.ca
January 13, 2023
Greek Alphabetlower
case capital name pronunciationlower case capital name pronunciationαAalpha (al-fah)νNnu (new)
βBbeta (bay-tah)ξΞxi (zie)
γΓgamma (gam-ah)o Oomicron (om-e-cron)
δ∆delta (del-ta)πΠpi (pie)
εEepsilon (ep-si-lon)ρPrho (roe)
ζZzeta (zay-tah)σΣsigma (sig-mah)
ηHeta (ay-tah)τTtau (taw)
θΘtheta (thay-tah)υϒupsilon (up-si-lon)
ιIiota (eye-o-tah)φΦphi (fie)
κKkappa (cap-pah)χXchi (kie)
λΛlambda (lamb-dah)ψΨpsi (si)
µMmu (mew)ωΩomega (oh-may-gah)
I
1Integrals.. . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.1 Areas and Distances
1.2 The Definite Integral
1.3 The Fundamental Theorem of Calculus
1.4 The Net Change Theorem
1.5 The Substitution Rule
2Applications of Integration.. . . . . . . . . . 47
2.1 Areas Between Curves
2.2 Areas in Polar Coordinates
2.3 Volumes
2.4 Volumes by Cylindrical ShellsPart One: Introduction to the
Integral
1. IntegralsIn this chapter we lay down the foundations for this course. We introduce the two motivating
problems for integral calculus: the area problem, and the distance problem. We then define the integral and discover the connection between integration and differentiation.
14Chapter 1. Integrals1.1Ar easand Distances
(This lecture corresponds to Section 5.1 of Stewart"sCalculus.) One can never know for sure what a desertedarealooks like. (George Carlin, American stand-up Comedian, Actor and Author, 1937-2008)
BIG Question.What is the meaning of the wordarea?
Vocabulary.Cambridge dictionary:
areanoun (a) a particular part of a place, piece of land or country; (b) the size of a flat surf acecalculated by multiplying its length by its width; (c) a subject or acti vity,or a part of it. (d) (W ikipedia)- Area is a ph ysicalquantity e xpressingthe size
of a part of a surface.Example 1.1Find the area of the region in the coordinate plane bounded by the coordinate
axes and linesx=2 andy=3. Example 1.2Find the area of the region in the coordinate plane bounded by thex-axis and linesy=2xandx=3. Example 1.3Find the area of the region in the coordinate plane bounded by thex-axis and linesy=x2andx=3.
1.1 Areas and Distances 15Example 1.4Estimatethe area of the region in the coordinate plane bounded by thex-axis and
curvesy=x2andx=3.
16Chapter 1. IntegralsExample 1.5(Over- and under-estimates.)In the previous example, show that
lim n→∞Rn=9 and limn→∞Ln=9.
A more general formulation.
Ingredients: A functionfthat is continuous on a closed interval[a,b].
Letn∈N, and define∆x=b-an
. Let x 0=a x
1=a+∆x
x
2=a+2∆x
x
3=a+3∆x
... x n=a+n∆x=b.
Define
R n=f(x1)·∆x+f(x2)·∆x+...+f(xn)·∆x.
("R" stands for "right-hand", since we are using the right hand endpoints of the little rectangles.)Definition 1.1.1 - Area.TheareaAof the regionSthat lies under the graph of the continuous
functionfover and interval[a,b]is the limit of the sum of the areas of approximating rectangles
Rn. That is,
A=limn→∞Rn=limn→∞[f(x1)+f(x2)+...+f(xn)]∆x. The more compactsigma notationcan be used to write this as
A=limn→∞Rn=limn→∞
n∑ i=1f(xi)! ∆x.
1.1 Areas and Distances 17
Example 1.6Find the area under the graph off(x) =100-3x2fromx=1 tox=5. From the definition of area, we haveA=limn→∞ n∑ i=1f(xi)! ∆x.Distance Problem.Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.Reminderdistance = velocity·time
18Chapter 1. IntegralsAdditional Notes:
1.2 The Definite Integral 19
1.2
The Definite Integral
(This lecture corresponds to Section 5.2 of Stewart"sCalculus.)After years of finding mathematics easy, I finally reached integral calculus and came up against
a barrier. I realized that this was as far as I could go, and to this day I have never successfully gone beyond it in any but the most superficial way. (Isaac Asimov, Russian-born American author and biochemist, best known for his works of science fiction, 1920-1992) Definition 1.2.1 - The Definite Integral.Supposefis a continuous function defined on the closed interval[a,b], we divide[a,b]intonsubintervals of equal width∆x= (b-a)/n. Let x
0=a,x1,x2, ...,xn=b
be the end points of these subintervals. Let x ∗1,x∗2,...,x∗n be anysample pointsin these subintervals, sox∗ilies in theith subinterval[xi-1,xi]. Then thedefinite integral of f from a to bis written asZ b af(x)dx, and is defined as follows: Zb af(x)dx=limn→∞n∑ i=1f(x∗i)∆x
20Chapter 1. IntegralsThe definite integral: some terminology
Z b af(x)dx=limn→∞n∑ i=1f(x∗i)∆x Z is theintegral sign
f(x)is theintegrand
aandbare thelimits of integration:
•a-lower limit •b-upper limit The procedure of calculating an inte gralis called integration.
n∑
i=1f(x∗i)∆xis called aRiemann sum (named after the German mathematician Bernhard Riemann,1826-1866)
Four Facts.
(a)
If f(x)>0 on[a,b]thenZ
b af(x)dx>0.
Iff(x)<0 on[a,b]thenZ
b af(x)dx<0. (b)
F ora general function f,
Z b af(x)dx=(signed area of the region) = (area abovex-axis) - (area belowx-axis) (c) F ore veryε>0 there exists a numberN∈Nsuch that Z b af(x)dx-n∑ i=1f(x∗i)∆x <ε for everyn>Nand every choice ofx∗1,x∗2,...,x∗n. (d)Letfbe continuous on[a,b]and leta=x0
1.2 The Definite Integral 21 Some facts you just have to know.
1 (a) n∑ i=1i=n(n+1)2 (b) n∑ i=1i2=n(n+1)(2n+1)6 (c) n∑ i=1i3=n(n+1)2 2 (d) n∑ i=1c=cn (e) n∑ i=1(cai) =cn∑ i=1a i (f) n∑ i=1(ai±bi) =n∑ i=1a i±n∑ i=1b i1For visual proofs of (a) and (b) see [Gol02]: Goldoni, G. (2002)A visual proof for the sum of the first n squares
and for the sum of the first n factorials of order two. The Mathematical Intelligencer 24 (4): 67-69. You can access the
Mathematical Intelligencer through the SFU Library web site:http://cufts2.lib.sfu.ca/CJDB/BVAS/journal/
150620.
22Chapter 1. IntegralsExample 1.7EvaluateZ2
0(x2-x)dx.Example 1.8Express the limit
lim n→∞n∑ i=1(1+xi)cosxi∆x as a definite integral on the interval[π,2π].Example 1.9ProveZ2 0p4-x2dx=π.
1.2 The Definite Integral 23Theorem 1.2.1- Choosing a good sample point . ...Midpoint Rule.To approximate an
integral it is usually better to choosex∗ito be the midpointx iof the interval[xi-1,xi]: Z b af(x)dx≈n∑ i=1f(x i)∆x=∆x[f(x 1)+f(x
2)+...+f(x
n)] Recall the midpoint of an interval[xi-1,xi]is given byx i=12 (xi-1+xi).Example 1.10Use the Midpoint Rule withn=4 to approximate the integralZ 5 1dxx 2. 24Chapter 1. IntegralsTheorem 1.2.2- T woSpecial Pr opertiesof the Integral.
(a) If a>bthenZb
af(x)dx=-Z a bf(x)dx. (b) If a=bthenZb
af(x)dx=0. Some More Properties of the Integral.
