6034 fundamental theorem of calculus (part 2)
DESCRIPTION
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 PresentationTRANSCRIPT
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
π΄ (π₯ )=sin (π‘ )+π
3=sin ( π2 )+π3=1+π
2=ππ΄ (π₯ )=sin( π2 )+2
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 (π‘)|π₯0ΒΏ sin (π₯ )β sin(0)
π΄ (π₯ )=sin(π₯)
0
0
cos (π‘ ) ππ‘=0
0
π6
cos (π‘ ) ππ‘=12
0
π4
cos (π‘ ) ππ‘=β22
0
π3
cos (π‘ ) ππ‘=β32
1
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
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 :
βaβ is a constant
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
βsin (π₯ )β1
Example:
Find and verify:
2
1
1xdt dt
dx
2 1x
this
Not this
(π₯2+1 ) (1 ) π‘3
3+π‘|π₯1
=
π΄β² (π₯ )=(π₯2+1 )
β( 13+1)=β 4
3
Example:
Find without Integrating:
2
3
1xdt dt
dx
βπ₯2+1 (1 )
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)β¦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
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 uβ
Demonstration: < The composite function >
Find:
In Words:
3
4
cos( )xd
t dtdx
3 2cos( )*(3 )x x
ΒΏcos (π₯3 ) 3π₯2
Verify π¦=sin π₯3β sinπ4
=sin π₯3β β22
π¦ β²=cos π₯3 (3 π₯2)Sub in for t and multiply by the derivative of
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
Example: Rewriting the Integral
2
5
(2 5)x
t dt
Find without integrating: Show middle stepdy
dx
β5
π₯2
(2π‘β5 )ππ‘
π¦ β²=β ( 2π₯2β5 ) (2π₯ )
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
βπ‘+1ππ‘+ 0
sin π₯
βπ‘+1ππ‘
β 0
cos π₯
βπ‘+1ππ‘+ 0
sin π₯
βπ‘+1ππ‘
ββcosπ₯+1 (βsin π₯ )+βsin π₯+1 (cosπ₯ )
Last Update:
β’ 1/25/11
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