1
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