(a) If cis a constant, thenZ
b acdx=c(b-a) (b) Z b a[f(x)±g(x)]dx=Z b af(x)dx±Z b ag(x)dx (c) If cis a constant, thenZ
b acf(x)dx=cZ b af(x)dx (d) Z c af(x)dx+Z b cf(x)dx=Z b af(x)dxExample 1.11EvaluateZ 3 0 2x-3p9-x2
dx. 1.2 The Definite Integral 25
Example 1.12EvaluateZ
3 0f(x)dxiff(x) =1-xifx∈[0,1]
-p1-(x-2)2ifx∈(1,3] More Properties of the definite integral.
(a) If f(x)≥0 fora≤x≤b, thenZ
b af(x)dx≥0.(b)If f(x)≥g(x)fora≤x≤b, thenZ b af(x)dx≥Z b ag(x)dx.(c)If mandMare constants, andm≤f(x)≤Mfora≤x≤b, then m(b-a)≤Z b af(x)dx≤M(b-a) 26Chapter 1. IntegralsExample 1.13Prove
1e 4≤Z
2 1e-x2dx≤1e
Example 1.14(a)If fis continuous on[a,b], show that Z b af(x)dx≤Z b a|f(x)|dx. (b) Sho wthat if fis continuous on[0,2π]then
Z 2π 0f(x)sin2xdx≤Z
2π 0|f(x)|dx.
1.2 The Definite Integral 27
Additional Notes:
28Chapter 1. Integrals1.3The Fundamental Theor emof Calculus
(This lecture corresponds to Section 5.3 of Stewart"sCalculus.)All of my fundamental principles that were instilled in me in my home, from my childhood, are
still with me. (Hakeem Abdul Olajuwon, a former NBA player,1963-) Motivating ProblemDoes every continuous functionfhave an antiderivative? That is, does there exist a functionFsuch that F ′(x) =f(x)?Motivating ProblemWhat is the antiderivative off(x) =sinxx ? Theorem 1.3.1
- The Fundamental Theor emof Calculus ,P art1. Iffis a continuous on [a,b], then the functiongdefined by g(x) =Z x af(t)dt,a≤x≤b is continuous on[a,b]and differentiable on(a,b), and g ′(x) =f(x). 1.3 The Fundamental Theorem of Calculus 29Example 1.15Apply the Fundamental Theorem of Calculus, Part 1, to find the derivative of
the following functions: (a)g(x) =Z x 1sintt
dt (b)g(x) =Z x2 0 sint dt (c)g(x) =Z h(x) 0f(t)dt
(d)g(x) =Z ex -3xln(1+t2)dt 30Chapter 1. IntegralsTheorem 1.3.2- The Fundamental Theor emof Calculus ,P art2. Iffis continuous on[a,b],
thenZb af(x)dx=F(b)-F(a) whereFis any antiderivative off. That is, a function such thatF′=f. 1.3 The Fundamental Theorem of Calculus 31
Example 1.16Evaluate the following integrals:
(a)Z 1 0xdx (b) Z 3 2exdx (c) Z π 0sinx dx
(d) Z 1 0dx1+x2
32Chapter 1. IntegralsExample 1.17A Piecewise Example.Let
f(x) = 0 ifx<0 xif 0≤x≤1 2-xif 1 0 ifx>2
and letg(x) =Z x 0f(t)dt.
(a) Find an e xpressionfor g(x)similar to the one forf(x). (b) Sk etchthe graphs of fandg.
(c) Where is fdifferentiable? Where isgdifferentiable? 1.3 The Fundamental Theorem of Calculus 33
Additional Notes:
34Chapter 1. Integrals1.4The Net Change Theor em
(This lecture corresponds to Section 5.4 of Stewart"sCalculus.)At the end of some indefinite distance there was always a confused spot, into which her dream
died. (Gustave Flauber, French novelist, 1821-1880)ReminderThe Fundamental Theorem of Calculus, Part 2: Iffis continuous on[a,b], then
Z b af(x)dx=F(b)-F(a) whereFis any antiderivative off, that is a function such thatF′=f. Motivating ProblemSo, to be able to evaluate an integral, we need a way to find any antiderivative F of the given function f . How do we find antiderivatives? A new name for an old idea...
Definition 1.4.1The symbolRf(x)dxis called anindefinite integral, and it represents an antiderivative off. That is, Z f(x)dx=F(x)meansF′(x) =f(x) Warning!It could be confusing: The notationRf(x)dxis used to represent the setof all antiderivatives off Z f(x)dx={F:F′=f} a single functionthat is an antiderivative off. 1.4 The Net Change Theorem 35
Integrals you should know:
Z cf(x)dx=cZ f(x)dxZ [f(x)+g(x)]dx=Z f(x)dx+Z g(x)dxZ kdx=kx+C Z x ndx=xn+1n+1+C(n̸=-1)Zdxx =ln|x|+CZ e xdx=ex+CZ a xdx=axlna+CZ sinxdx=-cosx+CZ cosxdx=sinx+CZ sec 2xdx=tanx+CZ
csc 2xdx=-cotx+CZ
secxtanxdx=secx+CZ cscxcotxdx=-cscx+CZdxx 2+1=tan-1x+CZdx√1-x2=sin-1x+C
36Chapter 1. IntegralsExample 1.18Find the following indefinite integrals:
(a)Z x -2/3dx (b)Z t 2(3-4t5)dt
(c)Z (u-1)(u2+3)du (d)Z (4ev-sec2v)dv (e)Zcosz1-cos2zdz 1.4 The Net Change Theorem 37Theorem 1.4.1- The Net Change Theor em.The integral of a rate of change is the net change:
Zb aF′(x)dx=F(b)-F(a) Example 1.19Iff(x)is the slope of a hiking trail at a distance ofxmiles from the start of the trail, what doesZ 4 2f(x)dxrepresent?
Example 1.20 - Linear Motion of a Particle.A particle is moving along a line with the acceleration (in m/s2)a(t) =2t+3and the initial velocityv(0) =-4m/s with0≤t≤3. Find (a) the v elocityat time t, (b) the distance tra veledduring the gi ventime interv al. 38Chapter 1. IntegralsAdditional Notes:
1.5 The Substitution Rule 39
1.5 The Substitution Rule
(This lecture corresponds to Section 5.5 of Stewart"sCalculus.) Persuasion is often more effectual than force.
