1
Fundamental theorem of calculus II
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)
πππ ( β«
π₯=π
π₯=π
π (π₯ )ππ₯)|π=π₯0= π (π₯0 )
Fundamental theorem of calculus I Change of variables
Integrals
h (π‘ )π hππ‘
Area under the curve
Integrate
Area under the curve
2
π₯0
π (π₯ )
ππ
3
π₯0
π (π₯ )
ππ
π (π+β π₯ )
β π₯π+βπ₯
β π΄1β π΄2β π΄3β π΄4
STOPVerify that this sum makes sense. There are values of Dx that break this picture. What are they?
π΄β βπ=1
πβπβ π₯
π (π+(πβ1 )β π₯ )β βπ₯β π΄
Area under the curve
πππ₯
0
π (π₯ )π (π₯ )
4
π΄β limβπ₯β0
βπ=1
πβπβπ₯
π (π+ (πβ1 )β π₯ ) ββπ₯
β
π΄= β«π₯=π
π₯=π
π (π₯ ) ππ₯
βDefinite integralβ
STOPπ?ππ₯= lim
β π₯β0
Ξ ?Ξπ₯
We wrote a differential. What is coordinately shrinking with ?
π΄β βπ=1
πβπβ π₯
π (π+(πβ1 )β π₯ )β βπ₯
Area under the curve
β π΄
π΄= β«π₯=π
π₯=π
π (π₯ ) ππ₯
π₯0
π (π₯ )
5
ππ
π (π₯ )=2π₯
2π
2π
2πβ2π
π΄= (2π ) (πβπ )+ 12(2πβ2π ) (πβπ )
π΄= (π+π) (πβπ )π΄=π2βπ2
STOPπ π΄ππ|
π=π₯=2 π₯
If we hold a in place, the derivative of A βhappensβ to be
Differentiation βundoesβ integration. Do you remember why?
Example: Area under a line
Fundamental theorem of calculus II
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)
πππ ( β«
π₯=π
π₯=π
π (π₯ )ππ₯)|π=π₯0= π (π₯0 )
Fundamental theorem of calculus I Change of variables
Integrals
6
h (π‘ )π hππ‘
Area under the curve
Integrate
FToC: Differentiation βundoesβ integration
7
π΄= β«π₯=π
π₯=π
π (π₯ ) ππ₯
π (π₯ )
π
π΄ (π₯0+β π₯ )=Area of
π΄ (π₯0 )=Area of
π΄ (π₯0+β π₯ )β π΄ (π₯0 )=Area of
limβ π₯β0
π΄ (π₯0+β π₯ )β π΄ (π₯0 )βπ₯
Want
π₯0 π₯0π₯0+β π₯
FToC: Differentiation βundoesβ integration
8
π (π₯ )
π
limβ π₯β0
π΄ (π₯0+β π₯ )β π΄ (π₯0 )βπ₯
Want
π₯0π₯0+β π₯
π΄ (π₯0+β π₯ )β π΄ (π₯0 )=Area of
π₯0
9
π (π₯ )
π
limβ π₯β0
π΄ (π₯0+β π₯ )β π΄ (π₯0 )βπ₯
Want
π₯0π₯
0 π₯0+β π₯
π΄ (π₯0+β π₯ )β π΄ (π₯0 )=Area of
π΄ (π₯0+β π₯ )β π΄ (π₯0 )β Area of
π (π₯0 )
β π₯
π΄ (π₯0+β π₯ )β π΄ (π₯0 )β π (π₯0 )β π₯
π΄ (π₯0+β π₯ )βπ΄ (π₯0 )β π₯ β π (π₯0 )
π π΄ππ|
π=π₯0= π (π₯0 )
π΄
FToC: Differentiation βundoesβ integration
Fundamental theorem of calculus II
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)
πππ ( β«
π₯=π
π₯=π
π (π₯ )ππ₯)|π=π₯0= π (π₯0 )
Fundamental theorem of calculus I Change of variables
Integrals
10
Area under the curve
h (π‘ )π hππ‘ Integrate
π (π₯ )
π₯0
FToC: Integration βundoesβ differentiation
