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Chapter 7

antiderivatives and the area problem

Let me begin by dening the terms in the title:

1. an an tiderivativeof fis another functionFsuch thatF0=f. 2. th earea problem i s:"nd the ar eaof a s hapein the plane"

This chapter is concerned with understanding the area problem and then solving it through the fundamental

theorem of calculus(FTC).

We begin by discussing antiderivatives. At rst glance it is not at all obvious this has to do with the area

problem. However, antiderivatives do solve a number of interesting physical problems so we ought to consider

them if only for that reason. The beginning of the chapter is devoted to understanding the type of question

which an antiderivative solves as well as how to perform a number of basic indenite integrals. Once all of

this is accomplished we then turn to the area problem.

To understand the area problem carefully we'll need to think some about the concepts of nite sums, se-

quences and limits of sequences. These concepts are quite natural and we will see that the theory for these is

easily transferred from some of our earlier work. Once the limit of a sequence and a number of its basic prop-

erties are established we thendene areaand the denite integral. Finally, the remainder of the chapter is

devoted to understanding the fundamental theorem of calculus and how it is applied to solve denite integrals.

I have attempted to be rigorous in this chapter, however, you should understand that there are superior

treatments of integration(Riemann-Stieltjes, Lesbeque etc..) which cover a greater variety of functions in a

more logically complete fashion. The treatment here is more or less typical of elementary calculus texts.

245

246CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

7.1 indenite integration

Don't worry, the title of this section will make sense later.

7.1.1 why antidierentiate?

The antiderivative is the opposite of the derivative in the following sense:

Denition 7.1.1.antiderivative.IffandFare functions such thatF0=fthen we say thatFis anantiderivativeoff.Example 7.1.2.Supposef(x) =xthen an antiderivative offis a functionFsuch thatdFdx

=x. We could tryx2but thenddx (x2) = 2xhas an unwanted factor of2. What to do? Just adjust our guess a little: try

F(x) =12

x2. Note thatddx (12 x2) =12 ddx (x2) =12 (2x) =x: Example 7.1.3.Letkbe a constant. Supposeg(t) =ektthen we guessG(t) =1k ektand note it works; ddt (1k ekt) =ektthereforeg(t) =ekthas antiderivativeG(t) =1k ekt. Example 7.1.4.Supposeh() = cos(). GuessH() = sin()and note it works;dd (sin()) = cos(). Obviously these guesses are not random. In fact, these are educated guesses. We simply have to think

about how we dierentiated before and just try to think backwards. Simple enough for now. However, we

should stop to notice that the antiderivative is far from unique. You can easily check thatF(x) =12 x2+c1,

G(t) =1k

ekt+c2andH() = sin() +c3are also antiderivatives for any constantsc1;c2;c32R.

Proposition 7.1.5.antiderivatives dier by at most a constant.Iffhas antiderivativesF1andF2then there existsc2Rsuch thatF1(x) =F2(x) +c.Proof:We are given thatdF1dx

=f(x) anddF2dx =f(x) thereforedF1dx =dF2dx . Hence, by Proposition 6.1.10 we ndF1(x) =F2(x) +c. To understand the signicance of this constant we should consider a physical question.

Example 7.1.6.Suppose that the velocity of a particle at positionxis measured to be constant. In particular,

suppose thatv(t) =dxdt andv(t) = 1. The conditionv(t) =dxdt means thatxshould be an antiderivative of

v. Forv(t) = 1the form of all antiderivatives is easy enough to guess:x(t) =t+c. The value forccannot

be determined unless we are given additional information about this particle. For example, if we also knew

that at time zero the particle was atx= 3then we could t thisinitial datato pick a value forc: x(0) = 0 +c= 3)c= 3)x(t) =t+ 3:

For a given velocity function each antiderivative gives a possible position function. To determine the precise

position function we need to know both the velocity and some initial position. Often we are presented with

a problem for which we do not know the initial condition so we'd like to have a mathematical device to leave

open all possible initial conditions.

7.1. INDEFINITE INTEGRATION247

Denition 7.1.7.indenite integral.Iffhas an antiderivativeFthen theindenite integraloffis given by:

Z f(x)dx=fG(x)jG0(x) =f(x)g=fF(x) +cjc2Rg: However, we will customarily drop the set-notation and simply write Z

f(x)dx=F(x) +cwhereF0(x) =f(x):The indenite integral includes all possible antiderivatives for the given function. Technically the indenite

integral is not a function. Instead, it is a family of functions each of which is an antiderivative off.

Example 7.1.8.Consider the constant acceleration problem1; we are given thata=gwhereg= 9:8m=s2 anda=dvdt . We can take the indenite integral of the equation: dvdt =g)v(t) =Z g dt=gt+c1:

Furthermore, ifv=dydt

then dydt =gt+c1)y(t) =Z gt+c1dt=12 gt2+c1t+c2: Therefore, we nd the velocity and position are given by formulas v(t) =c1gt y(t) =c2+c1t12 gt2: If we know the initial velocity isvoand the initial position isyothen v(0) =vo=c10)v(t) =vogty(0) =yo=c200)y(t) =yo+vot12

gt2These formulas were derived by Galileo without the benet of calculus. Instead, he used experiment and a

healthy skepticism of the philosophical nonsense of Aristotle. The ancient Greek's theory of motion said that

if something was twice as heavy then it falls twice as fast. This is only true when the objects compared have

air friction clouding the dynamics. The equations above say the objects' motion is independent of the mass.1

hereF=maismg=masoa=gbut that's physics, I supply the equation of motion in calculus. You just have to do the math.

248CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Remark 7.1.9.redundant comment (again).The indenite integral is a family of antiderivatives:

Rf(x) =F(x) +cwhereF0(x) =f(x). The

following equation shows how indenite integration is undone by dierentiation: ddx Z f(x)dx=f(x) the functionfis called theintegrandand the variable of indenite integration isx. Notice the

constant is obliterated by the derivative in the equation above. Leibniz' notation intentionally makes

you think of cancelling thedx's as if they were tiny quantities. Newton called them uxions. In fact calculus was sometimes called thetheory of uxionsin the early 19-th century. Newton had in mind thatdxwas the change inxover a tiny time, it was a uctuation with respect to a time implicit. We no longer think of calculus in this way because there are easier ways to think about foundations of

calculus. That said, it is still an intuitive notation and if you are careful not to overextend intuition

it is a powerful mnemonic. For example, the chain rule dfdx =dfdu dudx . Is the chain rule just from

multiplying by one? No. But, it is a nice way to remember the rule.A dierential equation is an equation which involves derivatives. We have solved a number of dierential

equations in this section via the process of indenite integration. The example that follows doesn't quite t

the same pattern. However, I will again solve it by educated guessing 2.

Example 7.1.10.A simple model of population growth is that the rate of population growth should be directly

proportional to the size of the populationP. This means there existsk2Rsuch that dPdt =kP: Fortunately, we just did Example 7.1.3 where we observed that Z e ktdt=1k ekt+c

So we know that one solution is given byP(t) =1k

ekt. Change variables by substitutingu= ln(P)so dudt =1P dPdt thusdPdt =Pdudt . Hence we can solvePdudt =kPordudt =kinstead. This we can antidierentiate to ndu(t) =kt+c1. Thus,ln(P) =kt+c1henceP(t) =ekt+c1=ec1ekt. If the initial population is given

to bePothen we ndP(0) =Po=ec1thusP(t) =Poekt:The same mathematics govern simple radioactive decay, continuously compounded interest, current or voltage

in an LR or RC circuit and a host of other simplistic models in the natural sciences. Real human population

growth involves many factors beyond just raw population, however for isolated systems this type of model

does well. For example, growth of bacteria in a petri dish.

Remark 7.1.11.why antidierentiate?We antidierentiate to solve simple dierential equations. When one variable (sayv) is the instan-

taneous rate of change of another (sayssov=dsdt ) then we can reverse the process of dierentiation to discover the formula ofsif we are given the formula forv. However, because constants are lost in dierentiation we also need an initial condition if we wish to uniquely determine the formula fors.

I have emphasized the utility of the concept of antidierentiation as it applies to physics, but that

was just my choice.2

Actually, the method I use here is rather unusual but the advanced reader will recognize the idea from dierential

equations. The easier way of solving this is called separation of a variables, but we discuss that method much later

7.1. INDEFINITE INTEGRATION249

Notice, I have yet to even discuss the area problem. We already see that indenite integration is an important

skill to master. The methods I have employed in this section are ad-hoc. We would like a more systematic

method. I oer organization for guessing in the next section.

