Lesson 25: Areas and Distances; The Definite Integral

. . . . . .

Section5.1–5.2AreasandDistancesTheDefiniteIntegral

V63.0121.034, CalculusI

November30, 2009

Announcements

I Quiz5thisweekinrecitationon4.1–4.4, 4.7I FinalExam, December18, 2:00–3:50pm

. . . . . .

Outline

AreathroughtheCenturiesEuclidArchimedesCavalieri

GeneralizingCavalieri’smethodAnalogies

DistancesOtherapplications

Thedefiniteintegralasalimit

Propertiesoftheintegral

. . . . . .

EasyAreas: Rectangle

DefinitionThe area ofarectanglewithdimensions ℓ and w istheproductA = ℓw.

..ℓ

.w

Itmayseemstrangethatthisisadefinitionandnotatheorembutwehavetostartsomewhere.

. . . . . .

EasyAreas: Parallelogram

Bycuttingandpasting, aparallelogramcanbemadeintoarectangle.

..b

.b

.h

SoA = bh

. . . . . .



..b

.b

.h

SoA = bh

. . . . . .



.

.b .b

.h

SoA = bh

. . . . . .



.

.b

.b

.h

SoA = bh

. . . . . .



.

.b

.b

.h

SoA = bh

. . . . . .

EasyAreas: Triangle

Bycopyingandpasting, atrianglecanbemadeintoaparallelogram.

..b

.h

SoA =

12bh

. . . . . .

EasyAreas: Triangle


..b

.h

SoA =

12bh

. . . . . .

EasyAreas: Triangle


..b

.h

SoA =

12bh

. . . . . .

EasyAreas: Polygons

Anypolygoncanbetriangulated, soitsareacanbefoundbysummingtheareasofthetriangles:

.

. . . . . .

HardAreas: CurvedRegions

.

???

. . . . . .

Meetthemathematician: Archimedes

I Greek(Syracuse), 287BC –212BC (afterEuclid)

I GeometerI Weaponsengineer

. . . . . .




. . . . . .




. . . . . .

.

Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.

A =

1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·

. . . . . .

.

.1


A = 1

+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·

. . . . . .

.

.1.18 .18


A = 1+ 2 · 18

+ 4 · 164

+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·

. . . . . .

.

.1.18 .18

.164 .164

.164 .164


A = 1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·

. . . . . .

.

.1.18 .18

.164 .164

.164 .164


A = 1+ 2 · 18+ 4 · 1

64+ · · ·

= 1+14+

116

+ · · ·+ 14n

+ · · ·

. . . . . .

Wewouldthenneedtoknowthevalueoftheseries

1+14+

116

+ · · ·+ 14n

+ · · ·

Butforanynumber r andanypositiveinteger n,

(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43

as n → ∞.

. . . . . .


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43

as n → ∞.

. . . . . .


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4

→ 13/4

=43

as n → ∞.

. . . . . .


1+14+

116

+ · · ·+ 14n

+ · · ·


(1− r)(1+ r+ · · ·+ rn) = 1− rn+1

So

1+ r+ · · ·+ rn =1− rn+1

1− r

Therefore

1+14+

116

+ · · ·+ 14n

=1− (1/4)n+1

1− 1/4→ 1

3/4=

43

as n → ∞.

. . . . . .

Cavalieri

I Italian,1598–1647

I Revisitedtheareaproblemwithadifferentperspective

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:

L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.12


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.13

.

.23


L2 =18

L3 =

127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.13

.

.23


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.14

.

.24

.

.34


L2 =18

L3 =127

+427

=527

L4 =

164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.14

.

.24

.

.34


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.15

.

.25

.

.35

.

.45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =

1125

+4

125+

9125

+16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1

.

.15

.

.25

.

.35

.

.45


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1

125+

4125

+9

125+

16125

=30125

Ln =?

. . . . . .

Cavalieri’smethod

.

.y = x2

..0

..1.

.


L2 =18

L3 =127

+427

=527

L4 =164

+464

+964

=1464

L5 =1

125+

4125

+9

125+

16125

=30125

Ln =?

. . . . . .

Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth

1n.

Therectangleoverthe ithintervalandundertheparabolahasarea

1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

TheArabsknewthat

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

TheArabsknewthat

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

TheArabsknewthat

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

TheArabsknewthat

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3

→ 13

as n → ∞.

. . . . . .


1n.


1n·(i− 1n

)2

=(i− 1)2

n3.

