6034 fundamental theorem of calculus (part 2) ab calculus

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

6034 Fundamental Theorem of Calculus (Part 2)

AB CALCULUS

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

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 (π‘₯ )=sin (𝑑 )+𝑐

3=sin ( πœ‹2 )+𝑐3=1+𝑐

2=𝑐𝐴 (π‘₯ )=sin( πœ‹2 )+2

Page 3: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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.

ΒΏ sin (𝑑)|π‘₯0ΒΏ sin (π‘₯ )βˆ’ sin(0)

𝐴 (π‘₯ )=sin(π‘₯)

0

0

cos (𝑑 ) 𝑑𝑑=0

0

πœ‹6

cos (𝑑 ) 𝑑𝑑=12

0

πœ‹4

cos (𝑑 ) 𝑑𝑑=√22

0

πœ‹3

cos (𝑑 ) 𝑑𝑑=√32

1

Page 4: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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

Page 5: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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 :

β€œa” is a constant

Page 6: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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

βˆ’sin (π‘₯ )βˆ—1

Page 7: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

Example:

Find and verify:

2

1

1xdt dt

dx

2 1x

this

Not this

(π‘₯2+1 ) (1 ) 𝑑3

3+𝑑|π‘₯1

=

𝐴′ (π‘₯ )=(π‘₯2+1 )

βˆ’( 13+1)=βˆ’ 4

3

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

Example:

Find without Integrating:

2

3

1xdt dt

dx

√π‘₯2+1 (1 )

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

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) AB CALCULUS

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’

Page 11: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

Demonstration: < The composite function >

Find:

In Words:

3

4

cos( )xd

t dtdx

3 2cos( )*(3 )x x

ΒΏcos (π‘₯3 ) 3π‘₯2

Verify 𝑦=sin π‘₯3βˆ’ sinπœ‹4

=sin π‘₯3βˆ’ √22

𝑦 β€²=cos π‘₯3 (3 π‘₯2)Sub in for t and multiply by the derivative of

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

Example :

Find without Integrating:

If , solve for

2

22

1( )

x

Q x dtt

/ ( )Q x

1

(π‘₯2)2(2π‘₯ )=2 π‘₯

π‘₯4 =2

π‘₯3

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

Example: Rewriting the Integral

2

5

(2 5)x

t dt

Find without integrating: Show middle stepdy

dx

βˆ’5

π‘₯2

(2π‘‘βˆ’5 )𝑑𝑑

𝑦 β€²=βˆ’ ( 2π‘₯2βˆ’5 ) (2π‘₯ )

Page 14: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

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 osπ‘₯

0

βˆšπ‘‘+1𝑑𝑑+ 0

sin π‘₯

βˆšπ‘‘+1𝑑𝑑

βˆ’ 0

cos π‘₯

βˆšπ‘‘+1𝑑𝑑+ 0

sin π‘₯

βˆšπ‘‘+1𝑑𝑑

βˆ’βˆšcosπ‘₯+1 (βˆ’sin π‘₯ )+√sin π‘₯+1 (cosπ‘₯ )

Page 15: 6034 Fundamental Theorem of Calculus (Part 2) AB CALCULUS

Last Update:

β€’ 1/25/11

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