7.1.2 properties of indenite integration

In this section we list all the basic building blocks for indenite integration. Some of these we already guessed

in specic examples. If you need to see examples you can skip ahead to the section that follows this one.

Proposition 7.1.12.basic properties of indenite integration.Supposef;gare functions with antiderivatives andc2Rthen

Z [f(x) +g(x)]dx=Z f(x)dx+Z g(x)dx Z cf(x)dx=cZ f(x)dxProof:SupposeRf(x)dx=F(x) +c1andRg(x)dx=G(x) +c2note that ddx [F(x) +G(x)] =ddx [F(x)] +ddx [G(x)] =f(x) +g(x) hence R[f(x) +g(x)]dx=F(x) +G(x) +c3=Rf(x)dx+Rg(x)dxwhere the constantc3is understood to be included in either theRf(x)dxor theRg(x)dxintegral as a matter of custom. Proposition 7.1.13.power rule for integration. supposen2Randn6=1thenZ x ndx=1n+ 1xn+1+c:Proof: ddx [1n+1xn+1] =n+1n+1xn+11=xn. Note thatn+ 16= 0 sincen6=1. Note that the special case ofn=1 stands alone. You should recall thatddx ln(x) =1x providedx >0. In the casex <0 then by the chain rule applied to the positive case:ddx ln(x) =1x(1) =1x . Observe then that for allx6= 0 we haveddx lnjxj=1x . Therefore the proposition below follows: Proposition 7.1.14.reciprocal function is special case.Z 1x

dx= lnjxj+c:Note that it is common to move the dierential into the numerator of such expressions. We could just as

well have written thatRdxx = lnjxj+c. I leave the proof of the propositions in the remainder of this section to the reader. They are not dicult.

250CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Proposition 7.1.15.exponential functions. supposea >0anda6= 1,Z a xdx=1ln(a)ax+cin particular:Z e xdx=ex+cThe exponential function has basea=eand ln(e) = 1 so the formulas are consistent.

Proposition 7.1.16.trigonometric functions.Z

sin(x)dx=cos(x) +cZ cos(x)dx= sin(x) +c Z sec

2(x)dx= tan(x) +cZ

sec(x)tan(x)dx= sec(x) +c Z csc

2(x)dx=cot(x) +cZ

csc(x)cot(x)dx=csc(x) +c:You might notice that many trigonometric functions are missing. For example, how would you calculate

3Rtan(x)dx? We do not have the tools for that integration at this time. For now we are simply cataloguing

the basic antiderivatives that stem from reading basic derivative rules backwards.

Proposition 7.1.17.hyperbolic functions.Z

sinh(x)dx= cosh(x) +cZ

cosh(x)dx= sinh(x) +cNaturally there are also basic antiderivatives forsech2(x);sech(x)tanh(x);csch2(x) andcsch(x)coth(x) how-

ever I omit them for brevity and also as to not antagonize the struggling student at this juncture. Proposition 7.1.18.special algebraic and rational functionsZ dx1 +x2= tan1(x) +cZdxp1x2= sin1(x) +c: Zdxpx

21= cosh1(x) +cZdxp1 +x2= sinh1(x) +c:

Zdx1x2= tanh1(x) +c:One can replace the expressions above with natural logs of certain algebraic functions. These identities are

explored on page 466 of Stewart's 6-th edition. Page 488 has a nice summary of these basic integrals that

we ought to memorize (although we have not covered tan(x) and cot(x) at this point)3 the answer is lnjsec(x)j+cif you're curious and impatient.

7.1. INDEFINITE INTEGRATION251

7.1.3 examples of indenite integration

Example 7.1.19.

Z dx=Z x

0dx=x+cExample 7.1.20.

Zpx+13

px dx=Z x 12 dx+Z x 13 dx=2 3 x32 +32
x23 +cExample 7.1.21.

Zp13x7dx=p13

Z x 72
dx=p13 9 2 x92 +c=2 p13 9 x4px+cExample 7.1.22.

Zdx3x2=13

Z x 2dx=13 x1=13x+cExample 7.1.23.

Z2xdxx

2= 2Zdxx

= 2lnjxj+c=ln(x2) +cNote thatjxj=xthusjxj2= (x)2=x2so it was logical to drop the absolute value bars after bringing in

the factor of two by the propertyln(Ac) =cln(A).

Example 7.1.24.

Z

3ex+2dx= 3Z

e

2exdx= 3e2Z

e xdx= 3e2(ex+c1) =3ex+2+cExample 7.1.25. Z (x+ 2)2dx=Z (x2+ 4x+ 4)dx = Z x

2dx+ 4Z

xdx+ 4Z dx =1 3 x3+ 2x2+ 4x+cExample 7.1.26. Z (2x3+ 3)dx=24 x4+ 3x+c=1 2 x4+ 3x+cExample 7.1.27. Z (2 x+ 3cosh(x))dx=Z 2 xdx+ 3Z cosh(x)dx=1 ln(2)

2x+ 3sinh(x) +cExample 7.1.28.

Z2x3+ 3x

dx=Z 2x3x +3x  dx=Z

2x2dx+ 3Zdxx

=2 3 x3+ 3lnjxj+c

252CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Example 7.1.29.

Zx21 +x2dx=Z1 +x211 +x2dx=Z111 +x2dx=xtan1(x) +cExample 7.1.30. Z sin(x+ 3)dx=Zsin(x)cos(3) + sin(3)cos(x)dx = cos(3) Z sin(x)dx+ sin(3)Z cos(x)dx =cos(3)[cos(x) +c1] + sin(3)[sin(x) +c2] = sin(3)sin(x)cos(3)cos(x) +c

=cos(x+ 3) +cIncidentally, we nd a better way to do this later with the technique ofu-substitution.

Example 7.1.31.

Z1cos

2(x)dx=Z

sec

2(x)dx=tan(x) +cExample 7.1.32.

Zdxx

2+ cos2(x) + sin2(x)=Zdxx

2+ 1=tan

1(x) +cEvery example in this section is easily checked by dierentiation.

Problems

Problem 7.1.1.hope to add problems in the future..

7.1. INDEFINITE INTEGRATION253

.

254CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

7.2 area problem

The area of a general shape in the plane can be approximately calculated by dividing the shape into a bunch

of rectangles or triangles. Since we know how to calculate the area of a rectangle [A=lw] or a triangle

[A=12

bh] we simply add together all the areas to get an approximation of the total area. In the special case

that the shape has at sides then we can nd the exact area since any shape with at sides can be subdivided

into a nite number of triangles. Generally shapes have curved edges so no nite number of approximating

rectangles or triangles will capture the exact area. Archimedes realized this some two milennia ago in ancient

Syracuse. He argued that if you could nd two approximations of the area one larger than the true area

and one smaller than the true area then you can be sure that the exact area is somewhere between those

approximations. By such squeeze-theorem type argumentation he was able to demonstrate that the value of

must be between22371 and227 (in decimals 3:1408< u3:1416<3:1429 ). In Apostol's calculus text he

discusses axioms for area and he uses Archimedes' squeezing idea to dene both area and denite integrals.

Our approach will be less formal and less rigorous.

Our goal in this section is to careful construct a method to calculate the area bounded by a function on

some interval [a;b]. Since the function could take on negative values in the interval we actually are working

on a method to calculate signed area under a graph. Area found beneath thex-axis is counted negative

whereas area above thex-axis is counted positive. Shapes more general than those described by the graph

of a simple function are treated in the next chapter.

7.2.1 sums and sequences in a nutshell

A sequence is function which corresponds uniquely to an ordered list of values. We consider real-valued

sequences but the concept extends to many other objects 4.

