6034 Fundamental Theorem of Calculus (Part 2)

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6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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6034 Fundamental Theorem of Calculus (Part 2). AB CALCULUS. The Indefinite Integral (Antiderivative) finds a Family of Functions whose derivative is given. Given an Initial Condition we find the Particular Function. The Definite Integral as a Particular Function :. - PowerPoint PPT Presentation

Transcript of 6034 Fundamental Theorem of Calculus (Part 2)

Page 1: 6034  Fundamental Theorem of Calculus  (Part 2)

6034 Fundamental Theorem of Calculus (Part 2)

AB CALCULUS

Page 2: 6034  Fundamental Theorem of Calculus  (Part 2)

The Indefinite Integral (Antiderivative) finds a Family of Functions whose derivative is given.

( ) cos( )A x t dt

Given an Initial Condition we find the Particular Function

( ) 32

f

𝐴 (𝑥 )=sin (𝑡 )+𝑐

3=sin ( 𝜋2 )+𝑐3=1+𝑐

2=𝑐𝐴 (𝑥 )=sin( 𝜋2 )+2

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The Definite Integral as a Particular Function:

0

( ) cos( )x

A x t dtEvaluate at 0, , ,

6 4 3x

Evaluate the Definite Integral for each of these points.

The Definite Integral is actually finding points on the Accumulation graph.

Evaluate the definite integral.

¿ sin (𝑡)|𝑥0¿ sin (𝑥 )− sin(0)

𝐴 (𝑥 )=sin(𝑥)

0

0

cos (𝑡 ) 𝑑𝑡=0

0

𝜋6

cos (𝑡 ) 𝑑𝑡=12

0

𝜋4

cos (𝑡 ) 𝑑𝑡=√22

0

𝜋3

cos (𝑡 ) 𝑑𝑡=√32

1

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Since A(x) is a function, what then is the rate of change of that function?

0

( ) cos( )x

A x t dt( ) sin( )A x x

( ) cos( )A x x

In words, integration and differentiation are inverse operations

Take derivative

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2nd Fundamental Theorem of Calculus

 

Given: , we want to find

Note: a is a constant, u is a function of x; and the order matters!

( ) ( )x

a

A x f t dt/ ( )A x

( ) ( )u

a

df t dt f u u

dx

2nd Fundamental Theorem of Calculus: If f is continuous on an open interval, I, containing a point, a,

then for every x in I :

“a” is a constant

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Demonstration: < function x only >

find

2

( ) sin( )x

A x t dt

( ( ))dA x

dx

2

( ( )) sin( ) ]xd d

A x t dtdx dx

In Words:Sub in the function u and multiply by derivative of u

−sin (𝑥 )∗1

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Example:

Find and verify:

2

1

1xdt dt

dx

2 1x

this

Not this

(𝑥2+1 ) (1 ) 𝑡3

3+𝑡|𝑥1

=

𝐴′ (𝑥 )=(𝑥2+1 )

−( 13+1)=− 4

3

Page 8: 6034  Fundamental Theorem of Calculus  (Part 2)

Example:

Find without Integrating:

2

3

1xdt dt

dx

√𝑥2+1 (1 )

Page 9: 6034  Fundamental Theorem of Calculus  (Part 2)

THE COMPOSITE FUNCTION 

If g(x) is given instead of x:

 

 

In words: Substitute in g(x) for t and then multiply by the derivative of g(x)…exactly the chain rule

(derivative of the outside * derivative of the inside)

( )/ ( ( )) ( )

g x

a

dQ g x f t dt

dx

( )[ ( )]g xadF t

dx

( ( )) ( )dF g x F a

dx

( ( ))* '( ) ( ( ))* '( )F g x g x or f g x g x

Page 10: 6034  Fundamental Theorem of Calculus  (Part 2)

THE COMPOSITE FUNCTION

If , (a composite function)

then

( )u g x

/( ) ( )*u

a

df t dt f u u

dx

In Words:Sub u in for t and multiply by u’

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Demonstration: < The composite function >

Find:

In Words:

3

4

cos( )xd

t dtdx

3 2cos( )*(3 )x x

¿cos (𝑥3 ) 3𝑥2

Verify 𝑦=sin 𝑥3− sin𝜋4

=sin 𝑥3− √22

𝑦 ′=cos 𝑥3 (3 𝑥2)Sub in for t and multiply by the derivative of

Page 12: 6034  Fundamental Theorem of Calculus  (Part 2)

Example :

Find without Integrating:

If , solve for

2

22

1( )

x

Q x dtt

/ ( )Q x

1

(𝑥2)2(2𝑥 )=2 𝑥

𝑥4 =2

𝑥3

Page 13: 6034  Fundamental Theorem of Calculus  (Part 2)

Example: Rewriting the Integral

2

5

(2 5)x

t dt

Find without integrating: Show middle stepdy

dx

−5

𝑥2

(2𝑡−5 )𝑑𝑡

𝑦 ′=− ( 2𝑥2−5 ) (2𝑥 )

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Example: Rewriting the Integral - Two variable limits:

Find without Integrating:

 

break into two parts . . . . . chose any number in domain of for a and rewrite into required form

.

sin( )

cos( )

1x

x

dt dt

dx

1t

c os𝑥

0

√𝑡+1𝑑𝑡+ 0

sin 𝑥

√𝑡+1𝑑𝑡

− 0

cos 𝑥

√𝑡+1𝑑𝑡+ 0

sin 𝑥

√𝑡+1𝑑𝑡

−√cos𝑥+1 (−sin 𝑥 )+√sin 𝑥+1 (cos𝑥 )

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Last Update:

• 1/25/11

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