fundamental theorem of calculus ii
DESCRIPTION
Integrals. Integrate. Area under the curve. Fundamental theorem of calculus I. Change of variables. Fundamental theorem of calculus II. Area under the curve. 0. Area under the curve. Verify that this sum makes sense. There are values of D x that break this picture. What are they?. 0. - PowerPoint PPT PresentationTRANSCRIPT
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