Denition 7.2.1.sequence of real numbers.IfUZhas a smallest member and the property thatn2Uimpliesn+ 12Uthen a function

f:U!Ris asequence. Moreover, we may denote the sequence by listing its values f=ff(u1);f(u2);f(u3);:::g=ffu1;fu2;fu3;:::g=ffujg1j=1 TypicallyU=NorU=N[ f0gand we study sequences of the form

fajg1j=0=fa0;a1;a2;:::g fbng1n=1=fb1;b2;b3:::gExample 7.2.2.Sequences may dened by a formula:an=nfor alln2Ngives

fang1n=1=f1;2;3;:::g: Or by an iterative rule:f1= 1;f2= 1thenfn=fn1+fn2for alln3denes the Fibonacci sequence: ffng1n=1=f1;1;2;3;5;8;13;21;:::g:

Beyond this we can add, subtract and sometimes divide sequences because a sequence is just a function with

a discrete domain.4 sequences of functions, matrices or even spaces are studied in modern mathematics

7.2. AREA PROBLEM255

Denition 7.2.3.nite sum notation.Supposeaj2Rforj2N. Then dene: 1 X j=1a j=a1n X j=1a j=n1X j=1a j+an forn2. This iterative denition gives us the result that n X j=1a j=a1+a2+


+an: The variablejis called thedummy index of summation. Moreover, sums such as j NX j=j1a j=aj1+aj2+


+ajN|{z}

Nsummands

can be carefully dened by a similar iterative formula.Example 7.2.4.Sums can give particularly interesting sequences. Consideran=Pn

j=1jforn= 1;2:::. fang1n=1=f1;1 + 2;1 + 2 + 3;1 + 2 + 3 + 4;:::g=f1;3;6;10;:::g:

The greatest mathematician of the 19-th century is generally thought to be Gauss. As a child Gauss was tasked

with computinga100. The story goes that just as soon as the teacher asked for the children to calculate the

sum Gauss wrote the answer5050on his slate. How did he know how to calculate the sum1+2+3+


+50 with such ease? Gauss understood that generally n X j=1j=n(n+ 1)2

For example,

a

1=1(1 + 1)2

= 1; a2=2(2 + 1)2 = 3; a3=3(3 + 1)2 = 6; a

4=4(4 + 1)2

= 10; :::;a100=(100)(101)2 = 50(101) = 5050:

What method of proof is needed to prove results such as this? The method is called "proof by mathematical

induction". We discuss it in some depth in the Math 200 course. In short, the idea is this: you prove the

result you interested in is true forn= 1 then you prove that ifnis true thenn+ 1 is also true for an arbitraryn2N. Let's see how this plays out for the preceding example: Proof of Gauss' Formula by induction:note thatn= 1 is clearly true sincea1= 1. Assume thatPn j=1j=n(n+1)2 (?) is valid and consider that, by the recursive denition of the nite sum, n+1X j=1j=nX j=1j+n+ 1 =n(n+ 1)2 |{z} using?+n+ 1 =12 (n2+ 3n+ 2) =([n+ 1])([n+ 1] + 1)2

256CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

which is precisely the claim forn+1. Therefore, by proof by mathematical induction, Gauss' formula is true

for alln2N.

Formulas for simple sums such as

P1;Pn;Pn2;Pn3are also known and can be proven via induction.

Let's collect these results for future reference:

Proposition 7.2.5.special formulas for nite sums.n X k=11 =n;nX k=1k=n(n+ 1)2 ;nX k=1k

2=n(n+ 1)(2n+ 1)6

;nX k=1k

3=n2(n+ 1)24

:The following results are less surprising but are even more useful: Proposition 7.2.6.nite sum properties. supposeak;bk;c2Rfor allkand letn;m2Nsuch thatm < n,(i:)nX k=1a k+nX k=1b k=nX k=1(ak+bk); (ii:)nX k=1ca k=cnX k=1a k; (iii:)nX k=1a k=mX k=1a k+nX k=m+1a k:Proof:begin with (i:). The proof is by induction onn. Note that (i:) is true forn= 1 sinceP1 k=1ak+P1 k=1bk=a1+b1=P1 k=1(ak+bk). Suppose that (i:) is true fornand consider n+1X k=1a k+n+1X k=1b k=nX k=1a k+an+1+nX k=1b k+bn+1by defn. ofX = nX k=1(ak+bk) +an+1+bn+1by induction hypothesis forn = n+1X k=1(ak+bk) by defn. ofX

Therefore, (i:) true fornimplies (i:) is true forn+1 hence by proof by mathematical induction we conclude

(i:) is true for alln2N. The proof for (ii:) is similar. We leave the proof of (iii:) to the reader.

We would like to have sums with many terms in the sections that follow. In fact, we will want to letn! 1.

The denition that follows is essentially the same we gave previously for functions of a continuous variable.

The main dierence is that only integers are considered in the limiting process.

7.2. AREA PROBLEM257

Denition 7.2.7.limit of a sequence.We say the sequencefangconverges toL2Rand denote lim n!1an=L

i for each >0 there existsN2Nsuch that for alln > Nwe ndjanLj< The skills you developed in studying functions of a continuous variable transfer to the study of sequential

limits because of the following fundamental lemma:

Lemma 7.2.8.correspondence of limits of functions onRand sequences.Supposefangis a sequence andfis a function such thatf(n) =anfor alln2dom(fang). If

lim

x!1f(x) =L2Rthen limn!1an=L.Proof:assume limx!1f(x) =L2Randf(n) =anfor alln2N. Let >0 and note that by the given

limit there existsM2Rsuch thatjf(x)Lj< for allx > M. ChooseNto be the next integer beyond MsoN2NandN > M. Suppose thatn2Nandn > Nthenjf(n)Lj=janLj< . Therefore, lim n!1an=L.

The converse is not true. You could extend a sequence so that it gave a function of a continuous variable

which diverged. Just imagine a function which oscillates wildly between the natural numbers.

Denition 7.2.9.innite sum.1

X k=1a k= limn!1n X k=1a

k:Given a particular formula forakit is generally not an easy matter to determine if the limit above exists.

These sums without end are calledseries. In particular, we deneP1 k=1ak=a1+a2+a3+


to converge i the limit lim n!1Pn k=1akconverges to a real number. We discuss a number of various criteria to analyze

this question in calculus II. I believe this amount of detail is sucient for our purposes in solving the area

problem. Our focus will soon shift away from explicit calculation of these sums.

7.2.2 left, right and midpoint rules

We aim to calculate the signed-area bounded byy=f(x) foraxb. In this section we discuss three

methods to approximate the signed-area. To begin we should settle some standard notation which we will

continue to use for several upcoming sections. Denition 7.2.10.partition of[a;b].Supposea < bthen [a;b]R. Dene
x=ban forn2Nand letxj=a+j
xforj= 0;1;:::;n.

In particular,xo=aandxn=b.The closed interval [a;b] is a union ofn-subintervals of length
x. Note that the closed interval [a;b] =

[xo;x1][[x1;x2][


[[xn1;xn]. The following rule is an intuitively obvious way to calculate the signed-area.

258CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Denition 7.2.11.left endpoint rule (Ln).Suppose that [a;b]dom(f) then we dene L n=n1X

j=0f(xj)
x= [f(x0) +f(x1) +


+f(xn1)]
x:Example 7.2.12.Letf(x) =x2and estimate the signed-area bounded byfon[1;3]by the left-endpoint

rule. To keep things simple I'll just illustrate the calculation withn= 4. Note
x=314 = 0:5thus x o= 1;x1= 1:5;x2= 2;x3= 2:5andx4= 3. L

4= [f(1) +f(1:5) +f(2) +f(2:5)]
x= [1 + 2:25 + 4 + 6:25](0:5) = 6:75

It's clear from the picture below thatL4underestimates the true area under the curve.Denition 7.2.13.right endpoint rule (Rn).Suppose that [a;b]dom(f) then we dene

R n=nX

j=1f(xj)
x= [f(x1) +f(x2) +


+f(xn)]
x:Example 7.2.14.Letf(x) =x2and estimate the signed-area bounded byfon[1;3]by the right end-point

rule. To keep things simple I'll just illustrate the calculation withn= 4. Note
x=314 = 0:5thus x o= 1;x1= 1:5;x2= 2;x3= 2:5andx4= 3. R

4= [f(1:5) +f(2) +f(2:5) +f(3)]
x= [2:25 + 4 + 6:25 + 9](0:5) = 10:75

It's clear from the picture below thatR4overestimates the true area under the curve.