(Aesop, Greek fabulist, 6th century BC)Motivating ProblemFindZ -2xe-x2dxHint.What if we think of the "dx" above as a differential? Ifu=e-x2, what is the differential
du? Theorem 1.5.1
- The Substitution Rule. Ifu=g(x)is a differentiable function whose range is an intervalIandfis continuous onI, then Z f(g(x))g′(x)dx=Z f(u)du. Notes:
(a) This rule can be proved using the Chain Rule for differentiation. In this sense, it is a reversal of the Chain Rule. (b) The substitution rule says that we can work with "dx" and "du" that appear after the Z symbols as if they were differentials.Example 1.21Find the following indefinite integrals: (a)Z x 2(x3+5)9dx
40Chapter 1. Integrals(b)
Zdt√3-5t
(c) Z sin3t dt (d) Zduu(lnu)2
(e) Zsin(π/v)v
2dv 1.5 The Substitution Rule 41
(f) Zz2√1-zdzComputers are ideal for computing integrals, and Wolfram|Alpha (www.wolframalpha.com) gives
you easy access to this computing power. Use it as a tool to help you study. But be warned:you still have to understand how to do these computations yourself, since Wolfram|Alpha won"t be with you for quizzes and exams. 42Chapter 1. IntegralsTheorem 1.5.2- Substitution Rule f orDefinite Integrals .Ifg′is continuous on[a,b]and iff
is continuous on the range ofu=g(x), then Z b af(g(x))g′(x)dx=Z g(b) g(a)f(u)du. Notes:
(a) When we make the substitutionu=g(x), then the interval[a,b]on thex-axis becomes the interval[g(a),g(b)]on theu-axis. (b) Writing Zb
af(g(x))g′(x)dx=Z b af(u)du=Z g(b) g(a)f(u)du wouldNOTbe right. Make the substitutionANDchange the limits of integration at the same time! Example 1.22Evaluate the following definite integrals: (a)Z 2π πcos3t dt
(b) Z e2 e(lnu)2duu 1.5 The Substitution Rule 43
Again, use Wolfram|Alpha to check your answer.Even or Odd?Leta>0 and letfbe continuous on[-a,a]. If fisoddthenZa
-af(x)dx=0 If fiseventhenZa
-af(x)dx=2Z a 0f(x)dx
44Chapter 1. IntegralsExample 1.23Evaluate the following definite integrals:
(a)Z 3 -3(2x4+3x2+4)dx (b) Z e -ee -u2sinu duu 2+10 1.5 The Substitution Rule 45
Additional Notes:
2. Applications of IntegrationIn the last chapter we learned how to compute the area between a curve and thex-axis, this gave
rise to theintegral. In this chapter the topics we will cover are:
the area between tw ocurv es, the area bounded by a curv egi venin polar coordinates, the disk and w ashermethods for computing the v olumeof a solid. 48Chapter 2. Applications of Integration2.1Ar easBetw eenCur ves
(This lecture corresponds to Section 6.1 of Stewart"sCalculus.)Mathematics knows no races or geographic boundaries; for mathematics, the cultural world is
one country. (David Hilbert, German mathematician, 1862-1943)Motivating ProblemFind the area bounded by parabolas y=2-x2andy=x2. Theorem 2.1.1
- Ar eaBetw eenCur ves.Supposefandgare continuous andf(x)≥g(x) for allx∈[a,b]. The areaAbounded by the curvesy=f(x),y=g(x), and the linesx=aand x=b, is given by A=Z b af(x)-g(x)dx. 2.1 Areas Between Curves 49
Example 2.1Find the area bounded by parabolas
y=2-x2andy=x2.Example 2.2Find the area of the region bounded by the liney=xand the parabolay=6-x2. Example 2.3Find the area of the region bounded by the liney=x/2and the parabolay2=8-x. 50Chapter 2. Applications of IntegrationTheorem 2.1.2- Doing this ar eacalcula tionalong the y-axis ....Suppose the areaAis
bounded by the curvesx=f(y),x=g(y), and the linesy=c,y=d, wherefandgare continuous andf(y)≥g(y)for ally∈[c,d]. Then the area is given by A=Z d c[f(y)-g(y)]dy. Example 2.4Findtheareaoftheregionboundedbytheliney=x/2andtheparabolay2=8-x. 2.1 Areas Between Curves 51
Additional Notes:
52Chapter 2. Applications of Integration2.2Ar easin P olarCoor dinates
(This lecture corresponds to Section 10.4 of Stewart"sCalculus.) These [equations] are your friends. Use them, know them, love them. (Donna Pierce, American astrophysicist, 1975-)Motivating ProblemSketch the curve and find the area that it encloses:
r=1+cosθ.Theorem 2.2.1- Ar eabounded b ypolar cur ves.The area of a polar regionRbounded by the curver=f(θ), forθ∈[a,b], is given by A=Z b a12 [f(θ)]2dθ=Z b a12 r2dθ. AREAS AND LENGTHS IN POLAR COORDINATES
In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle where,as in Figure 1,is the radius and is the radian measure of the central angle. Formula 1 follows from the fact that the area of a sector is proportional to its central angle: . (See also Exercise 35 in Section 7.3.) Let be the region,illustrated in Figure 2,bounded by the polar curv e and by the rays and ,where is a positiv e continuous function and where . We divide the interval into subintervals with endpoints ,, ,...,and equal width . The rays then di vide into smaller regions with central angle . If we choose in the subinterval ,then the area of the th region is approximated by the area of the sector of a circle with central angle and radius . (See Figure 3.) Thus from Formula 1 we have
and so an approximation to the total area of is It appears from Figure 3 that the approximation in (2) improves as . But the sums in (2) are Riemann sums for the function ,so It therefore appears plausible (and can in fact be proved) that the formula for the area of the polar region is Formula 3 is often written as
with the understanding that . Note the similarity between Formulas 1 and 4. When we apply Formula 3 or 4,it is helpful to think of the area as being swept out by a rotating ray through that starts with angle and ends with angle . EXAMPLE 1Find the area enclosed by one loop of the four-leaved rose . SOLUTIONThe curve was sketched in Example 8 in Section 10.3. Notice from Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from r?cos 2? r?cos 2?V baO r?f??? A? y b a 1 2 r 2 d?4 A? y b a 1 2 ?f???? 2 d?3 ? A lim nl" ? n i?1 1 2 ?f?? i * ?? 2 #?? y b a 1 2 ?f???? 2 d? t???? 1 2 ?f???? 2 nl" A? ? n i?1 1 2 ?f?? i * ?? 2 #?2 ?A #A i ? 1 2 ?f?? i * ?? 2 #? f?? i * ?#? i#A i ?? i$1 , ? i ?ith? i * #??? i $? i$1 n???? i #?? n ? 2 ? 1 ? 0 ?a, b?0%b$a&2' f??b??a r?f???? A????2'?'r
2 ? 1 2 r 2 ? ?r A? 1 2 r 2 ?1 10.4 650||||CHAPTER 10PARAMETRIC EQUATIONS AND POLAR COORDINATES
¨ r FIGURE 1
FIGURE 2
O ¨=b
b ¨=a
r=f(¨) a ? O ¨=b
¨=a
¨=¨
i-1 ¨=¨
i Ψ
f(¨ i * ) FIGURE 3
AREAS AND LENGTHS IN POLAR COORDINATES
In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the area of a sector of a circle where,as in Figure 1,is the radius and is the radian measure of the central angle. Formula 1 follows from the fact that the area of a sector is proportional to its central angle: . (See also Exercise 35 in Section 7.3.) Let be the region,illustrated in Figure 2,bounded by the polar curv e and by the rays and ,where is a positiv e continuous function and where . We divide the interval into subintervals with endpoints ,, ,...,and equal width . The rays then di vide into smaller regions with central angle . If we choose in the subinterval ,then the area of the th region is approximated by the area of the sector of a circle with central angle and radius . (See Figure 3.) Thus from Formula 1 we have
and so an approximation to the total area of is It appears from Figure 3 that the approximation in (2) improves as . But the sums in (2) are Riemann sums for the function ,so It therefore appears plausible (and can in fact be proved) that the formula for the area of the polar region is Formula 3 is often written as
with the understanding that . Note the similarity between Formulas 1 and 4. When we apply Formula 3 or 4,it is helpful to think of the area as being swept out by a rotating ray through that starts with angle and ends with angle . EXAMPLE 1Find the area enclosed by one loop of the four-leaved rose . SOLUTIONThe curve was sketched in Example 8 in Section 10.3. Notice from Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from r?cos 2? r?cos 2?V baO r?f??? A? y b a 1 2 r 2 d?4 A? y b a 1 2 ?f???? 2 d?3 ? A lim nl" ? n i?1 1 2 ?f?? i * ?? 2 #?? y b a 1 2 ?f???? 2 d? t???? 1 2 ?f???? 2 nl" A? ? n i?1 1 2 ?f?? i * ?? 2 #?2 ?A #A i ? 1 2 ?f?? i * ?? 2 #? f?? i * ?#? i#A i ?? i$1 , ? i ?ith? i * #??? i $? i$1 n???? i #?? n ? 2 ? 1 ? 0 ?a, b?0%b$a&2' f??b??a r?f???? A????2'?'r
2 ? 1 2 r 2 ? ?r A? 1 2 r 2 ?1 10.4 650||||CHAPTER 10PARAMETRIC EQUATIONS AND POLAR COORDINATES
¨ r FIGURE 1
FIGURE 2
O ¨=b
b ¨=a
r=f(¨) a ? O ¨=b
¨=a
¨=¨
i-1 ¨=¨
i Ψ
f(¨ i * ) FIGURE 3
2.2 Areas in Polar Coordinates 53
Example 2.5Find the area enclosed byr=1+cosθ.Example 2.6Find the area of the region enclosed by the 3-leaved roser=4sin(3θ).
