5.2: The Definite Integral - WordPress.com · 2017. 9. 5. · Properties of the Definite Integral...
7
The funny looking "S" ∫ ⬚ is called the integral sign. The "S" stands for (represents) "sum" because an integral is a limit of sums. • The function () is called the integrand. • The values and are the (upper and lower) limits of integration. • does not have any meaning by it self, but it is a necessary piece of the notation; an integral without a does not make sense. For now, you can think of it as the "period" at the end of the sentence; it tells you what the variable is that you are integrating with respect to. • ∑ ( J ∗ )∆ M JNO is called a "Riemann sum" after the German mathematician Bernhard Riemann. 5.2: The Definite Integral
Transcript of 5.2: The Definite Integral - WordPress.com · 2017. 9. 5. · Properties of the Definite Integral...
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
Thefunnylooking"S"∫⬚�� iscalledtheintegralsign.The"S"standsfor(represents)"sum"becauseanintegralisalimit ofsums.
•
Thefunction𝑓(𝑥) iscalledtheintegrand.•
Thevalues𝑎 and𝑏 arethe(upperandlower)limitsofintegration.•
𝑑𝑥 doesnothaveanymeaningbyitself,butit isanecessarypieceofthenotation;anintegralwithout a𝑑𝑥 doesnotmakesense.Fornow,youcanthinkofitasthe"period"attheendofthesentence;it tellsyouwhatthevariableisthatyouareintegratingwithrespectto.
•
∑ 𝑓(𝑥J∗)∆𝑥MJNO iscalleda"Riemannsum"aftertheGermanmathematician
BernhardRiemann.
∫ 𝑓(𝑥)𝑑𝑥ST istheresultoftakingthelimit of∑ 𝑓(𝑥J∗)∆𝑥M
JNO as𝑛 → ∞.
Theareaunderthecurvebetween𝑎 and𝑏 istheareaunderthecurvebetween0 and𝑏 minustheareafrom0 to𝑎.Ifwethinkofintegralsstrictlyasareas,wecanwritethefollowing:
[ 𝑓(𝑥)𝑑𝑥
S
T
= 𝐴O −𝐴_
Withsomefunctions,partoftheareacreatedbythecurvecomesfromfindingnegative𝑦-values.Butanareaisstillapositivevalue,sowhenthishappens,wecompensatebyusingtheoppositeofthevaluefoundwhencalculatingthispartofthearea.
TheareaintheyellowregionwouldneedtobecalculatedusingaRiemannsumwithvalueof𝑓(𝑥) thatarenegative.
Inotherwords,ifafunctioniscontinuous,thenitispossibletotakeitsintegral(theintegralexists);ifafunctionisintegrable,thentheintegralisdefinedbybeingthelimit oftheRiemannsum.
Thisstatementisimportantbecausenotallfunctionsareintegrable!
Example:Evaluatetheintegralbyinterpretingit intermsofareas:
[(𝑥 − 1)𝑑𝑥
f
g
Thegraphof𝑦 = 𝑥 − 1 showsusthatpartofthefunctionisnegativeandpartispositive.
Toevaluatethisintegralasthesumofareas,weneedtofindthetwoareasandsubtract𝐴_ from𝐴O.Fortunately,botharetriangles,andtheirareaseasytocompute:
𝐴O =2 ⋅ 22⎯⎯⎯⎯
= 2𝐴_ =1 ⋅ 12⎯⎯⎯⎯
=12⎯⎯
𝐴O − 𝐴_ = 2−12⎯⎯=32⎯⎯
Thearea"under"thecurveisf_⎯.
PropertiesoftheDefiniteIntegral
LiketheLimitLawsinCalculus1,therearesomepropertiesofintegralsthatcanbeuseful.
[ 𝑐𝑑𝑥
S
T
= 𝑐(𝑏 − 𝑎)
where𝑐 isanyconstant
[[𝑓(𝑥) ± 𝑔(𝑥)]𝑑𝑥
S
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
± [𝑔(𝑥)𝑑𝑥
S
T
[ 𝑐 ⋅ 𝑓(𝑥)𝑑𝑥
S
T
= 𝑐 ⋅ [ 𝑓(𝑥)𝑑𝑥
S
Twhere𝑐 isanyconstant
[ 𝑓(𝑥)𝑑𝑥
T
T
= 0
[ 𝑓(𝑥)𝑑𝑥
S
T
= −[𝑓(𝑥)𝑑𝑥
T
S
[ 𝑓(𝑥)𝑑𝑥
s
T
= [ 𝑓(𝑥)𝑑𝑥
S
T
+[𝑔(𝑥)𝑑𝑥
s
S
Example:
[ 𝑓(𝑥)𝑑𝑥
Og
g
= 17,[𝑓(𝑥)𝑑𝑥
v
g
= 12
Find
[ 𝑓(𝑥)𝑑𝑥
Og
v
A:Since
[ 𝑓(𝑥)𝑑𝑥
Og
v
= [ 𝑓(𝑥)𝑑𝑥
Og
g
−[ 𝑓(𝑥)𝑑𝑥
v
g
[ 𝑓(𝑥)𝑑𝑥
Og
v
= 17− 12 = 5
Example:Showthat∫ (𝑥_ − 4𝑥 + 4)𝑑𝑥z
g ≥ 0 without integrating.
A:Sincethefunction𝑓(𝑥) = 𝑥_ − 4𝑥 + 4 isgreaterthanorequalto0between𝑥 = 0 and𝑥 = 4,byComparisonProperty6,∫ f(x)𝑑𝑥z
g ≥ 0
Homework:p382:#3,9,11,17,19,21,23,25
5.2:TheDefiniteIntegral
![The Definite Integral - Math - The University of Utah · 2013. 1. 4. · 25B Definite Integral 3 Definition of the Definite Integral Let f be a function that is defined on [a,b].](https://static.fdocuments.net/doc/165x107/60a2bfa42907903993233547/the-definite-integral-math-the-university-of-2013-1-4-25b-definite-integral.jpg)