7.2. AREA PROBLEM259

Denition 7.2.15.midpoint rule (Mn).Suppose that [a;b]dom(f) and denote the midpoints by xk=12 (xk+xk1) and dene M n=nX

j=1f(xj)
x= [f(x1) +f(x2) +


+f(xn)]
x:Example 7.2.16.Letf(x) =x2and estimate the signed-area bounded byfon[1;3]by the midpoint

rule. To keep things simple I'll just illustrate the calculation withn= 4. Note
x=314 = 0:5thus x1= 1:25;x2= 1:75;x3= 2:25andx4= 2:75. M

4= [f(1:25) +f(1:75) +f(2:25) +f(2:75)]
x= [1:5625 + 3:0625 + 5:0625 + 7:5625](0:5) = 8:625

ClearlyL4< M4< R4and if you study the errors you can seeL4< M4< A < R4.Notice that the size of the errors will shrink if we increasen. In particular, it is intuitively obvious that

asn! 1we will obtain the precise area bounded by the curve. Moreover, we expect that the distinction betweenLn;RnandMnshould vanish asn! 1. Careful proof of this seemingly obvious claim is beyond the scope of this course. Example 7.2.17.Letf(x) =x2and calculate the signed-area bounded byfon[1;3]by the right end-point rule. To perform this calculation we need to set upRnfor arbitrarynand then take the limit asn! 1. Notexk= 1 +k
xand
x= 2=nthusxk= 1 + 2k=n. Calculate, f(xk) =

1 +2kn

 2 = 1 +4kn +4k2n 2 thus, R n=nX k=1f(xk)
x = nX k=1

1 +4kn

+4k2n 2 2n = 2n n X k=11 +8n 2n X k=1k+8n 3n X k=1k 2 = 2n n+8n

2n(n+ 1)2

+8n

3n(n+ 1)(2n+ 1)6

= 2 + 4 1 +1n  +86
 2 +3n +1n 2

260CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Note that

1n and1n

2clearly tend to zero asn! 1thus

lim n!1Rn= 2 + 4 +166 =263 u8:6667:

Challenge: showLnandMnalso have limit263

asn! 1. Notice that the error inM4is simplyE= 8:66678:625 = 0:0417 which is within 0:5% of the true area. I

will not attempt to give an quantitative analysis of the error inLn;RnorMnat this time. Stewart discusses

the issue inx8:7. Qualitatively, if the function is monotonic then we should expect that the area is bounded

betweenLnandRn.

7.2.3 Riemann sums and the denite integral

In the last section we claimed that it was intuitively clear that asn! 1all the dierent approximations

of the signed-area converge to the same value. You could construct other rules to select the height of the

rectangles. Riemann's denition of the denite integral is made to exploit this freedom in the limit. Again,

it should be mentioned that this begs an analytical question we are unprepared to answer. For now I have

to ask you to trust that the following denition is meaningful. In other words, you have to trust me that

it doesn't matter the details of how the point in each subinterval is chosen. Intuitively this is reasonable as

x!0 asn! 1. Therefore, the subinterval [xj;xj+
x]! fxjgso the choice between the left, right and

midpoints is lost in the limit. Actually, special functions which are very discontinuous could cause problems

to the intuitive claim I just made. For that reason we insist that the function below is continuous on [a;b]

in order that we avoid certain pathologies.

Denition 7.2.18.Riemann sum and the denite integral of continuous function on[a;b].Suppose thatfis continuous on [a;b] supposexk2[xk1;xk] for allk2Nsuch that 1kn

then ann-th Riemann sum is dened to be R n=nX

j=1f(xk)
x= [f(x1) +f(x2) +


+f(xn)]
x:Notice that no particular restriction is placed on the sample pointsxk. This means a Riemann sum could

be a left, right or midpoint rule. This freedom will be important in the proof of the Fundamental Theorem

of Calculus I oer in a later section.

7.2. AREA PROBLEM261

Denition 7.2.19.denite integrals.Suppose thatfis continuous on [a;b], thedenite integralofffromatobis dened to be

lim n!Rnin particular we denote: Z b a f(x)dx= limn!Rn= limn!1 nX j=1f(xk)
x : The functionfis called theintegrand. The variablexis called thedummy variable of integra- tion. We sayais thelower boundandbis theupper bound. The symboldxis themeasure.

We also dene fora < b

Z a b f(x)dx=Z b a f(x)dxandZ a a f(x)dx= 0: Thesigned-areabounded byy=f(x) foraxbis dened to beRb af(x)dx.The integral above is known as the Riemann-integral. Other denitions are possible 5.

Iffis continuous on the intervals (a1;a2);(a2;a3);:::(ak;ak+1) and each discontinuity is a nite-jump

discontinuity then the denite integral offon [a1;ak+1] is dened to be the sum of the integrals: Z ak+1 a

1f(x)dx=kX

j=1Z aj+1 a jf(x)dx:

Technically this leaves something out since we have only carefully dened integration over a closed interval

and here we need the concept of integration over a half-open or open interval. To be careful one has the

limit of the end points tending to the points of discontinuity. We discuss this further in Calculus II when we

studyimproper integrals

In the graph ofy=f(x) below I have shaded the positive signed-area green and the negative signed-area

blue for the region4x3. The total signed-area is calculated by the denite integral and can also be

found from the sum of the three regions: 11:61:3 + 8:7 = 19:0 =R3 4f(x)dx.5

the Riemann-Stieltjes integral or Lesbesque are generalizations of this the basic Riemann integral. Riemann-

Stieltjes integral might be covered in some undergraduate analysis courses whereas Lesbesque's measure theory is

typically a graduate analysis topic.

262CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Example 7.2.20.Supposef(x) = sin(x). Set-up the denite integral from[0;]. We chooseR=Rnfor convenience. Note
x==nand the typical sample point isxj=j=n. Thus R n=nX j=1sin(xj)
x=nX j=1sinjn  n )Z  0 sin(x)dx= limn!1n X j=1sinjn  n :

At this point, most of us would get stuck. In order to calculate the limit above we need to nd some identity

to simplify sums such as sin n  + sin2n  +


+ sin(n1)n  = ?:

If you gure it out please show me.

Symmetry can help integrate. Note that by the symmetry of the sine function it is clear that R

0sin(x)dx=R0

sin(x)dxand consequently the signed area bounded byy= sin(x) on [;] is simply zero.7.2.4 properties of the denite integral

As we just observed a particular Riemann integral can be very dicult to calculatedirectlyeven if the

integrand is a relatively simple function. That said, there are a number of intuitive properties for the

denite integral whose proof is easier in general than the preceding specic case.

Proposition 7.2.21.algebraic properties of denite integration.Supposef;gare continuous on [a;b] anda < c < b,2R

(i:)Z b a [f(x) +g(x)]dx=Z b a f(x)dx+Z b a g(x)dx; (ii:)Z b a f(x)dx=Z b a f(x)dx; (iii:)Z b a f(x)dx=Z c a f(x)dx+Z b c f(x)dx:

7.2. AREA PROBLEM263Proof:sincef;gare continuous it followsf+gis likewise continuous hencef;g;f+gare all bounded on

[a;b] and consequently their denite integrals exist (the limit of the Riemann sums must converge to a real

value). Consider then, Z b af(x) +g(x)dx= limn!1 nX j=1 f(xk) +g(xk)
x = lim n!1 nX j=1f(xk)
x+nX j=1g(xk)
x = lim n!1 nX j=1f(xk)
x +limn!1 nX j=1g(xk)
x = Z b a f(x)dx+Z b a g(x)dx

We used the linearity properties of nite sums and the linearity properties of sequential limits in the calcu-

lation above. In the case= 1 we obtain a proof for (i:). In the caseg= 0 we obtain a proof for (ii:). The

proof of (iii:) will require additional thinking. We need to think about a partition of [a;b] and split it into

two partitions, one for [a;c] and the other for [c;b]. Sincea < c < bthe value ofcmust appear somewhere

in the partition: x o=a < x1< x2<


< xjcxj+1<


< xn=a+n
x=b: for somej < n. Notexk=a+k
xand
x=ban fork= 1;2;:::;n. Note that asn! 1the following ratios hold (ifxj=cthen these are exact, however clearlyxj!casn! 1):
x=ban =caj =bcnj

these simply express the fact that the partition of [a;b] has equal length in each region. In what follows the

264CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

x jis the particular point in each partition of [a;b] close to the midpointc: Z b a f(x)dx= limn!1 nX k=1f(xk)
x = lim n!1 jX k=1f(xk)
x+nX k=j+1f(xk)
x = lim j!1 jX k=1f(zk)caj  + lim p!1 pX l=1f(yl)bcp  wherezk=xkandyl=xl+jforjncabaand we have replaced the limit ofn! 1with that of p=nj! 1which is reasonable sincejncabagivesnjnncaba=nbac+aba=nbcbahence n! 1impliesnj! 1asb > candb > aby assumption. Likewise, we replacedn! 1withj! 1 for the rst sum. This substitution is again justied sincec > aandb > athusjncabasuggestsn! 1 impliesj! 1. Finally, denote
y=caj and
z=bcp to obtain Z b a f(x)dx= limj!1 jX k=1f(zk)
z + lim p!1 pX l=1f(yl)
y = Z c a f(z)dz+Z b c f(y)dy = Z c a f(x)dx+Z b c f(x)dx:

This concludes the proof of (iii:).