54Chapter 2. Applications of IntegrationAdditional Notes:
2.3 Volumes 55
2.3 V olumes
(This lecture corresponds to Section 6.2 of Stewart"sCalculus.)I shall now recall to mind that the motion of the heavenly bodies is circular, since the motion
appropriate to a sphere is rotation in a circle. (Nicolaus Copernicus, mathematician, astronomer, jurist, physician, classical scholar, governor, administrator, diplomat, economist, and soldier, 1473-1543) Recall some classic volume formulas:
Motivating ProblemHow do we prove these formulas? Moreover, how do we define the volume of a solid object?Definition 2.3.1 - Definition of Volume...simple beginnings. (i) The volume of a general cylinder with cross sectional areaAand heighthis defined to be Ah. (ii) The v olumeof a general solid is defined using inte grals(calculus). Surprisingly, it turns out (byCavalieri"s principle) that these cross-sectional "area slices" can be rearranged and still give the same total volume. 56Chapter 2. Applications of IntegrationDefinition of Volume...the technique.Computing the volume of a general solidS.
2.3 Volumes 57Definition 2.3.2 - Definition of Volume.LetSbe a solid that lies betweenx=aandx=b. If
the cross-sectional area ofSin the planePx, throughxand perpendicular to thex-axis, isA(x), whereAis a continuous function, then thevolumeofSis V=Z b aA(x)dx. Example 2.7Find the volume of a pyramid whose base is a square with sideband whose height ish. Definition 2.3.3 - Solid of Revolution.A solid of revolution is a solid (volume) obtained by revolving a region (or area) in the plane about a line. In this case the cross-sections are disks or annuli (a.k.a disks or washers), so the volume formula V=Rb aA(x)dxis known as thewasher method. Example 2.8Some regions in the plane are shown below. Draw the resulting solid if these regions are rotated about the x-axis? 58Chapter 2. Applications of IntegrationExample 2.9Find the volume of the solid obtained by rotating the region bounded by the
curves y=sinx,x=π2 ,andy=0 about thex-axis. Example 2.10Find the volume of the solid obtained by rotating the region bounded by the curves y=√x,y=1,andx=0 about they-axis. 2.3 Volumes 59Example 2.11Find the volume of the solid obtained by rotating the regionR, which is enclosed
by the curvesy=xandy=x3in the first quadrant, about the line (a)y=3 (b)x=2 60Chapter 2. Applications of IntegrationExample 2.12
(a) Set up an inte gralfor the v olumeof a toruswith inner radiusrand outer radiusR. (b) By interpreting the inte gralas an area, find the v olumeof the torus. 2.3 Volumes 61
Additional Notes:
62Chapter 2. Applications of Integration2.4V olumesb yCylindr icalShells
(This lecture corresponds to Section 6.3 of Stewart"sCalculus.) Your pain is the breaking of the shell that encloses your understanding. (Kahlil Gibran, Lebanese born American philosophical essayist, novelist and poet, 1883-1931)Motivating ProblemConsider the region in thexy-plane bounded by the curvesy=3x2-x3
andy=0. Imagine this region rotated about they-axis. How do we find the volume of the resulting solid?Exercise your imagination! Let0≤a2.4 Volumes by Cylindrical Shells 63
Almost there...Note that each point of the solidSbelongs to only one cylindrical shellCx, for somex∈[a,b]. So
we can imagine thatSis obtained by gluing all cylindrical shells together. Each cylindrical shell contributes its surface (or "skin"!) to the volume ofS, or, in other words, the volume is the "sum" of all surfaces. Eachx∈[a,b]gives one shellCxwith a surface areaAx, and so the "sum" of all of them is given byV=Z b aAxdx=2πZ b axf(x)dx. This is known as thecylindrical shells method(or simply theshell method) for computing the volume of a solid of revolution. Example 2.13Find the volume of the solid obtained by rotating about they-axis the region bounded by curves y=3x2-x3andy=0. 64Chapter 2. Applications of IntegrationExample 2.14Find the volume of the solid that remains after you bore a circular hole of radius
athrough the center of a solid sphere of radiusb>a. 2.4 Volumes by Cylindrical Shells 65Example 2.15Consider the region in the first quadrant bounded by the curvesy2=xand
y=x3. Use the method of cylindrical shells to compute the volume of the solid obtained by revolving this region around (a)y-axis.(b)x-axis.(c)line x=1.Summary:A general guideline for which method to use is the following:
If the area section (strip) is parallelto the axis of rotation, use theshell method. If the area section (strip) is perpendicularto the axis of rotation, use thewasher method. 66Chapter 2. Applications of IntegrationAdditional Notes:
II 3Techniques of Integration and Applications
69
3.1 Integration By Parts
3.2 Trigonometric Integrals
3.3 Trigonometric Substitutions
3.4 Integration of Rational Functions by Partial Fractions 3.5 Strategy for Integration
3.6 Approximate Integration
3.7 Improper Integrals
4Further Applications of Integration.. . 109
4.1 Arc Length
4.2 Area of a Surface of Revolution
4.3 Calculus with Parametric CurvesPart Two: Integration
Techniques and Applications
3. Techniques of Integration and Applications
In this chapter we develop techniques for computing integrals (antiderivatives). Topics we will cover are:
inte grationby parts (i.e. undoing the product rule from dif ferentiation), trigonometric inte grals, substitutions with trigonometric functions, inte grationof rational functions by partial fractions, approximation of inte grals, improper inte grals. 70Chapter 3. Techniques of Integration and Applications3.1Integra tionBy P arts
(This lecture corresponds to Section 7.1 of Stewart"sCalculus.) Warning: this material is for a mature calculus audience.Disclaimeron the web pageThe absolutely outrageous CALCULUS IS COOL webpageby
Jochen Denzler, http://www.math.utk.edu/∼denzler/CalculusND/index.html (Jochen Denzler, German-born mathematician, 1963-)Motivating ProblemIntegrateZ xe xdx.Theorem 3.1.1- Integra tionBy P arts.Letfandgbe differentiable functions. Then Z f(x)g′(x)dx=f(x)g(x)-Z g(x)f′(x)dx. Here is an easier way to remember this: foru=f(x)andv=g(x) Z udv=uv-Z vdu.Example 3.1IntegrateZ xe xdx. 3.1 Integration By Parts 71
Example 3.2Integrate
(a)Z lnx dx (b)Z arcsinx dx (c)Z x 2e-xdx
(d)Z e 2xcos3x dx
72Chapter 3. Techniques of Integration and ApplicationsCheck answers with Wolfram|Alpha:Example 3.3(a)Pro vethe reduction formula
Z cos nx dx=1n cosn-1xsinx+n-1n Z cos n-2x dx 3.1 Integration By Parts 73
(b) Use part (a) to e valuate
Z cos 2x dx (c) Use parts (a) and (b) to e valuate
Z cos 4x dx 74Chapter 3. Techniques of Integration and ApplicationsExample 3.4EvaluateZ√3
1arctan1x
dx. 3.1 Integration By Parts 75
Additional Notes:
76Chapter 3. Techniques of Integration and Applications3.2T rigonometricIntegrals
(This lecture corresponds to Section 7.2 of Stewart"sCalculus.)Today, I am giving two exams...one in trig and the other in honesty. I hope you will pass them
both. If you must fail one, fail trig. There are many good people in the world who can"t pass trig, but there are no good people who cannot pass the exam of honesty. (Madison Sarratt, Dean and then Vice-Chancellor at Vanderbilt University, 1891-1978)Example 3.5Integrate the following:
(a)Z sin 2(3x)dx=
(b) Z cot 2(3x)dx=
3.2 Trigonometric Integrals 77
Products of Sines and Cosines.
