Indefinite Integrals and the Net Change Theorem · 05/09/2017 · Be careful to distinguish...
7
Indefinite Integrals: ∫ ( ) ’ ( can be evaluated as ( ) − (). But what does ∫ ( ) mean? A ( ) means that F(x), the anti-derivative of () has a derivative equal to (). In other words, A ( ) = () means L ( ) = ( ) For example, A N = O 3 ⎯ ⎯ ⎯ + because ⎯ ⎯ ⎯ T O 3 ⎯ ⎯ ⎯ + U = N Indefinite Integrals and the Net Change Theorem
Transcript of Indefinite Integrals and the Net Change Theorem · 05/09/2017 · Be careful to distinguish...
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
IndefiniteIntegrals:
∫ 𝑓(𝑥)𝑑𝑥'( canbeevaluatedas𝐹(𝑏) − 𝐹(𝑎).
Butwhatdoes∫ 𝑓(𝑥)𝑑𝑥�� mean?
A 𝑓(𝑥)𝑑𝑥
�
�
meansthatF(x),theanti-derivativeof𝑓(𝑥) hasaderivativeequalto𝑓(𝑥).Inotherwords,
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
means
𝐹L(𝑥) = 𝑓(𝑥)
Forexample,
A 𝑥N𝑑𝑥
�
�
=𝑥O
3⎯⎯⎯+ 𝐶
because𝑑𝑑𝑥⎯⎯⎯T𝑥O
3⎯⎯⎯+ 𝐶U = 𝑥N
Becarefultodistinguishdefiniteandindefiniteintegrals:
Thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
'
(isanumber;
thedefiniteintegral
A 𝑓(𝑥)𝑑𝑥
�
�isafunction.
Someindefiniteintegrals(anti-derivatives)andsomeproperties:
Examples:Findthegeneralintegrals:
A(10𝑥^ − 2𝑠𝑒𝑐N𝑥)𝑑𝑥
�
�
A𝑐𝑜𝑠𝜃𝑠𝑖𝑛N𝜃⎯⎯⎯⎯⎯ 𝑑𝜃
�
�(Hint:useatrigonometric identity)
A(𝑥O − 6𝑥)O
j
𝑑𝑥
TheNetChangeTheorem:Since
A 𝑓(𝑥)𝑑𝑥
�
�
= 𝐹(𝑥)
and
A 𝑓(𝑥)𝑑𝑥
'
(
= 𝐹(𝑏) − 𝐹(𝑎)
wecanwriteinstead,
𝐹′(𝑥) representsthe"slope"(thechangeinywithrespecttothechangein𝑝)ofourfunctionatapoint𝑥 = 𝑎,and𝐹(𝑏) − 𝐹(𝑎) isthenetchangeinthefunctionbetween𝑥 = 𝑎 and𝑥 = 𝑏.Netchangeisjustthetotalchangebetweenthetwovalues;if𝐹 isincreasing,thenthenetchangeisthesameastheaveragechange,butif𝐹 increasesandthendecreasesagainovertheintervalofinterest,thenetchangeistheaccumulatedpositiveandnegativechangeoverthatinterval.
Sincevelocity isthechangeinpositionwithrespecttothechangeintime,𝑣(𝑡) = 𝑠′(𝑡) (𝑠 representsposition).
Therefore,
A 𝑣(𝑡)𝑑𝑡
'
(
= 𝑉(𝑏) − 𝑉(𝑎)
or
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑉(𝑡N)− 𝑉(𝑡v)= 𝑠(𝑡N)− 𝑠(𝑡v)
Inotherwords,tofindthenetchangeinposition ofaparticlebetween𝑡vand𝑡N,evaluatetheintegralofitsvelocityfunctionoverthatinterval.
Iftherearetimesthattheposition oftheparticleismovinginanegativedirection, itschangeinpositionwillstillbeapositivenumericalvalue,soweintegratetheabsolutevalueofthevelocityfunction:
∫ |𝑣(𝑡)|𝑑𝑡stsu
= totaldistancetraveled.
Example:Aparticlemovesalongalinesothatitsvelocityattime𝑡is𝑣(𝑡) = 𝑡N − 𝑡 −6 (measuredinmeterspersecond).(a)Findthedisplacementoftheparticleduringthetimeperiod1 ≤ 𝑡 ≤ 4.(b)Findthedistancetraveledduringthistimeperiod.
Answer:(a)
A 𝑣(𝑡)𝑑𝑡
st
su
= 𝑠(𝑡N)− 𝑠(𝑡v)
becomes
A(𝑡N−𝑡− 6)𝑑𝑡
^
v
= 𝑠(1) − 𝑠(4) =𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
^
= |4O
3⎯⎯⎯−4N
2⎯⎯⎯− 6(4)} − |
1O
3⎯⎯⎯−1N
2⎯⎯⎯− 6(1)} = −
92⎯⎯
Theparticlehasmovedtotheleft(negative)fromitsoriginalposition 4.5meters.
(b)Thetotaldistancetraveledisthesumofthedistancemovedtotherightandtotheleft.Notethatthepositionfunction𝑡N − 𝑡 − 6 = (t − 3)(t+ 2) ispositiveontheinterval[3,4] andnegativeontheinterval[1,3],sothedistancetraveledis
A |𝑣(𝑡)|𝑑𝑡
^
v
= A(𝑡N−𝑡 − 6)𝑑𝑡 +A(𝑡N−𝑡 − 6)𝑑𝑡
^
O
O
v
=𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
v
O
+𝑡O
3⎯⎯⎯−𝑡N
2⎯⎯⎯− 6𝑡{
O
^
= �v�⎯⎯≈ 10.17m.
Homework:#1-11odd,15,17,21-43odd,51,59
IndefiniteIntegralsandtheNetChangeTheorem
