lesson 25: areas and distances; the definite integral
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. . . . . .
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
. . . . . .
EasyAreas: Parallelogram
Bycuttingandpasting, aparallelogramcanbemadeintoarectangle.
..b
.b
.h
SoA = bh
. . . . . .
EasyAreas: Parallelogram
Bycuttingandpasting, aparallelogramcanbemadeintoarectangle.
.
.b .b
.h
SoA = bh
. . . . . .
EasyAreas: Parallelogram
Bycuttingandpasting, aparallelogramcanbemadeintoarectangle.
.
.b
.b
.h
SoA = bh
. . . . . .
EasyAreas: Parallelogram
Bycuttingandpasting, aparallelogramcanbemadeintoarectangle.
.
.b
.b
.h
SoA = bh
. . . . . .
EasyAreas: Triangle
Bycopyingandpasting, atrianglecanbemadeintoaparallelogram.
..b
.h
SoA =
12bh
. . . . . .
EasyAreas: Triangle
Bycopyingandpasting, atrianglecanbemadeintoaparallelogram.
..b
.h
SoA =
12bh
. . . . . .
EasyAreas: Triangle
Bycopyingandpasting, atrianglecanbemadeintoaparallelogram.
..b
.h
SoA =
12bh
. . . . . .
EasyAreas: Polygons
Anypolygoncanbetriangulated, soitsareacanbefoundbysummingtheareasofthetriangles:
.
. . . . . .
HardAreas: CurvedRegions
.
???
. . . . . .
Meetthemathematician: Archimedes
I Greek(Syracuse), 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
Meetthemathematician: Archimedes
I Greek(Syracuse), 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
Meetthemathematician: Archimedes
I Greek(Syracuse), 287BC –212BC (afterEuclid)
I GeometerI Weaponsengineer
. . . . . .
.
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A =
1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
. . . . . .
.
.1
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1
+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
. . . . . .
.
.1.18 .18
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1+ 2 · 18
+ 4 · 164
+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
→ 13/4
=43
as n → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
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
.
.13
.
.23
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
.
.13
.
.23
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
.
.14
.
.24
.
.34
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
.
.14
.
.24
.
.34
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
.
.15
.
.25
.
.35
.
.45
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
.
.15
.
.25
.
.35
.
.45
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
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.
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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
→ 13
as n → ∞.
. . . . . .
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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 → ∞.
. . . . . .
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)
]2
So
Ln =n2(n− 1)2
4n4→ 1
4as 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 → ∞.
. . . . . .
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
AreathroughtheCenturiesEuclidArchimedesCavalieri
GeneralizingCavalieri’smethodAnalogies
DistancesOtherapplications
Thedefiniteintegralasalimit
Propertiesoftheintegral
. . . . . .
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
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)
I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Analogies
TheTangentProblem(Ch. 2–4)
I Wanttheslopeofacurve
I Onlyknowtheslopeoflines
I Approximatecurvewithaline
I Takelimitoverbetterandbetterapproximations
TheAreaProblem(Ch. 5)I Wanttheareaofacurvedregion
I Onlyknowtheareaofpolygons
I Approximateregionwithpolygons
I Takelimitoverbetterandbetterapproximations
. . . . . .
Outline
AreathroughtheCenturiesEuclidArchimedesCavalieri
GeneralizingCavalieri’smethodAnalogies
DistancesOtherapplications
Thedefiniteintegralasalimit
Propertiesoftheintegral
. . . . . .
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
AreathroughtheCenturiesEuclidArchimedesCavalieri
GeneralizingCavalieri’smethodAnalogies
DistancesOtherapplications
Thedefiniteintegralasalimit
Propertiesoftheintegral
. . . . . .
Thedefiniteintegralasalimit
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
. . . . . .
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
. . . . . .
Notation/Terminology
∫ b
af(x)dx
I∫
— integralsign (swoopy S)
I f(x) — integrand
I a and b — limitsofintegration (a isthe lowerlimit and bthe upperlimit)
I dx —??? (aparenthesis? aninfinitesimal? avariable?)I Theprocessofcomputinganintegraliscalled integration or
quadrature
. . . . . .
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
. . . . . .
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 orquadrature
. . . . . .
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
. . . . . .
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
AreathroughtheCenturiesEuclidArchimedesCavalieri
GeneralizingCavalieri’smethodAnalogies
DistancesOtherapplications
Thedefiniteintegralasalimit
Propertiesoftheintegral
. . . . . .
Propertiesoftheintegral
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.
. . . . . .
Propertiesoftheintegral
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.
. . . . . .
Propertiesoftheintegral
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.
. . . . . .
Propertiesoftheintegral
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
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