To evaluateZ
sin nxcosmxdx, there are only two possibilities: (a) At least one of the numbers nandmisodd.Example 3.6 Z sin 3xcos2x dx=
(b) Both nandmareeven.Example 3.7
Z sin 2xcos2x dx=Example 3.8IntegrateZ
cos 5x dx 78Chapter 3. Techniques of Integration and ApplicationsExample 3.9Integrating Other Trig Functions:Tangent, Cotangent, Secant, and Cosecant.
(a)Z tanx dx= (b) Z cotx dx= (c) Z secx dx= (d) Z cscx dx= 3.2 Trigonometric Integrals 79
Additional Notes:
80Chapter 3. Techniques of Integration and Applications3.3T rigonometricSubstitutions
(This lecture corresponds to Section 7.3 of Stewart"sCalculus.) There is no harm in patience, and no profit in lamentation. (Abu Bakr, The First Caliph, 573-634)Here our goal is to use trig functions to try to simplify the integrand, hopefully converting it to one
that is easier to integrate.Motivating ProblemAssuming that|x| ≤a, evaluate Z pa 2-x2dx.Integration by Substitution (using Trigonometric Functions
If the integral involvesthen substituteand use the identity a 2-u2u=asinθ1-sin2θ=cos2θa
2+u2u=atanθ1+tan2θ=sec2θu
2-a2u=asecθsec
2θ-1=tan2θExample 3.10IntegrateZp1-x2dx,assuming|x|<1
3.3 Trigonometric Substitutions 81
Additional Notes:
82Chapter 3. Techniques of Integration and Applications3.4Integra tionof Ra tionalFunctions b yP artialFractions
(This lecture corresponds to Section 7.4 of Stewart"sCalculus.) It does not matter how slowly you go so long as you do not stop. (Confucius, Chinese Philosopher, 551-479 BC)Motivating ProblemEvaluateZx-1x 2-5x+6dx.Algorithm- Integra tingRa tionalFunctions .
Problem.EvaluateZP(x)Q(x)dx, wherePandQare polynomials. If degP≥degQthen (by long division) there are polynomialsq(x)andr(x)such that P(x)Q(x)=q(x)+r(x)Q(x)and eitherr(x)is identically0ordegr remainder produced by the long division process. Ifr(x) =0, thenP(x)Q(x)is really just a polynomial, so we can ignore that case here. Now ZP(x)Q(x)dx=Z
q(x)dx+Zr(x)Q(x)dx. We can easily integrate the polynomialq, so the general problem reduces to the problem of integrating a rational functionr(x)Q(x)with degrSo, for the purposes of investigating how to integrate a rational function we can supposef(x)=P(x)Q(x)
with degP(x)Theorem 3.4.1 - F actAbout Ev eryP olynomialQ. Qcan be factored as a product of linear factors (i.e. of the formax+b) and / or irreducible quadratic forms (i.e. of the formax2+bx+c, whereb2-4ac<0). 3.4 Integration of Rational Functions by Partial Fractions 83
Our strategy to integrate the rational functionf(x)is as follows: F actorQ(x)into linear and irreducible quadratic factors Write f(x)as a sum ofpartial fractions, where each fraction is of the form K(ax+b)sorLx+M(ax2+bx+c)t.
Inte grateeach partial fraction in the sum.
Question.How do we findK,L, andM?
Let"s look at some examples.Example 3.11Integrate
(a)Z4x2-3x-4x 3+x2-2xdx
84Chapter 3. Techniques of Integration and Applications(b)
Zx3-4x-1x(x-1)3dx
(c) Z5x3-3x2+2x-1x
4+x2dx
3.4 Integration of Rational Functions by Partial Fractions 85
(d) Z1x(x2+1)2dx
86Chapter 3. Techniques of Integration and ApplicationsExample 3.12Find the volume of the solid obtained by revolving the regionRbetween the
curve y=x-9x 2-3x and thex-axis over the interval 1≤x≤2, around they-axis. 3.4 Integration of Rational Functions by Partial Fractions 87
The steps to integrate a rational function f - A Technical Look Supposef(x) =P(x)Q(x)with degP Step 1:First factorQ(x)into its linear and irreducible quadratic pieces. If there arendistinct linear factors andmdistinct quadratic factors, then Q(x) = (a1x+b1)r1...(anx+bn)rn(c1x2+d1x+e1)s1...(cmx2+dmx+em)sm Step 2:Thef(x)can be written as a sum ofpartial fractionsas follows P(x)Q(x)=A1,1a
1x+b1+A1,2(a1x+b1)2+...+A1,r1(a1x+b1)r1+
. .. + An,1a nx+bn+An,2(anx+bn)2+...+An,rn(anx+bn)rn+ + B1,1x+C1,1c
1x2+d1x+e1+B1,2x+C1,2(c1x2+d1x+e1)2+...+B1,s1x+C1,s1(c1x2+d1x+e1)s1+
. .. + Bm,1x+Cm,1c
mx2+dmx+em+Bm,2x+Cm,2(cmx2+dmx+em)2+...+Bm,smx+Cm,sm(cmx2+dmx+em)sm Step 3:Integrate each partial fraction in the sum. 88Chapter 3. Techniques of Integration and ApplicationsAdditional Notes:
3.5 Strategy for Integration 89
3.5 Stra tegyf orIntegra tion
(This lecture corresponds to Section 7.5 of Stewart"sCalculus.)A math student"s best friend is BOB (the Back Of the Book), but remember that BOB doesn"t
come to school on test days. (Joshua Folb, High School Teacher, Winchester, Virginia)Table of Integration Formulas:Constants of integration have been omitted.
You should know this table!
Z x ndx=xn+1n+1,(n̸=-1)Zdxx =ln|x| Z e xdx=exZ a xdx=axlna Z sinx dx=-cosxZ cosx dx=sinx Z sec 2x dx=tanxZ
csc 2x dx=-cotxZ
secxtanx dx=secxZ cscxcotx dx=-cscxZ secx dx=ln|secx+tanx|Z cscx dx=ln|cscx-cotx|Z tanx dx=ln|secx|Z cotx dx=ln|sinx|Z sinhx dx=coshxZ coshx dx=sinhx Z dxx 2+a2=1a
arctanxa Z dx√a 2-x2=arcsinxa
Z dxx 2-a2=12alnx-ax+a
Z dx√x 2±a2=lnx+px
2±a2
90Chapter 3. Techniques of Integration and ApplicationsExample 3.13Integrate
(a)Zx+4x 3+xdx (b) Z π/2
0sin4xcos3xdx
(c)Z e xsin(2x)dx 3.5 Strategy for Integration 91
Example 3.14Integrate
(a)Z e 1lnxx dx (b)Z cos 2(5x)dx
(c)Z x 3lnxdx
(d)Z xsec(x2)tan(x2)dx (e) Z3x+1x(x+1)dx
92Chapter 3. Techniques of Integration and ApplicationsExample 3.15Integrate
(a)Z x 2(lnx)2dx
(b) Z π/2
0cos3xsin(2x)dx
(c) Z3x -1/2(x3/2-x1/2)dx (d) Z√x
2-1x dxhint: Use the substitutionx=secθ. (e) Z 3 0dxx 2-3x-4
3.5 Strategy for Integration 93
Example 3.16Integrate
(a)Z x 5e-x3dx
(b) Z 5 1p-x2+6x-5dx
(c) Z √1+x√1-xdx (d) Zcosx4-sin2xdx
(e) Zdx√1+x2hint: Use the substitutionx=tanθ.
94Chapter 3. Techniques of Integration and ApplicationsAdditional Notes:
3.6 Approximate Integration 95
3.6 Appr oximateIntegra tion
All exact science is dominated by the idea of approximation. (Bertrand Russell, English Logician and Philosopher 1872-1970) And now that you don"t have to be perfect, you can be good. (John Steinbeck, American Author, 1902-1968)Motivating ProblemEvaluateZ 1 0e-x2dx.ReminderIffis continuous on[a,b]and if[a,b]is divided intonsubintervals
[a=x0,x1],[x1,x2],...,[xn-1,xn=b] of equal length∆x=b-an then Z b af(x)dx≈n∑ i=1f(x∗i)∆x wherex∗iis any point in[xi-1,xi]. Various ways of choosing the sample pointsx∗i:Endpoint Approximation.The left-point approximationLnand the right-point approximationRntoRb
af(x)dxwith∆x=b-an are L n=n∑ i=1f(xi-1)∆x and R n=n∑ i=1f(xi)∆x. 96Chapter 3. Techniques of Integration and ApplicationsMidpoint Approximation.
