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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 Presentation
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Fundamental theorem of calculus II
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑𝑑𝑏 ( ∫
𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 )𝑑𝑥)|𝑏=𝑥0= 𝑓 (𝑥0 )
Fundamental theorem of calculus I Change of variables
Integrals
h (𝑡 )𝑑 h𝑑𝑡
Area under the curve
Integrate
Area under the curve
2
𝑥0
𝑓 (𝑥 )
𝑏𝑎
3
𝑥0
𝑓 (𝑥 )
𝑏𝑎
𝑓 (𝑎+∆ 𝑥 )
∆ 𝑥𝑎+∆𝑥
∆ 𝐴1∆ 𝐴2∆ 𝐴3∆ 𝐴4
STOPVerify that this sum makes sense. There are values of Dx that break this picture. What are they?
𝐴≅ ∑𝑘=1
𝑏−𝑎∆ 𝑥
𝑓 (𝑎+(𝑘−1 )∆ 𝑥 )∙ ∆𝑥∆ 𝐴
Area under the curve
𝑏𝑎𝑥
0
𝑓 (𝑥 )𝑓 (𝑥 )
4
𝐴≔ lim∆𝑥→0
∑𝑘=1
𝑏−𝑎∆𝑥
𝑓 (𝑎+ (𝑘−1 )∆ 𝑥 ) ∙∆𝑥
≔
𝐴= ∫𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 ) 𝑑𝑥
“Definite integral”
STOP𝑑?𝑑𝑥= lim
∆ 𝑥→0
Δ ?Δ𝑥
We wrote a differential. What is coordinately shrinking with ?
𝐴≅ ∑𝑘=1
𝑏−𝑎∆ 𝑥
𝑓 (𝑎+(𝑘−1 )∆ 𝑥 )∙ ∆𝑥
Area under the curve
∆ 𝐴
𝐴= ∫𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 ) 𝑑𝑥
𝑥0
𝑓 (𝑥 )
5
𝑏𝑎
𝑓 (𝑥 )=2𝑥
2𝑎
2𝑏
2𝑏−2𝑎
𝐴= (2𝑎 ) (𝑏−𝑎 )+ 12(2𝑏−2𝑎 ) (𝑏−𝑎 )
𝐴= (𝑎+𝑏) (𝑏−𝑎 )𝐴=𝑏2−𝑎2
STOP𝑑 𝐴𝑑𝑏|
𝑏=𝑥=2 𝑥
If we hold a in place, the derivative of A “happens” to be
Differentiation “undoes” integration. Do you remember why?
Example: Area under a line
Fundamental theorem of calculus II
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑𝑑𝑏 ( ∫
𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 )𝑑𝑥)|𝑏=𝑥0= 𝑓 (𝑥0 )
Fundamental theorem of calculus I Change of variables
Integrals
6
h (𝑡 )𝑑 h𝑑𝑡
Area under the curve
Integrate
FToC: Differentiation “undoes” integration
7
𝐴= ∫𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 ) 𝑑𝑥
𝑓 (𝑥 )
𝑎
𝐴 (𝑥0+∆ 𝑥 )=Area of
𝐴 (𝑥0 )=Area of
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )=Area of
lim∆ 𝑥→0
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )∆𝑥
Want
𝑥0 𝑥0𝑥0+∆ 𝑥
FToC: Differentiation “undoes” integration
8
𝑓 (𝑥 )
𝑎
lim∆ 𝑥→0
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )∆𝑥
Want
𝑥0𝑥0+∆ 𝑥
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )=Area of
𝑥0
9
𝑓 (𝑥 )
𝑎
lim∆ 𝑥→0
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )∆𝑥
Want
𝑥0𝑥
0 𝑥0+∆ 𝑥
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )=Area of
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )≅ Area of
𝑓 (𝑥0 )
∆ 𝑥
𝐴 (𝑥0+∆ 𝑥 )− 𝐴 (𝑥0 )≅ 𝑓 (𝑥0 )∆ 𝑥
𝐴 (𝑥0+∆ 𝑥 )−𝐴 (𝑥0 )∆ 𝑥 ≅ 𝑓 (𝑥0 )
𝑑 𝐴𝑑𝑏|
𝑏=𝑥0= 𝑓 (𝑥0 )
𝐴