It's interesting that what is intuitively obvious is not necessarily so intuitive to prove. Another example of

this pattern is the Jordan curve lemma from complex variables. Basically the lemma simply states that you

can divide the plane into two regions, one inside the curve and one outside the curve. The proof isn't typically

oered until the graduate course on topology. It's actually a technically challenging thing to prove precisely.

This is one of the reasons that rigor is so important to mathematics: what is intuitive maybe be wrong.

Historically, appeal to intuition has trapped us for centuries with wrong ideas. However, without intuition

we'd probably not advance much either. My personal belief is that for good mathematics to progress we need

many dierent types of mathematicians working in concert. We need visionaries to forge ahead sometimes

without proof (Edward Witten is probably the most famous example of this type currently) and then we

need careful analytical types to make sure the visionaries are not just going in circles. In this modern age it

is no longer feasible to expect all major progress be made by people like Gauss who both propose the idea

and provide the proof at levels of rigor sucient to convince the whole mathematical community. In any

event, whether you are a math major or not, I hope this course helps you understand what mathematics is

about. By now you should be convinced it's not just about secret formulas and operations on equations.

7.2. AREA PROBLEM265

Proposition 7.2.22.inequalities of denite integration.Supposef;gare continuous on [a;b] andm;M2R, (i:) iff(x)0 for allx2[a;b] thenZ b a f(x)dx0; (ii:) iff(x)g(x) for allx2[a;b] thenZ b a f(x)dxZ b a g(x)dx; (iii:) ifmf(x)Mfor allx2[a;b] thenm(ba)Z b a

f(x)dxM(ba):Proof:sincef;gare continuous we can be sure that the limits dening the denite integrals exist. We need

the existence of the limits in order to apply the limit laws in the arguments that follow. Begin with (i:),

assumef(x)0 and partition [a;b] as usuala=xo;b=xnandxk=a+
x. Sample pointsxkare chosen from each subinterval [xk1;xk]. Consider, for any particularn2Nit is clear that: f(xk)0 and
x=ban >0)nX k=1f(xk)
x0

Consequently,Rn=Pn

k=1f(xk)
x0 for alln2Nhence by comparison property for sequential limits, lim n!1Rnlimn!1(0) = 0 and (i:) follows immediately. To prove (ii:) constructh(x) =f(x)g(x) and notef(x)g(x) for allx2[a;b] impliesh(x) =f(x)g(x)

0 for allx2[a;b]. We apply (i:) to the clearly continuous functionhand obtain:

Z b a h(x)dx0)Z b af(x)g(x)dx0)Z b a f(x)dxZ b a g(x)dx0 and (ii:) clearly follows. Proof of (iii:) follows from observing that iffis bounded bymf(x)Mfor allx2[a;b] then mf(xk)Mfor eachxk2[a;b]. Hence, n X k=1mnX k=1f(xk)nX k=1M:

Butm;M2Rso the summations on the edges are easy:

mnnX k=1f(xk)Mn:

Finally, we can multiply by
x=ban

to obtain mn ban nX k=1f(xk)
xMnban )m(ba)nX k=1f(xk)
xM(ba): Apply the sequential limit squeeze theorem and take the limit asn! 1to nd m(ba)limn!1 nX k=1f(xk)
x M(ba)

266CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

This proves (iii:).

One easy fact to glean from the proof of (iii:) is the following:

Corollary 7.2.23.integral of a constant. Letm2R,Z

b a

mdx=m(ba):Given that the denite integral was constructed to calculate area this result should not be surprising. Note

0ymforaxbdescribes a rectangle of widthbaand heightm.

Problems

Problem 7.2.1.hope to add problems in the future..

7.3. FUNDAMENTAL THEOREM OF CALCULUS267

7.3 fundamental theorem of calculus

In the preceding section we detailed a careful procedure for calculating the signed area betweeny=f(x)

andy= 0 foraxb. Unless the function happened to be very simple or enjoyed some obvious

symmetry it was dicult to actually calculate the area. We can write the limits but we typically have no

way of simplifying the sum to evaluate the limit. In this section we will prove the Fundamental Theorem

of Calculus (FTC) which amazingly shows us how to calculate signed-areas without explicit simplication

of the Riemann sum or evaluation of the limit. I begin by studying area functions. I show how the FTC

part I is seen naturally for both the rectangular and triangular area functions. These two simple cases are

discussed to help motivate why we would even expect to nd such a thing as the FTC. Then we regurgitate

the standard arguments found in almost every elementary calculus text these days to prove "FTC part I"

and "FTC part II". Finally, I oer a constructive proof of FTC part II and I argue why FTC part I follows

intuitively.

7.3.1 area functions and FTC part I

In that discussion the endpointsaandbwere given and xed in place. We now shift gears a bit. We study

area functionsin this section. The idea of an area function is simply this: if we are given a functionf

then we can dene an area function forfonce we pick some base pointa. ThenA(x) will be dened to be the signed-area bounded byy=f(t) foratx. I usetin the place ofxsince we wish to usexin a less general sense in the pictures that follow here. Denition 7.3.1.area function.Givenfand a pointawe dene thearea functionoffrelative toaas follows:

A(x) =Z

x a f(t)dt:

We say thatA(x) is the signed-area bounded byfon [a;x].We would like to look for patterns about area functions. We've seen already that direct calculation is dif-

cult. However, we know two examples from geometry where the area is easily calculated without need of

calculus. Area function of rectangle:letf(t) =cthen the area bounded betweent= 0 andt=xis simply length (x) times height (c). By geometry we have thatA(x) =Rx

0cdt=cx, see the picture below:

268CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

If we positioned the rectangle atatxthen length becomes (xa) and the height is still (c). Therefore,

by geometry,A(x) =Rx acdt=c(xa) =cxca:Again, see the picture below where I have pictured a particularxbut I have graphedy=A(t) for manytbesidesx. You can imagine other choices ofxand you

should nd the area function agrees with the area under the curve.Area function of triangle:I begin with a triangle formed at the origin with thet-axis and the liney=mt

andt=x. For a particularx, we have base lengthxand heighty=mxthus the area of the triangle is given by geometry:A(x) =Rx

0mtdt=12

mx2:I picture the function (y=mtin red) as well as the area function (y=12

mt2in green) in the picture below:We calculate the area bounded byy=mtforatxby subtracting the area of the small triangle from

0tafrom the area of the larger triangle 0txas pictured below. Thus from geometry we nd

A(x) =Rx

amtdt=12 mx212 ma2:

7.3. FUNDAMENTAL THEOREM OF CALCULUS269The area under a parabola could also be calculate without use of further theory. We could work out from

the special summation formulas that the area function fory=t2foratxis given byA(x) =Rx at2dt= 13 x313 a3(I might ask you to show this in a homework). I suspect this is beyond the scope of constructive geometry (compass/straight-edge and paper). We should notice a pattern:

1.A(x) =Rx

acdt=cxhasdAdx =c.

2.A(x) =Rx

amtdt=12 mx2hasdAdx =mx.

3.A(x) =Rx

at2dt=13 x3hasdAdx =x2.

We suspect that ifA(x) =Rx

af(t)dtthendAdx =f(x). Let's examine an intuitive graphical argument for

why this is true for an arbitrary function:Formally,dA=A(x+dx)A(x) =f(x)dxhencedA=dx=f(x). This proof made sense to you (if it

did) because you believe in Leibniz' notation. We should oer a rigorous proof since this is one of the most

important theorems in all of calculus.

270CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Theorem 7.3.2.Fundamental Theorems of Calculus part I (FTC I).Supposefis continuous on [a;b] andx2[a;b] then,

ddx Z x a f(t)dt=f(x):Proof:letA(x) =Rx af(t)dtand note that

A(x+h) =Z

x+h a f(t)dt=Z x a f(t)dt+Z x+h x f(t)dt=A(x) +Z x+h x f(t)dt Therefore, the dierence quotient for the area function is simply as follows:

A(x+h)A(x)h

=1h Z x+h x f(t)dt However, note that by continuity offwe can nd bounds forfonJ= [x;x+h] (ifh >0) orJ= [x+h;x] (ifh <0). By the extreme value theorem, there existu;v2Jsuch thatf(u)f(x)f(v) for allx2J. Therefore, ifh >0, we can apply the inequality properties of denite integrals and nd (x+hx)f(u)Z x+h x f(t)dt(x+hx)f(v))f(u)1h Z x+h x f(t)dtf(v) Ifh <0 then dividing byhreverses the inequalities hencef(v)1h R x+h xf(t)dtf(u). Finally, observe that lim h!0u=xand limh!0v=x. Therefore, by continuity off, limh!0f(u) =f(x) and limh!0f(v) =f(x).