So

Ln =1n3

+22

n3+ · · ·+ (n− 1)2

n3=

1+ 22 + 32 + · · ·+ (n− 1)2

n3

TheArabsknewthat

1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)

6

So

Ln =n(n− 1)(2n− 1)

6n3→ 1

3as n → ∞.

. . . . . .

Cavalieri’smethodfordifferentfunctions

Trythesametrickwith f(x) = x3. Wehave

Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)

=1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4

Theformulaoutofthehatis

1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2

So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .



Ln =1n· f(1n

)+

1n· f(2n

)+ · · ·+ 1

n· f(n− 1n

)=

1n· 1n3

+1n· 2

3

n3+ · · ·+ 1

n· (n− 1)3

n3

=1+ 23 + 33 + · · ·+ (n− 1)3

n4


1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)

]2So

Ln =n2(n− 1)2

4n4→ 1

4as n → ∞.

. . . . . .

Cavalieri’smethodwithdifferentheights

.

Rn =1n· 1

3

n3+

1n· 2

3

n3+ · · ·+ 1

n· n

3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1n4

[12n(n+ 1)

]2=

n2(n+ 1)2

4n4→ 1

4

as n → ∞.

Soeventhoughtherectanglesoverlap, westillgetthesameanswer.

. . . . . .

Cavalieri’smethodwithdifferentheights

.

Rn =1n· 1

3

n3+

1n· 2

3

n3+ · · ·+ 1

n· n

3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1n4

[12n(n+ 1)

]2=

n2(n+ 1)2

4n4→ 1

4

as n → ∞.Soeventhoughtherectanglesoverlap, westillgetthesameanswer.

. . . . . .

Outline






. . . . . .

Cavalieri’smethodingeneralLet f beapositivefunctiondefinedontheinterval [a,b]. Wewanttofindtheareabetween x = a, x = b, y = 0, and y = f(x).Foreachpositiveinteger n, divideuptheintervalinto n pieces.

Then ∆x =b− an

. Foreach i between 1 and n, let xi bethe nth

stepbetween a and b. So

..a .b. . . . . . ..x0 .x1 .x2 .xi.xn−1.xn

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

· · · · · ·

xi = a+ i · b− an

· · · · · ·

xn = a+ n · b− an

= b

. . . . . .

FormingRiemannsums

Wehavemanychoicesofhowtoapproximatethearea:

Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆x

Rn = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x

Mn = f(x0 + x1

2

)∆x+ f

(x1 + x2

2

)∆x+ · · ·+ f

(xn−1 + xn

2

)∆x

Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x

=n∑

i=1

f(ci)∆x

. . . . . .

FormingRiemannsums

Wehavemanychoicesofhowtoapproximatethearea:

Ln = f(x0)∆x+ f(x1)∆x+ · · ·+ f(xn−1)∆x

Rn = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x

Mn = f(x0 + x1

2

)∆x+ f

(x1 + x2

2

)∆x+ · · ·+ f

(xn−1 + xn

2

)∆x

Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x

=n∑

i=1

f(ci)∆x

. . . . . .

TheoremoftheDay

TheoremIf f isacontinuousfunctionon [a,b] orhasfinitelymanyjumpdiscontinuities, then

limn→∞

Sn = limn→∞

{f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x}

existsandisthesamevaluenomatterwhatchoiceof ci wemade.

. . . . . .

Analogies

TheTangentProblem(Ch. 2–4)

I Wanttheslopeofacurve

I Onlyknowtheslopeoflines

I Approximatecurvewithaline

I Takelimitoverbetterandbetterapproximations

TheAreaProblem(Ch. 5)

I Wanttheareaofacurvedregion

I Onlyknowtheareaofpolygons

I Approximateregionwithpolygons


. . . . . .

Analogies






TheAreaProblem(Ch. 5)

I Wanttheareaofacurvedregion




. . . . . .

Analogies






TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion




. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Analogies










. . . . . .

Outline






. . . . . .

Distances

Justlike area = length×width, wehave

distance = rate× time.

SohereisanotheruseforRiemannsums.

. . . . . .

ExampleA sailingshipiscruisingbackandforthalongachannel(inastraightline). Atnoontheship’spositionandvelocityarerecorded, butshortlythereafterastormblowsinandpositionisimpossibletomeasure. Thevelocitycontinuestoberecordedatthirty-minuteintervals.

Time 12:00 12:30 1:00 1:30 2:00Speed(knots) 4 8 12 6 4Direction E E E E W

Time 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E

Estimatetheship’spositionat4:00pm.

. . . . . .

