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