u-du : integrating composite functions
DESCRIPTION
u-du : Integrating Composite Functions. AP Calculus. Integrating Composite Functions (Chain Rule). Remember: Derivatives Rules Remember: Layman’s Description of Antiderivatives. *2 nd meaning of “ du ” du is the derivative of an implicit “ u ”. - PowerPoint PPT PresentationTRANSCRIPT
u-du: Integrating Composite Functions
AP Calculus
Integrating Composite Functions(Chain Rule)
( 1)( ) = n( ) *n nd u u udx
Remember: Derivatives Rules
Remember: Layman’s Description of Antiderivatives
( 1)( ) n nn u du u c
*2nd meaning of “du” du is the derivative of an implicit “u”
u-du SubstitutionIntegrating Composite Functions
(Chain Rule)Revisit the Chain Rule
If let u = inside function
du = derivative of the inside
becomes
2 3( 1)d xdx
2 3 2 2( 1) 3( 1) (2 )d x x xdx
3 2( ) = 3( ) *d u u dudx
Development
from the layman’s idea of antiderivative
“The Family of functions that has the given derivative”
must have the derivative of the inside in order to find
---------- the antiderivative of the outside
( ( )) '( ( ))* '( )d f g x f g x g xdx
( ( )) '( ( ))* '( )d f g x f g x g xdx
( ( )) '( ( ))* '( )f g x f g x g x dx
3( )d udx
23( ) * u du
A Visual Aid
USING u-du Substitution a Visual AidREM: u = inside function du = derivative of the inside
let u =
becomes now only working with f , the outside function
2 23( 1) *2x xdx23u du
Working With Constants: Constant Property of Integration
With u-du Substitution
REM: u = inside function du = derivative of the inside
Missing Constant?
2 2 2 23( 1) *2 = 3 ( 1) *2x xdx x xdx 23 u du
Worksheet - Part 1
5cos 5 cosx dx x dx
4(1 2 )x dx u = du =
4 4 42 1 1(1 2 ) = (1 2 ) 2 = ( )2 2 2
x dx x dx u du
Example 1 : du given
Ex 1: 2 3(5 1) *10x xdx
Example 2: du given
Ex 2:
1 22 33 ( 1)x x dx
Example 3: du given
Ex 3:
2
2 *1
x dxx
Example 4: du given
Ex 4:
2( ) sec ( )tan x x dx
Example 5: Regular Method
Ex 5:
2
cossin
x dxx
Working with Constants < multiplying by one>
Constant Property of Integration
ILL. let u =
du = and
becomes =
Or alternately = =
5cos 5 cosx dx x dx
4(1 2 )x dx (1 2 )x
4 1( )2
u du 41 ( )
2u du
2dx
42 (1 2 )2
x dx 41 ( )2
u du
12
du dx
41 (1 2 ) 22
x dx
Example 6 : Introduce a Constant - my method
2* 9x x dx
Example 7 : Introduce a Constant
2sec (3 )x dx
Example 8 : Introduce a Constant << triple chain>>
4sin (2 )cos(2 )x x dx
Example 9 : Introduce a Constant - extra constant
<< extra constant>
5(3 4)x dx
Example 10: Polynomial
2 4
3 1(3 2 1)
x dxx x
Example 11: Separate the numerator
2
2 11
x dxx
Formal Change of Variables << the Extra “x”>>
Solve for x in terms of u
ILL: Let
Solve for x in terms of u then
and becomes
2 6 *2x x dx (2 6)u x
62
u x
2du dx
6 * *2
u u du
Formal Change of Variables << the Extra “x”>>
Rewrite in terms of u - du
2 13
x dxx
Complete Change of Variables << Changing du >>
At times it is required to even change the du as the u is changed above.
1cos2
x dx u x du dxx
22
xdu dxu du dx
cos 2u u du
We will solve this later in the course.
Development
must have the derivative of the inside in order to find
the antiderivative of the outside
*2nd meaning of “dx” dx is the derivative of an implicit “x” more later if x = f then dx = f /
( ( )) '( ( ))* '( )d f g x f g x g xdx
( ( )) '( ( ))* '( )d f g x f g x g xdx
( ( )) [ '( ( ))* '( )]d f g x f g x g x dx
( ( )) '( ( ))* '( )f g x f g x g x dx