The midpoint approximationMnwith∆x=b-an
is M n=n∑ i=1f(x i)∆x wherex i=xi-1+xi2 .Trapezoid Rule. The trapezoidal approximation to
Z b af(x)dxwith∆x=b-an is T n=∆x2 [f(x0)+2f(x1)+2f(x2)+...+2f(xn-1)+f(xn)]. 3.6 Approximate Integration 97
Example 3.17Calculate an approximation to the integral Z 3 0x2dx withn=6 and∆x=0.5 by using (a) left-endpoint approximation (b) right-endpoint approximation (c) midpoint approximation (d) trapezoidal approximation 98Chapter 3. Techniques of Integration and ApplicationsErrors in Approximation:TheerrorEin using an approximation is defined to be the difference between the actual value and
the approximationA. That is, E=Z b af(x)dx-A It turns out that the size of the error depends on the second derivative of the functionf, which measures how much the graph is curved. The following fact is usually proved in a course onnumerical analysis(MACM316), so we just state it here. Theorem 3.6.1
- Err orbounds .Suppose that|f′′(x)| ≤Kforxin the interval[a,b]. IfETand EMare the errors in the Trapezoidal and Midpoint Rules then |ET| ≤K(b-a)312n2and|EM| ≤K(b-a)324n2. Note:|f′′(x)|"measures" how far the function is away from being a line; the length of the interval
is part of the estimate; and, in both cases the error (bound) is quadratic in step-size1/n, i.e., if you
double the number of points the error should be divided by approximately4.Example 3.18SinceZ2 1dxx =ln2 the Trapezoidal and Midpoint Rules could be used to approximateln2. Estimate the errors in the the Trapezoidal and Midpoint approximations of this integral by usingn=10 intervals. 3.6 Approximate Integration 99
Example 3.19
(a) Use the Midpoint Rule with n=10 to approximate the integralZ 1 0e-x2dx.
(b) Gi vean upper bound for the error in volvedin this approximation. (c)How large do we have to choosenso that the approximationMnto the integral in part (a) is accurate to within 0.00001? 100Chapter 3. Techniques of Integration and ApplicationsApproximation using parabolic segments:Letfbe continuous on[a,b]and divide the interval into aneven numbernsubintervals of equal
length∆x=b-an. Suppose the endpoints of these subintervals are, as usual,a=x0,x1,x2,...,xn=b. LetPibe the point(xi,f(xi)). For each even numberiLetfbe continuous on[a,b]and∆x=b-an
withneven. Then we can approximateZ
b af(x)dxby the sum S n=∆x3 [f(x0)+4f(x1)+2f(x2)+4f(x3)+...+2f(xn-2)+4f(xn-1)+f(xn)].Example 3.20ApproximateZ 3 0dx1+x4by Simpson"s Rule withn=6.
3.6 Approximate Integration 101
Error in Simpson"s Rule.
This time, the size of the error depends on thefourthderivative off.Theorem 3.6.2- Err orBound in Simpson" sRule. Suppose that|f(4)(x)| ≤Kfor allxin the
interval[a,b]. IfESis the error in using Simpson"s Rule, then |ES| ≤K(b-a)5180n4. For Simpson, the size of the error depends on thefourthderivative off; It measures how far awayfis from being a cubic polynomial. If you doublen, the number of subintervals, the error will be divided by approximately16now! Example 3.21How large should we takenin order to guarantee that the Simpson"s Rule approximation toZ 2 11x dxis accurate within 0.0001? 102Chapter 3. Techniques of Integration and ApplicationsAdditional Notes:
3.7 Improper Integrals 103
3.7 Impr operIntegrals
(This lecture corresponds to Section 7.8 of Stewart"sCalculus.)You and I are essentially infinite choice-makers. In every moment of our existence, we are in
that field of all possibilities where we have access to an infinity of choices. (Deepak Chopra, Indian ayurvedic Physician and Author, 1947-)Motivating ProblemEvaluate the area of the region bounded by the curves
y=1x 2,y=0,x=1.Improper Integral of Type I:
(a) If Zt af(x)dx exists for allt≥a, then Z∞
af(x)dx=limt→∞Z t af(x)dx provided that this limit exists (i.e. as a finite number). (b) If Zb tf(x)dx exists for allt≤b, then Zb -∞f(x)dx=limt→-∞Z b tf(x)dx provided that this limit exists (i.e. as a finite number). The improper integrals
Z∞
af(x)dx and Zb -∞f(x)dx are calledconvergentif the correspond- ing limit exists anddivergentif the limit does not exist. (c) If both Z
a -∞f(x)dxandZ ∞ af(x)dxare convergent, then we define Z ∞ -∞f(x)dx=Z a -∞f(x)dx+Z ∞ af(x)dx. 104Chapter 3. Techniques of Integration and ApplicationsExample 3.22Investigate the improper integrals.
(a)Z ∞ 1dxx (b)Z ∞ 1dxx 2 (c)Z 0 -∞dx√1-x (d)Z ∞ -∞dx1+x2 3.7 Improper Integrals 105
Motivating ProblemEvaluate the area of the region bounded by the curves y=1√x ,y=0,x=0,x=1.Improper Integral of Type II (a) If fis continuous on[a,b)and is discontinuous atb, then Z b af(x)dx=limt→b-Z t af(x)dx provided that this limit exists. (b) If fis continuous on(a,b]and is discontinuous ata, then Z b af(x)dx=limt→a+Z b tf(x)dx provided that this limit exists.The improper integral Zb af(x)dx is calledconvergentif the corresponding limit exists and divergentif the limit does not exist. (c) Iffhas a discontinuity atc, wherea Zc af(x)dx and Zb cf(x)dx are convergent, then we define Z b af(x)dx=Z c af(x)dx+Z b cf(x)dx. 106Chapter 3. Techniques of Integration and ApplicationsExample 3.23Investigate the improper integrals.
(a)Z 2 1dx(x-2)2
(b)Z 2 0dx(2x-1)2/3
(c)Z 1 0lnx dxTheorem 3.7.1- Compar isonTheor em.Suppose thatfandgare continuous functions with
0≤g(x)≤f(x)forx≥a.
(a) If Z ∞ af(x)dxis convergent thenZ ∞ ag(x)dxis convergent. (b) If Z ∞ ag(x)dxis divergent thenZ ∞ af(x)dxis divergent. 3.7 Improper Integrals 107Example 3.24Use the Comparison Theorem to determine if the following integrals are conver-
gent or divergent. (a)Z ∞ 4dxlnx-1
(b)Z ∞ 1e-x2/2dx
108Chapter 3. Techniques of Integration and ApplicationsAdditional Notes:
4. Further Applications of Integration
In this chapter we continue looking at more applications of integration. Topics we will cover are:
arc length, surf acearea, inte grationalong parametric curv es. 110Chapter 4. Further Applications of Integration4.1Ar cLength
(This lecture corresponds to Section 8.1 of Stewart"sCalculus.)As usual, Ronaldinho takes the free kick. He sent the ball whistling into the air with his right
foot. Just as ball looked to fly wide, it curled in a perfect arc and entered the net at the top right corner. (From http://hvdofts.wordpress.com/2006/11/21/controversy-in-camp-nou/) Motivating ProblemFind the length of the arc of the parabolay= (x-1)2between the points (0,1)and(3,4). Theorem 4.1.1
- The Ar cLength For mula.Iff′is continuous on[a,b], then the length of the curvey=f(x),a≤x≤bis L=Z b aq1+[f′(x)]2dx. 4.1 Arc Length 111
Example 4.1Find the length of the following arcs.