Remember,f(u)1h

R x+h xf(t)dtf(v) and apply the squeeze theorem to deduce: lim h!01h Z x+h x f(t)dt=f(x)

Consequently,

lim h!0A(x+h)A(x)h =f(x) Which, by denition of the derivative forA, givesdAdx =f(x).

The FTC part I is hardly a solution to the area problem. It's just a curious formula. The FTC part II takes

this curious formula and makes it useful. It is true there are a few functions dened as area functions hence

the dierentiation in the FTC I is physically interesting. However, such problems are fairly rare. You can

read about the Fresnel function in the text.

7.3. FUNDAMENTAL THEOREM OF CALCULUS271

Remark 7.3.3.a method to derive antiderivatives without guessing.Notice that the FTC I also gives us a method to calculate antiderivatives without guessing. But, I can

only derive a few very simple antiderivatives. For example, here is aderivationof the antiderivative off(x) = 3. I calculate thatR3dx= 3x+cwithout guessing:7.3.2 FTC part II, the standard arguments

The fact that

ddx R x af(t)dt=f(x) is just half of what we observed in our examination of the rectangular and triangular area functions. If the area was measured away from the origin on some regionatxthen we can observe another pattern:the area was given by the dierence of the antiderivative of the integrand at the end points 1.Rx acdt=cxca 2. Rx amtdt=12 mx212 ma2

This suggests the following theorem may be true:

Theorem 7.3.4.Fundamental Theorems of Calculus part II (FTC II).Supposefis continuous on [a;b] and has antiderivativeFthen

Z b a f(x)dx=F(b)F(a):Proof:consider the area function based ata:A(x) =Rx af(t)dt. The FTC I says thatAis an antiderivative off. SinceFis given to be another antiderivative we know thatF0(x) =A0(x) =f(x) which meansFand Adier by at most a constantc2R:F(x) =A(x) +c. SinceFandAare dierentiable on [a;b] it follows they are also continuous on [a;b] hence,

F(a) = lim

x!a+F(x) = lim x!a+[A(x) +c] =A(a) +c=Z a a f(t)dt+c=c and

F(b) = lim

x!bF(x) = lim x!b[A(x) +c] =A(b) +c=Z b a f(t)dt+c:

272CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Hence,F(b)F(a) =Rb

af(t)dt+cc=Rb af(t)dt. Of course,tis just the dummy variable of integration so we can change it toxat this point to complete the proof of the FTC part II. Example 7.3.5.We return to Example 7.2.20 where we were stuck due to an incalculable summation.

We wish to calculateR

0sin(x)dx. Observe thatF(x) =cos(x)hasF0(x) = sin(x)hence this is a valid

antiderivative for the given integrandsin(x). Apply the FTC part II to nd the area: Z  0 sin(x)dx=F()F(0) =cos() + cos(0) = 2: Obviously this is much easier than calculation from the denition of the Riemann integral.

Denition 7.3.6.evaluation notation.We dene the symbols below to denote evaluation of an expression:

F(x)b

a =F(b)F(a) In this notation the FTC part II is written as follows: Z b a f(x)dx=F(x)b a =F(b)F(a):7.3.3 FTC part II an intuitive constructive proof

Let me restate the theorem to begin:FTC II:Supposefis continuous on [a;b] and has antiderivativeFthen

Z b a f(x)dx=F(b)F(a):Proof:We seek to calculateRb af(x)dx. Use the usual partition for then-th Riemann sum offon [a;b]; x o=a;x1=a+
x;:::;xn=bwhere
x=ban . Suppose thatfhas an antiderivativeFon [a;b]. Recall the Mean Value Theorem fory=F(x) on the interval [xo;x1] tells us that there existsx12[xo;x1] such that F

0(x1) =F(x1)F(xo)x

1xo=F(x1)F(xo)
x

Notice that this tells us thatF0(x1)
x=F(x1)F(xo). But,F0(x) =f(x) so we have found that f(x1)
x=F(x1)F(xo). In other words, the area undery=f(x) forxoxx1is well approximated by

the dierence in the antiderivative at the endpoints. Thus we choose the sample points for then-th Riemann

sum by applying the MVT on each subinterval to selectxjsuch thatf(xj)
x=F(xj)F(xj1). With

7.3. FUNDAMENTAL THEOREM OF CALCULUS273

this construction in mind calculate: Z b a f(x)dx= limn!1 nX j=1f(xj)
x = lim n!1 nX j=1

F(xj)F(xj1)

= lim n!1F(x1)F(xo) +F(x2)F(x1) +


+F(xn)F(xn1)
= lim n!1F(xn)F(xo)
= lim n!1F(b)F(a)
=F(b)F(a):

This result clearly extends to piecewise continuous functions which have only nite jump discontinuities.

We can apply the FTC to each piece and take the sum of those results. This Theorem is amazing. We can

calculate the area under a curve based on the values of the antiderivative at the endpoints. Think about

that, ifa= 1 andb= 3 thenR3

1f(x)dxdepends only onF(3) andF(1). Doesn't it seem intuitively likely

that what valuef(2) takes should matter as well? Why don't we have to care aboutF(2) ? The values of

the function atx= 2 certainly went into the calculation of the area, if we calculate a left sum we would

need to take values of the function between the endpoints. The cancellation that occurs in the proof is the

root of why my naive intuition is bogus. Next, let me show you how to derive FTC I from FTC II

6. We have just proved that

Z b a f(t)dt=F(b)F(a): Supposeb=xand consider dierentiating with respect tox, ddx Z x a f(t)dt=ddx [F(x)F(a)] =dFdx =f(x): thus we obtain FTC I simply by dierentiating FTC II. Moreover, we can obtain a more general result without doing much extra work:

Theorem 7.3.7.dierentiation of integral with variable bounds. (FTC III for fun)Supposeu;vare dierentiable functions ofxandfis continuous where it is integrated,

ddx Z v(x) u(x)f(t)dt=f(v(x))dvdx f(u(x))dudx :Proof:letfhave antiderivativeFand apply FTC II at eachxto obtain: Z v(x) u(x)f(t)dt=F(v(x))F(u(x))6

note I didn't need to use FTC I in the argument for the FTC II in this section, instead I needed only assume

that there existed an antiderivative for the given integrand

274CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

now dierentiate with respect toxand apply the chain-rule, ddx Z v(x) u(x)f(t)dt=dFdx (u(x))dvdx dFdu (u(x))dudx But, dFdx =f(x) henceddx R v(x) u(x)f(t)dt=f(u(x))dvdx f(u(x))dudx . The examples based on FTC III are embarrassingly simple once you understand what's happening.

Example 7.3.8.

ddx Z x 3 cos(pt)dt= cos(px)d(px)dx cos(p3) d(3)dx =cos( px)12 px :Example 7.3.9. ddx Z x3 e xtanh(t2)dt= tanh((x3)2)d(x3)dx tanh((ex)2)d(ex)dx =3x2tanh(x6)extanh(e2x):Example 7.3.10.The functionSiis dened bySi(x) =Rx

0sin(t)t

dtforx6= 0andSi(0) = 0. This function arises in Electrical Engineering in the study of optics. ddx (Si(x)) =ddx Z x

0sin(t)t

dt=sin(x)x :Example 7.3.11. ddx Z x2+3 sin(x)ptdt=px

2+ 3d(x2+ 3)dx

psin(x)d(sin(x))dx =2xpx

2+ 3cos(x)psin(x):Example 7.3.12.Supposefis continuous onR. It follows thatfhas an antiderivative hence the FTC III

applies: ddx Z x x

2f(u)du=f(x)d(x)dx

f(x2)d(x2)dx =f(x)2xf(x2):Problems Problem 7.3.1.hope to add problems in the future..

7.3. FUNDAMENTAL THEOREM OF CALCULUS275

.

276CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

7.4 denite integration

Example 7.4.1.

Z1 0

2xdx=1ln(2)

2x1

0 =1ln(2) (2120) =1 ln(2) :Example 7.4.2.Leta;bbe constants, Z b a sinh(t)dt= cosh(t)b a =cosh(b)cosh(a):Example 7.4.3. Z 2 4dxx = lnjxj2

4= lnj 2j lnj 4j= ln(2)ln(4) =ln(1=2):If we had neglected the absolute value function in the antiderivative then we would have obtained an incorrect

result. The absolute value bars are important for this integral. Note the answer is negative here because

y= 1=xis under thex-axis in the region4x 2.