11
π₯π·0
π ππ π₯ |
π₯=π₯π·
π
π
π
π
β π΄= π ππ π₯|
π₯=π₯0β π₯
β π₯β π β π π
π π₯|π₯=π₯0
βπ₯
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)π₯0
π₯0
π ππ π₯ |
π₯=π₯0
β π₯
π (π₯ )
π₯0
π₯π·0
π ππ π₯ |
π₯=π₯π·
FToC: Integration βundoesβ differentiation
12
π
π
π
π
π (π)
π (π )
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)π₯0
π₯0
13
β«π₯ π·=π
π π ππ π₯ |π₯=π₯π·
ππ₯π·= π (π )β π (π)
π ππ π₯ |
π₯=π₯π·
=ππ₯π·πβ1
π (π₯ )=π₯π
β«π₯ π·=π
π
ππ₯π·πβ1ππ₯π·=π
πβππ
β«ππ₯πβ1ππ₯=π₯π+πΆ
β« cos (π )ππ=sin (π )+πΆ
β«β sin (π )ππ=cos (π )+πΆπ₯π₯
+πΆ
STOP π (stuff ΒΏbe differentiated )ππ₯ =result
Generic differentiation ruleNotion of anti-derivative: Instead of maligning the indefinite integral as the result of βforgettingβ to write down symbols in a definite integral, one often says that, in the context of an equation lacking beginning and end points, such as , the βcurvy Sβ indicates merely that taking the derivative of gives . This kind of use of language does not require discussion of the notion of area under a curve.
Example integral table
Fundamental theorem of calculus II
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)
πππ ( β«
π₯=π
π₯=π
π (π₯ )ππ₯)|π=π₯0= π (π₯0 )
Fundamental theorem of calculus I Change of variables
Integrals
14
Area under the curve
h (π‘ )π hππ‘ Integrate
β«π₯=π
π
π ( π (π₯ ) ) π ππ π₯|π₯ ππ₯= β«
π = π (π )
π (π )
π ( π ) π π
β π₯
15
π₯0
π
π
π (π₯ )
π π
β π β π β π ππ π₯|
π₯=π₯0βπ₯
βπ=1
πβπβπ₯
π ( π (π+(πβ1 )β π₯ ) ) π ππ π₯ |π₯=π+(πβ1)β π₯
β π₯
π₯0
π ( π (π₯ ) )
π (π(π₯0) )
π (π )
π (π)
β
Change of variables
Change of variables example: Trigonometric functions
16
β«π₯=π
π
π ( π (π₯ ) ) π ππ π₯|π₯ ππ₯= β«
π = π (π )
π (π )
π ( π ) π π
π₯0
π
π
π (π₯ )
π π
β ππ ( π (π₯ ) )
β«π=π
π
3 ( sin (π ) )2 cos (π ) ππ=?
π (π )=sin (π )Choose to identify π π
ππ|π=cos (π )
β«π=π
π
3 ( π (π ) )2 π πππ|πππ= β«
π =sin (π )
sin (π )
3 ( π )2π π
ΒΏ ( sin (π) )3β (sin (π ) )3ΒΏ π 3|π= sin (π )β π 3|π =sin (π )
β« 3 π 2π π= π 3+πΆFind in integration table:
Area under the curve
Fundamental theorem of calculus II
β«π₯ π·=π
π π ππ π₯ |
π₯=π₯π·
ππ₯π·= π (π )β π (π)
πππ ( β«
π₯=π
π₯=π
π (π₯ )ππ₯)|π=π₯0= π (π₯0 )
Fundamental theorem of calculus I Change of variables
Integrals
17
h (π‘ )π hππ‘ Integrate
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