SolutionWeestimatethatthespeedof4knots(nauticalmilesperhour)ismaintainedfrom12:00until12:30. Sooverthistimeintervaltheshiptravels (

4 nmihr

)(12hr)

= 2 nmi

Wecancontinueforeachadditionalhalfhourandget

distance = 4× 1/2+ 8× 1/2+ 12× 1/2

+ 6× 1/2− 4× 1/2− 3× 1/2+ 3× 1/2+ 5× 1/2

= 15.5

Sotheshipis 15.5 nmi eastofitsoriginalposition.

. . . . . .

Analysis

I Thismethodofmeasuringpositionbyrecordingvelocityisknownas deadreckoning.

I Ifwehadvelocityestimatesatfinerintervals, we’dgetbetterestimates.

I Ifwehadvelocityateveryinstant, alimitwouldtellusourexactpositionrelativetothelasttimewemeasuredit.

. . . . . .

OtherusesofRiemannsums

Anythingwithaproduct!I Area, volumeI Anythingwithadensity: Population, massI Anythingwitha“speed:” distance, throughput, powerI ConsumersurplusI Expectedvalueofarandomvariable

. . . . . .

Outline






. . . . . .


DefinitionIf f isafunctiondefinedon [a,b], the definiteintegralof f from ato b isthenumber∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

. . . . . .

Notation/Terminology

∫ b

af(x)dx

I∫

— integralsign (swoopy S)

I f(x) — integrandI a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)

I dx —??? (aparenthesis? aninfinitesimal? avariable?)I Theprocessofcomputinganintegraliscalled integration or

quadrature

. . . . . .


∫ b

af(x)dx

I∫




quadrature

. . . . . .


∫ b

af(x)dx

I∫


I f(x) — integrand

I a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)


quadrature

. . . . . .


∫ b

af(x)dx

I∫




quadrature

. . . . . .


∫ b

af(x)dx

I∫



I dx —??? (aparenthesis? aninfinitesimal? avariable?)

I Theprocessofcomputinganintegraliscalled integration orquadrature

. . . . . .


∫ b

af(x)dx

I∫




quadrature

. . . . . .

Thelimitcanbesimplified

TheoremIf f iscontinuouson [a,b] orif f hasonlyfinitelymanyjumpdiscontinuities, then f isintegrableon [a,b]; thatis, thedefinite

integral∫ b

af(x)dx exists.

TheoremIf f isintegrableon [a,b] then∫ b

af(x)dx = lim

n→∞

n∑i=1

f(xi)∆x,

where

∆x =b− an

and xi = a+ i∆x

. . . . . .

Thelimitcanbesimplified

TheoremIf f iscontinuouson [a,b] orif f hasonlyfinitelymanyjumpdiscontinuities, then f isintegrableon [a,b]; thatis, thedefinite

integral∫ b

af(x)dx exists.

TheoremIf f isintegrableon [a,b] then∫ b

af(x)dx = lim

n→∞

n∑i=1

f(xi)∆x,

where

∆x =b− an

and xi = a+ i∆x

. . . . . .

Outline






. . . . . .


Theorem(AdditivePropertiesoftheIntegral)Let f and g beintegrablefunctionson [a,b] and c aconstant.Then

1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.

. . . . . .



1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.

. . . . . .



1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.

. . . . . .



1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.

. . . . . .

MorePropertiesoftheIntegral

Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

Thisallowsustohave

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

bf(x)dx forall a, b, and c.

. . . . . .


Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

Thisallowsustohave

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


. . . . . .


Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

Thisallowsustohave

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c


. . . . . .

ExampleSuppose f and g arefunctionswith

I∫ 4

0f(x)dx = 4

I∫ 5

0f(x)dx = 7

I∫ 5

0g(x)dx = 3.

Find

(a)∫ 5

0[2f(x)− g(x)] dx

(b)∫ 5

4f(x)dx.

. . . . . .

SolutionWehave

(a) ∫ 5

0[2f(x)− g(x)] dx = 2

∫ 5

0f(x)dx−

∫ 5

0g(x)dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f(x)dx =

∫ 5

0f(x)dx−

∫ 4

0f(x)dx

= 7− 4 = 3

. . . . . .

SolutionWehave

(a) ∫ 5

0[2f(x)− g(x)] dx = 2

∫ 5

0f(x)dx−

∫ 5

0g(x)dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f(x)dx =

∫ 5

0f(x)dx−

∫ 4

0f(x)dx

= 7− 4 = 3

Education

Lesson 25: Areas and Distances; The Definite Integral