(a)y= (x-1)2between the points(0,1)and(3,4) (b)x=16 y3+12y, 1≤y≤2 (c)y=x3, 0≤x≤5 112Chapter 4. Further Applications of IntegrationThe Arc Length Function.Let a smooth curveChave the equationy=f(x),a≤x≤b. Lets(x)
be the distance alongCfrom the initial pointP0(a,f(a))to the pointQ(x,f(x)). (a) Find the formula for s(x).
(b) Find dsdx , as well as the differentialds. Example 4.2Find the arc length function for the curvey=2x3/2with starting pointP0(1,2). 4.1 Arc Length 113
Additional Notes:
114Chapter 4. Further Applications of Integration4.2Ar eaof a Surf aceof Re volution
(This lecture corresponds to Section 8.2 of Stewart"sCalculus.) Be like a duck. Calm on the surface, but always paddling like the dickens underneath. (Michael Caine, British Actor, 1933-)Motivating ProblemFind the surface area of the paraboloid which is obtained by revolving
the parabolic arcy=√x, 0≤x≤2, about thex-axis.Surface Area Let a smooth curveCbe given byy=f(x),x∈[a,b]. (a) The area of the surf aceobtained by rotating Cabout thex-axis is defined as S=Z b a2πf(x)q1+[f′x)]2dx. (b) The area of the surf aceobtained by rotating Cabout they-axis is defined as S=Z b a2πxq1+[f′(x)]2dx.xyy=√x 121
-1xy y=√x 121
4.2 Area of a Surface of Revolution 115
Area formulas for surfaces of revolution
Description ofRevolution aboutRevolution about
curveCx-axisy-axisy=f(x),x∈[a,b]Z b a2πf(x)q1+[f′(x)]2dxZ b a2πxq1+[f′(x)]2dxx=g(y),y∈[c,d]Z d c2πyq1+[g′(y)]2dyZ d c2πg(y)q1+[g′(y)]2dyExample 4.3Find the area of the surface obtained by rotating the given arc about the corre-
sponding axis. (a)y=√x, 0≤x≤2, about thex-axis (b)y=x3, 0≤x≤2, about thex-axis (c)y=x2, 0≤x≤√2, about they-axis 116Chapter 4. Further Applications of IntegrationExample 4.4Find the surface area of the torus.
4.2 Area of a Surface of Revolution 117
Additional Notes:
118Chapter 4. Further Applications of Integration4.3Calculus with P arametricCur ves
(This lecture corresponds to Section 10.2 of Stewart"sCalculus.)Contrary to common belief, the calculus is not the height of the so-called "higher mathematics."
It is, in fact, only the beginning.
(Morris Kline, American mathematician, 1908-1992)Motivating ProblemFind the arc length of one arch of the cycloid
x=r(t-sint),y=r(1-cost),0≤t≤2π. Also find the area under this arch.Some previous results in a new context ... Calculus with Parametric Curves:
Suppose the functiony(x), forx∈[a,b], is defined by the parametric equations x=f(t)andy=g(t)fort∈[α,β] and letCbe the corresponding parametric curve. (We assume thatfandgsatisfy all conditions that will guarantee that the functiony(x)has the necessary properties
that allow for the existence of all listed integrals.) 1. If fandgare differentiable withf′(t)̸=0, thendydx =dydt dxdt =g′(t)f ′(t). 2. If y(x)≥0, then theareaunder the curveCis given by A=Z b ay(x)dx=Z β αg(t)f′(t)dt
3. Iff′andg′are continuous on[α,β]andCis traversed exactly once astincreases fromαtoβ,
then thelength of the curveCis given by s=Z b as1+dydx 2 dx=Z β αq[f′(t)]2+[g′(t)]2dt
3. If g(t)≥0 then thearea of the surfaceobtained by rotatingCabout thex-axis is given by S=Z b a2πys1+dydx 2 dx=Z β α2πg(t)q[f′(t)]2+[g′(t)]2dt
4.3 Calculus with Parametric Curves 119Example 4.5Find the slope of the tangent to the astroidx=acos3θ,y=asin3θas a function
of the parameterθ. At what points is the tangent horizontal. Vertical? At what points does that tangent have slope 1. What about slope-1?Example 4.6Find the area under one arch of the cycloid: x=r(t-sint),y=r(1-cost),0≤t≤2π. Also find the arc length of this arch.
120Chapter 4. Further Applications of IntegrationExample 4.7Find the area of the surface obtained by rotating the curve
x=3t-t3,y=3t2,0≤t≤1 about thex-axis.Example 4.8Polar Coordinates are just parametric equations. Really!! Find the arc lengthsof the cardioid with polar equation r=1+cosθ. Also, find also the surface areaSgenerated by revolving the cardioid around thex-axis. 4.3 Calculus with Parametric Curves 121
Additional Notes:
III 5Infinite Sequences and Series.. . . . . . 125
5.1 Sequences
5.2 Series
5.3 The Integral Test and Estimates of Sums
5.4 The Comparison Test
5.5 Alternating Series
5.6Absolute Convergence and the Ratio and Root Test
5.7 Strategy for Testing Series
5.8 Power Series
5.9 Representation of Functions as Power Series
5.10 Taylor and Maclaurin Series
5.11 Applications of Taylor PolynomialsPart Three: Sequences and
Series
5. Infinite Sequences and SeriesIn this chapter we consider infinite sums of numbers, and how to represent functions as infinite
polynomials (power series). Topics we will cover are:
infinite sequences, infinite series, po werseries and T aylorseries 126Chapter 5. Infinite Sequences and Series5.1Sequences
(This lecture corresponds to Section 11.1 of Stewart"sCalculus.) Mensa Puzzle.What number comes next in this sequence? 1 3 8 19 42 ?
What is the 100th number in the sequence?Definition 5.1.1 - Sequence.Asequenceis a function whose domain is the setZ+=
{1,2,3,...}of positive integers. If the function iss:Z+→R, then the outputs(n)is usually written assn, we also write the whole sequence ass={sn}. Note:Sometimes the domain of a sequence may be taken asN=Z+∪{0}, in which case we write {sn}∞n=0.Example 5.1 (a) Write out the first fe wterms of the sequence
{cosnπ}∞n=2. Is it possible to write this sequence in a different form? (b) Graph the sequence n
1+(-1)nn
o . 5.1 Sequences 127
Definition 5.1.2 - Limit of a sequence.(Informal definition) A sequence{an}has thelimitLand we write
lim n→∞an=Loran→Lasn→∞ if we can make the termsanas close toLas we like by takingnsufficiently large.Iflimn→∞anexists, we say the sequenceconverges(or it isconvergent). Otherwise, we say the
sequencediverges(or isdivergent).Definition 5.1.3 - Limit of a sequence.(Formal or rigorous definition: "ε-N definition")
A sequence{an}has thelimitLand we write
lim n→∞an=Loran→Lasn→∞ if for everyε>0 there is a corresponding integerNsuch that |an-L|<εwhenevern>N.Example 5.2Is the sequence2nn+3 convergent or divergent?Theorem 5.1.1Consider the sequencef(n) =anwherenis an integer.