Example 7.4.4.

Z9

1dxp5x=1p5

Z 9 1dxpx =2pxp5  9 1 =2p9p5 2p1p5 =4p5 :Example 7.4.5.Letn >0and consider, Z ln(n+1)

ln(n)exdx=eln(n+1)eln(n)=n+ 1n=1:This is an interesting result. I've graphed a few examples of it below. Notice how asnincreases the distance

betweenln(n)andln(n+ 1)decreases, yet the exponential increases such that the bounded area still works

out to one-unit.

7.4. DEFINITE INTEGRATION277

7.4.1 area vs. signed-area

Example 7.4.6.Calculate the signed-area bounded byy= 3x23x6for0x2. Z 2 0 (3x23x6)dx= (x332 x26x)2 0 = 832 (4)12 = 818 =10:Here's an illustration of the calculation (the blue part):

The green area is calculated by

Z 4 2 (3x23x6)dx= (x332 x26x)4 2 = (6432

(16)24) + 10 = 6448 + 10 =26:Example 7.4.7.If we wanted to calculate the area bounded byy=f(x) = 3x23x6andy= 0for

0x4then we need to also count negative-signed-area as positive. This is nicely summarized by stating

we should integrate the absolute value of the function to obtain the area bounded between the function and the

x-axis. Generally analyzing an absolute value of a function takes some work, but given the previous example

it is clear how to break up the positive and negative cases: Z 4 0 j3x23x6jdx=Z 2 0 j3x23x6jdx+Z 4 2 j3x23x6jdx = Z 2 0 [(3x23x6)]dx+Z 4 2 (3x23x6)dx = 10 + 26 =36:Here's a picture of the function we just integrated. You can see how the absolute value ips the negative part of the original function up above thex-axis.

278CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Remark 7.4.8.absolute values and areas.To calculate the area bounded byy=f(x) foraxbwe may calculate

Area=Z

b a jf(x)jdx:Example 7.4.9.Calculate the area bounded byy= cos(x)on0x52 . Z 3 0 jcos(x)jdx=Z 2

0cos(x)dxZ

32 2 cos(x)dx+Z 52 32 cos(x)dx = sin(x)2

0sin(x)32

2 + sin(x)52 32 = (sin( 2 )sin(0))(sin(32 )sin(2 )) + (sin(52 sin(32 )) =5:7.4.2 average of a function

To calculate the average of nitely many things we can just add all the items together then divide by the

number of items. If you draw a bar chart and nd the area of all the bars and then divide by the number

of bars then that gives the average. A functionf(x) takes on innitely many values on a closed interval

so we cannot just add the values, however, we can calculate the area and divide by the length. This is the

continuous extension of the averaging concept:

Denition 7.4.10.average of a function over a closed interval.The average value offon [a;b] is dened by

f avg=1baZ b a f(x)dx:Example 7.4.11.Supposef(x) = 4x3. Find the average offon[0;2]. f avg=12 Z 2 0

4x3dx=12

x42 0 =8:Example 7.4.12.Supposef(x) = sin(x). Find the average offon[0;2]. f avg=12Z 2 0 sin(x)dx=12cos(x)2 0 =0:

7.4. DEFINITE INTEGRATION279

Example 7.4.13.In the case of constant accelerationa=gwe calculated thatv(t) =vogtwherevo;g were constants. Let's calculate the average velocity over some time interval[t1;t2], v avg=1t 2t1Z t2 t

1(vogt)dt

= 1t

2t1[votg2

t2]t 2 t 1 = 1t 2t1 v o(t2t1)g2 t22t21
 = 1t 2t1 v ot2g2 t22vot1+g2 t21 = y(t2)y(t1)t 2t1 where I have used a little imagination and a recollection thaty(t) =yo+votg2 t2. The result is comforting, we nd the average velocity is the average of the average velocity function.

There is a better way to calculate the last example. It will provide the rst example of the next topic.

7.4.3 net-change theorem

Combining FTC I and FTC II we nd a very useful result: the net-change theorem.

Theorem 7.4.14.net change theorem.Z

b adfdt dt=f(b)f(a):Example 7.4.15.Letv(t)be the instantaneous velocity wherev(t) =dydt then we can calculate the average velocity over some time interval[t1;t2], v avg=1t 2t1Z t2 t

1v(t)dt

= 1t 2t1Z t2 t 1dydt dt = 1t

2t1y(t2)y(t1)

= y(t2)y(t1)t 2t1: Notice we didn't even need to know the details of the velocity function.

We'll see more examples of like this one when we discuss innitesimal methods in the next Chapter. Your

text also has many more examples.

Problems

Problem 7.4.1.hope to add problems in the future..

280CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

.

7.5. U-SUBSTITUTION281

7.5 u-substitution

The integrations we have done up to this point have been elementary. Basically all we have used is lin-

earity of integration and our basic knowledge of dierentiation. We made educated guesses as to what the

antiderivative was for a certain class of rather special functions. Integration requires that you look ahead

to the answer before you get there. For example,Rsin(x)dx. To reason this out we think about our basic

derivatives, we note that the derivative of cos(x) givessin(x) so we need to multiply our guess by -1 to

x it. We conclude thatRsin(x)dx=cos(x) +c. The logic of this is essentially educated guessing. You

might be a little concerned at this point. Is that all we can do? Just guess? Well, no. There is more. But,

those basic guesses remain, They form the basis for all elementary integration theory. The new idea we look at in this section is called "u-substitution". It amounts to the reverse chain

rule. The goal of a properly posed u-substitution is to change the given integral to a new integral which

is elementary. Typically we go from an integration inxwhich seems incalculable to a new integration in

uwhich is elementary. For the most part we will make direct substitutions, these have the formu=g(x)

for some functionghowever, this is not strictly speaking the only sort of substitution that can be made.

Implicitly dened substitutions such asx=f() play a critical role in many interesting integrals, we will

deal with those more subtle integrations in a later chapter when we discuss trigonometric substitution.

Finally, I should emphasize that when we do a u-substitution we must be careful to convert each and

every part of the integral to the new variable. This includes both the integrand(f(x)) and the measure(dx)

in an indenite integralRf(x)dx. Or the integrand(f(x)), measure(dx) and upper and lower boundsa;b in a denite integralRb af(x)dx. I will provide a proof of the method at the conclusion of the section for a change of pace. Examples rst this time.

7.5.1u-substitution in indenite integrals

Example 7.5.1.

Z xe x2dx=Z xe udu2xletu=x2,dudx = 2xanddx=du2x= 12 Z e udusee how all thex's cancelled, this has to happen. = 12 eu+cnot done yet. =1 2 ex2+cdierentiate to check if in doubt.

Example 7.5.2.Leta;bbe constants. Ifa6= 0then,

Z (ax+b)13dx=Z u

13dualetu=ax+b,dudx

=aanddx=dua =

114au14+c

=1

14a(ax+b)14+c:Ifa= 0then clearlyR(ax+b)13dx=Rb13dx=b

13x+c:

282CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Example 7.5.3.

Z 5 x3 dx=Z 5 u(3du)letu=x3 ,dudx =13 anddx= 3du=

3ln(5)

5u+c =3 ln(5) 5x3 +c:Example 7.5.4. Z tan(x)dx=Zsin(x)cos(x)dx =

Zduuletu= cos(x),dudx

=sin(x)andsin(x)dx=du=ln(juj) +c =ln(jcos(x)j) +c:Notice thatlnjcos(x)j= lnjcos(x)j1= lnjsec(x)jhenceRtan(x)dx= lnjsec(x)j+c.

Example 7.5.5.

Z2x1 +x2dx=Zduuletu= 1 +x2,dudx

= 2xand2xdx=du= ln(juj) +c

=ln(1 +x2) +c:Notice thatx2+ 1>0for allx2Rthusjx2+ 1j=x2+ 1. We should only drop the absolute value bars if

we have good reason.

Example 7.5.6.

Z

3p13xdx=Z

3pu du3letu= 13x,dudx =3anddx=du3= 34
u43 +c =3 4 (13x)43 :Example 7.5.7.

Zdxx+b=Zduuletu=x+bthusdu=dx= lnjuj+c

=lnjx+bj+c:

7.5. U-SUBSTITUTION283

Example 7.5.8.supposex >0.