FToC: Differentiation “undoes” integration
Fundamental theorem of calculus II
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑𝑑𝑏 ( ∫
𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 )𝑑𝑥)|𝑏=𝑥0= 𝑓 (𝑥0 )
Fundamental theorem of calculus I Change of variables
Integrals
10
Area under the curve
h (𝑡 )𝑑 h𝑑𝑡 Integrate
𝑓 (𝑥 )
𝑥0
FToC: Integration “undoes” differentiation
11
𝑥𝐷0
𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑎
𝑎
𝑏
𝑏
∆ 𝐴= 𝑑 𝑓𝑑 𝑥|
𝑥=𝑥0∆ 𝑥
∆ 𝑥∆ 𝑓 ≅ 𝑑 𝑓
𝑑 𝑥|𝑥=𝑥0
∆𝑥
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)𝑥0
𝑥0
𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥0
∆ 𝑥
𝑓 (𝑥 )
𝑥0
𝑥𝐷0
𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
FToC: Integration “undoes” differentiation
12
𝑎
𝑎
𝑏
𝑏
𝑓 (𝑏)
𝑓 (𝑎 )
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)𝑥0
𝑥0
13
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
=𝑛𝑥𝐷𝑛−1
𝑓 (𝑥 )=𝑥𝑛
∫𝑥 𝐷=𝑎
𝑏
𝑛𝑥𝐷𝑛−1𝑑𝑥𝐷=𝑏
𝑛−𝑎𝑛
∫𝑛𝑥𝑛−1𝑑𝑥=𝑥𝑛+𝐶
∫ cos (𝜃 )𝑑𝜃=sin (𝜃 )+𝐶
∫− sin (𝜃 )𝑑𝜃=cos (𝜃 )+𝐶𝑥𝑥
+𝐶
STOP 𝑑 (stuff ¿be differentiated )𝑑𝑥 =result
Generic differentiation ruleNotion of anti-derivative: Instead of maligning the indefinite integral as the result of “forgetting” 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 “curvy S” 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
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑𝑑𝑏 ( ∫
𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 )𝑑𝑥)|𝑏=𝑥0= 𝑓 (𝑥0 )
Fundamental theorem of calculus I Change of variables
Integrals
14
Area under the curve
h (𝑡 )𝑑 h𝑑𝑡 Integrate
∫𝑥=𝑎
𝑏
𝑔 ( 𝑓 (𝑥 ) ) 𝑑 𝑓𝑑 𝑥|𝑥 𝑑𝑥= ∫
𝑓 = 𝑓 (𝑎 )
𝑓 (𝑏 )
𝑔 ( 𝑓 ) 𝑑 𝑓
∆ 𝑥
15
𝑥0
𝑓
𝑔
𝑓 (𝑥 )
𝑎 𝑏
∆ 𝑓 ∆ 𝑓 ≅ 𝑑 𝑓𝑑 𝑥|
𝑥=𝑥0∆𝑥
∑𝑘=1
𝑏−𝑎∆𝑥
𝑔 ( 𝑓 (𝑎+(𝑘−1 )∆ 𝑥 ) ) 𝑑 𝑓𝑑 𝑥 |𝑥=𝑎+(𝑘−1)∆ 𝑥
∆ 𝑥
𝑥0
𝑔 ( 𝑓 (𝑥 ) )
𝑔 (𝑓(𝑥0) )
𝑓 (𝑎 )
𝑓 (𝑏)
≅
Change of variables
Change of variables example: Trigonometric functions
16
∫𝑥=𝑎
𝑏
𝑔 ( 𝑓 (𝑥 ) ) 𝑑 𝑓𝑑 𝑥|𝑥 𝑑𝑥= ∫
𝑓 = 𝑓 (𝑎 )
𝑓 (𝑏 )
𝑔 ( 𝑓 ) 𝑑 𝑓
𝑥0
𝑓
𝑔
𝑓 (𝑥 )
𝑎 𝑏
∆ 𝑓𝑔 ( 𝑓 (𝑥 ) )
∫𝜃=𝑎
𝑏
3 ( sin (𝜃 ) )2 cos (𝜃 ) 𝑑𝜃=?
𝑓 (𝜃 )=sin (𝜃 )Choose to identify 𝑑 𝑓
𝑑𝜃|𝜃=cos (𝜃 )
∫𝜃=𝑎
𝑏
3 ( 𝑓 (𝜃 ) )2 𝑑 𝑓𝑑𝜃|𝜃𝑑𝜃= ∫
𝑓 =sin (𝑎 )
sin (𝑏 )
3 ( 𝑓 )2𝑑 𝑓
¿ ( sin (𝑏) )3− (sin (𝑎 ) )3¿ 𝑓 3|𝑓= sin (𝑏 )− 𝑓 3|𝑓 =sin (𝑎 )
∫ 3 𝑓 2𝑑 𝑓= 𝑓 3+𝐶Find in integration table:
Area under the curve
Fundamental theorem of calculus II
∫𝑥 𝐷=𝑎
𝑏 𝑑 𝑓𝑑 𝑥 |
𝑥=𝑥𝐷
𝑑𝑥𝐷= 𝑓 (𝑏 )− 𝑓 (𝑎)
𝑑𝑑𝑏 ( ∫
𝑥=𝑎
𝑥=𝑏
𝑓 (𝑥 )𝑑𝑥)|𝑏=𝑥0= 𝑓 (𝑥0 )
Fundamental theorem of calculus I Change of variables
Integrals
17
h (𝑡 )𝑑 h𝑑𝑡 Integrate
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