If limx→∞f(x) =Lthen limn→∞an=L. If you compare Definition 2 with Definition 2.6.7,you will see that the only difference between and is that is required to be an integer. Thus we have the following theorem,which is illustrated by Figure 6. THEOREMIf and when is an integer, then
. In particular, since we know that when (Theorem 2.6.5), we have if If becomes large as nbecomes large,we use the notation . The fol- lowing precise definition is similar to Definition 2.6.9. DEFINITIONmeans that for every positive number there is an integer such that ifthen If ,then the sequence is divergent but in a special way. We say that diverges to . The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar. If and are conver gent sequences and is a constant,then lim n l ? a n p ?[lim n l ? a n ] p if p?0 and a n ?0 lim n l ? a n b n ? lim nl? a n lim n l ? b n if lim n l ? b n ?0 lim n l ? ?a n b n ??lim n l ? a n ?lim n l ? b n lim n l ? c?c lim n l ? ca n ?c lim n l ? a n lim n l ? ?a n ?b n ??lim n l ? a n ?lim n l ? b n lim n l ? ?a n ?b n ??lim n l ? a n ?lim n l ? b n c?b n ??a n ? ??a n ? ?a n ?lim nl? a n ?? a n ?Mn?N N Mlim nl? a n ??5 lim nl? a n ??a n r?0lim nl? 1 n r ?04 r?0lim xl? ?1?x r ??0 FIGURE 6
2 0x y 134
L y=ƒ lim nl? a n ?L nf?n??a n lim xl? f?x??L3 nlim x l ? f?x??Llim n l ? a n ?L 678||||CHAPTER 11INFINITE SEQUENCES AND SERIES
LIMIT LAWS FOR SEQUENCES
128Chapter 5. Infinite Sequences and SeriesDefinition 5.1.4
limn→∞an=∞ means that for every positive numberMthere is an integerNsuch that a n>Mwhenevern>N.Theorem 5.1.2- F actsa boutsequences . If{an}and{bn}are convergent sequences andcis a constant, then (a) lim n→∞(an±bn) =limn→∞an±limn→∞bn (b) lim n→∞(can) =climn→∞an(in particular, this means that limn→∞c=c) (c) lim n→∞(anbn) =limn→∞an·limn→∞bn (d) lim n→∞a nb n=limn→∞anlim n→∞bn, as long as limn→∞bn̸=0 (e) lim n→∞(an)p= limn→∞an ponly forp>0 andan>0. (f) If lim
n→∞|an|=0, then limn→∞an=0. (g) If an≤cn≤bnfor alln≥N, and limn→∞an=limn→∞bn=L, then limn→∞cn=L.
(h) If lim
n→∞an=Land a functionfis continuous atL, then limn→∞f(an) =f(L)Example 5.3 (a) Sho wthat the sequence {n√n}converges to 1.
(b) Is the sequence an=sinnπ2
convergent or divergent? 5.1 Sequences 129
Example 5.4
(a) Does the sequence ncos(nπ)n
o converge or diverge? (b) F orwhat v aluesof ris the sequence{rn}convergent?Definition 5.1.5A sequence{an}is calledincreasingifan a 1 It is calleddecreasingifan>an+1for alln≥1.
It is calledmonotonicif it is either increasing or decreasing.Example 5.5Decide which of the following sequences is increasing, decreasing or neither.
(a)an=1+1n (b)bn=1-1n (c)cn=1+(-1)nn 130Chapter 5. Infinite Sequences and SeriesDefinition 5.1.6A sequence{an}isbounded aboveif there is a numberMsuch that
a n≤Mfor alln≥1. It isbounded belowif there is a numbermsuch that
m≤anfor alln≥1. If it is bounded above and below, then{an}is abounded sequence.Theorem 5.1.3- Monotonic Sequence Theor em.Every bounded, monotonic sequence is
convergent. EXAMPLE 12Show that the sequence is decreasing.
SOLUTION 1We must show that ,that is,
This inequality is equivalent to the one we get by cross-multiplication: Since ,we know that the inequality is true. Therefore and so is decreasing. SOLUTION 2Consider the function :
Thus is decreasing on and so . Therefore is decreasing.M DEFINITIONA sequence is bounded aboveif there is a number such that It is bounded belowif there is a number such that
If it is bounded above and below,then is a bounded sequence. For instance,the sequence is bounded belo w but not abov e. The sequence is bounded because for all . We know that not every bounded sequence is convergent [for instance,the sequence satisfies but is divergent from Example 6] and not every mono- tonic sequence is conver gent . But if a sequence is both bounded and monotonic,then it must be convergent. This fact is proved as Theorem 12,but intuitively you can understand why it is true by looking at Figure 12. If is increasing and for all ,then the terms are forced to crowd together and approach some number . The proof of Theorem 12 is based on the Completeness Axiomfor the set of real numbers,which says that if is a nonempty set of real numbers that has an upper bound (for all in ),then has a least upper bound. (This means that is an upper bound for , but if is an y other upper bound,then .) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line. b?MMS bbSSxx?MM S ? Ln a n ?M?a n ? ?a n ?nl"? #1?a n ?1a n ??#1? n n0$a n$1a n ?n??n%1? ?a n &0?a n ?n ?a n ? for all n'1m?a n m for all n'1a n ?M M?a n ?11 ?a n ?f?n?&f?n%1??1, "?f wheneverx 2 &1f(?x?? x 2 %1#2x 2 ?x 2 %1? 2 ? 1#x 2 ?x 2 %1? 2 $0 f?x?? x x 2 %1 ?a n ? a n%1$a n n 2 %n&1n'1 1$n 2 %n&? n 3 %n 2 %n%1$n 3 %2n 2 %2n&? ?n%1??n 2 %1?$n??n%1? 2 %1?&? n%1 ?n%1? 2 %1 $ n n 2 %1 n%1 ?n%1? 2 %1 $ n n 2 %1 a n%1$a n a n ? n n 2 %1 682||||CHAPTER 11INFINITE SEQUENCES AND SERIES
2 0 n a n 13 L M FIGURE 12Example 5.6Investigate the sequence{an}that is defined recursively by a 1=√6,an+1=p6+an,forn≥1.
5.1 Sequences 131
(example continued) 132Chapter 5. Infinite Sequences and SeriesAdditional Notes:
5.2 Series 133
5.2 Ser ies
(This lecture corresponds to Section 11.2 of Stewart"sCalculus.)Joke:An infinite crowd of mathematicians enters a bar. The first one orders a pint, the second
one a half pint, the third one a quarter pint ... "I understand," says the bartender - and pours two pints.Definition 5.2.1 - Series.Suppose{an}is a sequence of numbers. An expression of the form a 1+a2+a3+...+an+...
is called aninfinite seriesand it is denoted by the symbol ∞∑ n=1a nor∑an.Definition 5.2.2 - Partial Sum.If∞∑ i=1a iis a series then s n=n∑ i=1a i=a1+a2+...+an is called itsnthpartial sum. But does it make sense to "add infinitely many numbers"? Not directly, so we imagine adding finitely many terms, but more and more terms each time, and look at what happens to these cumulative sums. 134Chapter 5. Infinite Sequences and SeriesDefinition 5.2.3Given the series∑∞i=1ai=a1+a2+..., letsndenote itsnthpartial sum:
s n=n∑ i=1a i=a1+a2+...+an.If the sequence{sn}is convergent andlimn→∞sn=sexists as a real number, then the series∑anis
calledconvergentand we write ∞∑ i=1a i=a1+a2+...=s The numbersis called thesumof the series.
If the limit above does not exist, then the series is calleddivergent. Example 5.7The series1+12
+14 +18 +116
+...has partial sumss1=1,s2=1.5,s3=1.75, s4=1.875...and in general it turns out thatsn=2-12 n-1. Sincesn→2 asn→∞, the series is convergent and has sum 2. Example 5.8Show that thegeometric series
∞∑ i=1ari-1=a+ar+ar2+... is convergent if |r|<1 and its sum is ∞∑ i=1ari-1=a1-r,|r|<1 If|r| ≥1, the geometric series is divergent.
(Here we are assuminga̸=0, otherwise the series converges to 0 regardless of the value ofr.) 5.2 Series 135
Example 5.9Determine whether the given series converges or diverges. (a)∞∑ n=1 e10 n (b) ∞∑ n=1(-1)n3e nExample 5.10Express 0.5555...as a rational number.Example 5.11Show that the series∞∑ n=11(n+1)(n+2)is convergent and find its sum. 136Chapter 5. Infinite Sequences and SeriesExample 5.12Show that theharmonic series
∞∑ n=11n =1+12 +13 +... is divergent. Two Useful Results:Theorem 5.2.1- T estf orDiv ergence. (a)