Z x2dxpx

2x4=Zx2dxx

p1x2 =Zxdxp1x2 =Zdu2 puletu= 1x2thusdu=2 =xdx= 12 2pu+c =

p1x2+c:Notice, we start to see examples where educated guessing alone probably wouldn't have solved it. Of course

there are numerous software programs to assist with integration these days but unless you do a bunch of

these at some point in your life you'll never really understand what the computer is doing.

Example 7.5.9.supposex >0.

Z ln(x)dxx =Z uduletu= ln(x)thusdu=dx=x= 12 u2+c =1 2 ln(x)2+c:Example 7.5.10. Z sin(3)d=Z sin(u)du3letu= 3thusd=du3 = 13 cos(u) +c =13 cos(3) +c:Example 7.5.11. Z sin1(z)p1z2dz=Z uduletu= sin1(z)thusdu=dzp1z2= 12 u2+c =1 2 [sin1(z)]2+c:

284CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Example 7.5.12.

Z tcos(t2+)dt=12 Z cos(u)duletu=t2+thustdt=du2 = 12 sin(u) +c =1 2

sin(t2+) +c:If you understand the example below then you will be able to integrate any odd power of sine or cosine.

Example 7.5.13.

Z sin

3(x)dx=Z

sin

2(x)sin(x)dx

= Z (1cos2(x))sin(x)dx = Z (1u2)(du)letu= cos(x)thusdu=sin(x)dx= Z (u21)du = 13 u3u+c =1 3 cos3(x)cos(x) +c:Example 7.5.14.supposea6= 0 Zdxx

2+a2=1a

2Z dxx 2a 2+ 1 = 1a 2Z aduu

2+ 1letu=xa

thusadu=dx= 1a tan1(u) +c =1 a tan1xa  +c:Example 7.5.15.supposea6= 0 Z cos(aex+ 3)exdx=1a Z cos(u)duletu=aex+ 3thusdu=a=exdx= 1a sin(u) +c =1 a sin(aex+ 3) +c:

7.5. U-SUBSTITUTION285

Example 7.5.16.

Z sin

2()d=Z12



1sin2()

dby trigonmetry. = 12 Z d12 Z cos(2)d = 2 14

sin(2) +c:In the preceding example I omitted au-substitution because it was fairly obvious. In the next example I

demonstrate a notation which allows you to performu-substitution without even stating explicitly theu.

You will not nd this notation in many American textbooks.

Example 7.5.17.

Z 4sinh

2(x)dx= 4Z

12 exex
2 dxby denition ofsinh(x). = Z  (ex)22exex+ (ex)2 dx = Z  e

2x2 +e2x

dx = Z e

2xdx2Z

dx+Z e 2xdx = 12 Z e

2xd(2x)2x12

Z e 2xd(2x) = 12 e2x2x12 e2x+c: =sinh(2x)2x+c:Interesting, if you trust my calculation then we may deduce 4sinh

2(x) =ddx

[sinh(2x)2x] = 2cosh(2x)2 thussinh2(x) =12 [cosh(2x)1].

7.5.2u-substitution in denite integrals

There are two ways to do these. I expect you understand both methods. 1. Fi ndthe an tiderivativevia u-substitution and then use the FTC to evaluate in terms of the given upper and lower bounds inx. (see Example 7.5.18 below) 2.

Do the u-substitution and change the bounds all at once, this means you will use the FTC and evaluate

the upper and lower bounds inu. (see Example 7.5.19 below)

I will deduct points if you write things like a denite integral is equal to an indenite integral ( just leave

o the bounds during the u-substitution). The notation is not decorative, it is necessary and important to

use correct notation.

286CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

Example 7.5.18.We previously calculated thatRtcos(t2+)dt=12 sin(t2+)+c:We can use this together with the FTC to calculate the following denite integral: Z p 2

0tcos(t2+)dt=12

sin(t2+)p 2 0 = 12 sin(2 +)12 sin() =12

:This illustrates method (1.) we nd the antiderivative o to the side then calculate the integral using the

FTC in thex-variable. Well, thet-variable here. This is a two-step process. In the next example I'll work

the same integral using method (2.). In contrast, that is a one-step process but the extra step is that you

need to change the bounds in that scheme. Generally, some problems are easier with both methods. Also,

sometimes you may be faced with an abstract question which demands you understand method 2.).

Example 7.5.19.

Zp 2

0tcos(t2+)dt=12

Z 32 cos(u)duletu=t2+thustdt=du2 = 12 sin(u)32 alsou2
=32 andu(0) == 12 sin(32 )12 sin() =12 :Example 7.5.20. Z 92

42sin(px)dxpx

=Z 3

2sin(u)(2du)letu=pxthus2du=dxpx

=2cos(u)3

2alsou(92) =p92= 3andu(42) =p42= 2=2cos(3) + 2cos(2)

=4:7.5.3 theory ofu-substitution

In the past 20 examples we've seen how the technique ofu-substitution works. To summarize, you take an

integrand and measure in terms ofx(sayg(f(x))dx) and propose a new variableu=f(x) for some function f. Then we dierentiatedudx =f0(x) and solve fordx=duf

0(x)which gives us

Z g(f(x))dx=Z g(u)duf 0(x) and if our choice ofuis well thought out then the expressiong(u)f

0(x)can be simplied into a nice elementary

integrable functionh(u) (meaningRh(u)duwas on our list of elementary integrals). In a nutshell, that is

what we did in each example. Let's me raise a couple questions to criticize the method:

7.5. U-SUBSTITUTION287

1. what in the w orlddo I mean b ydx=duf

0(x)? This sort of division is not rigorous.

2.

what if f0(x) = 0? Especially if we were doing an integration with bounds, is it permissible to have a

point in the domain of integration where the substitution seems to indicate division by zero? Question (1.) is not too hard to answer. Let me propose the formal result as a theorem.

Theorem 7.5.21.change of variables in integration.Supposegis continuous on the connected intervalJwith endpointsf(a) andf(b) andfis dieren-

tiable ona;bthen 1. Z g(u)du u=f(x)=Z g(f(x))dfdx dx 2. Zf(b) f(a)g(u)du=Z b a g(f(x))dfdx dx:Proof:Note thatgcontinuous indicates the existence of an antiderivativeGonJ. Letu=f(x) and apply the chain-rule to dierentiateG(u), ddx [G(u)] =G0(u)dudx =g(u)dfdx =g(f(x))dfdx At this stage we have already proved the indenite integral substitution rule:

G(f(x)) =

Z g(u)du u=f(x)=Z g(f(x))dfdx dx=H(x) +c: Use the result above and FTC II to see why (2:) is true: Z b a g(f(x))dfdx dx=H(b)H(a) =G(f(b))G(f(a)) =Z f(b) f(a)g(u)du:

I assumed continuity for simplicity of argument. One could prove a more general result for piecewise con-

tinuous functions. Furthermore, note we never really divided byf0(x) thusf0(x) = 0 does not rule out the

applicability of this theorem. Example 7.5.22.Consider the following problem: calculate Z 2 0 esin(x)cos(x)dx:

In this case we should identifyu=f(x) = sin(x)andg(u) =eu. Clearly the hypotheses of the theorem above

are met. Moreover,f(0) = sin(0) = 0andf(2) = sin(2) = 0hence Z 2 0 esin(x)cos(x)dx=Z 2 0 esin(x)d(sin(x))dx dx=Z 0 0 eudu= 0:

288CHAPTER 7. ANTIDERIVATIVES AND THE AREA PROBLEM

For whatever reason, using the notation above seems unnatural to most people so we instead think about

substituting formulas withuinto the integrand. Same calculation, but this time with our usual approach:

Z 2 0 esin(x)cos(x)dx=Z 0 0 eucos(x)ducos(x)letu= sin(x)thusdx=ducos(x)= Z 0 0

eudualsou(0) = sin(0)andu(2) = sin(2).=0:The apparent division by zero was just a sloppy way of communicating application of the theorem for variable

change. This phenomenon of the bounds collapsing to a point will only occur if dudx = 0 somewhere alongaxb.

Otherwise,

dudx

6= 0 henceuis strictly monotonic on [a;b] hence eitheru(a)< u(b) oru(b)> u(a).

Remark 7.5.23.geometric meaning ofu-substitution.The geometric meaning of substitution is an interesting topic that this current version of my notes

does not address. You are free to read the text on that topic and it is probable I will spend a few

minutes of lecture contemplating the geometry ofu-substitution.Well, that'