4001 antiderivatives and indefinite integration bc calculus
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4001 ANTIDERIVATIVES AND INDEFINITE INTEGRATION
BC Calculus
ANTIDERIVATIVES AND INDEFINITE INTEGRATION
Rem:
DEFN: A function F is called an Antiderivative of the function f, if for
every x in f: F /(x) = f(x)
If f (x) = then F(x) =
or since
If f / (x) = then f (x) =
2
2
2
3 7
3 12
3
y x x
y x x
y x x
2 3y x
y
y
23x
3x3 2( ) 3
dx x
dx
23x
3x
Notation:
Differential Equation
Differential Form (REM: A Quantity of change)
Integral symbol =
Integrand =
Variable of Integration =
( )dy
f xdx
( )dy f x dx
( )y f x dx ( )f x
dx
The Variable of Integration
( )y f x dx
1 22
Gm m
drr
Newton’s Law of gravitational attraction
NOW: dr tells which variable is being integrated r
Will have more meanings later!
ANTIDERIVATIVES
Layman’s Idea:
A) What is the function that has f (x) as its derivative?.
-Power Rule:
-Trig:
B) The antiderivative is never unique, all answers must include a
+ C (constant of integration)
1
( ) ( )1
nn x
f x x F xn
( ) cos ( ) sinf x x F x x ( ) sin ( ) cosf x x F x x
The Family of Functions whose derivative is given.
Verify the statement by showing the derivative of the right side equals the integral of the left side.
4 3
32
9 3
sin( ) cos( )3
dx cx x
tt t dt t c
Family of Graphs +C
cos( )dy
xdx
The Family of Functions whose derivative is given.
Notation:
Differential Equation
Differential Form
( REM: A Quantity of change)Increment of change
Antiderivative or Indefinite Integral
Total (Net) change
3 4dy
xdx
dy
y
dy
General Solution
A) Indefinite Integration and the Antiderivative are the same thing. General Solution _________________________________________________________ ILL:
( ) ( )f x dx F x c 3
3
1
1
2
xdx
dxx x
dxx
General Solution: EX 1.
General Solution: The Family of Functions ( ) ( )f x dx F x c
3
1dx
xEX 1:
General Solution: EX 2.
General Solution: The Family of Functions ( ) ( )f x dx F x c
(2sin )x dxEX 2:
General Solution: EX 3.
General Solution: The Family of Functions ( ) ( )f x dx F x c / 1( )f x
xEX 3:
Careful !!!!!
Special Considerations
4
2
2
3
( 3)
3 1
1
x dx
x dx
x xdx
x
Initial Condition Problems:
B) Initial Condition Problems:Particular solution < the single graph of the Family –
through a given point> ILL: through the point (1,1)
-Find General solution
-Plug in Point < Initial Condition >and solve for C
2 1dy
xdx
through the point (1,1)
Initial Condition Problems: EX 4.
B) Initial Condition Problems:Particular solution < the single graph of the Family –
through a given point.> Ex 4:
/ 1( )
2f x x
11
2f
Initial Condition Problems: EX 5.
B) Initial Condition Problems:Particular solution < the single graph of the Family –
through a given point.> Ex 5:
/ ( ) cos( )f x x ( ) 13
f
Initial Condition Problems: EX 6.
B) Initial Condition Problems:
A particle is moving along the x - axis such that its acceleration is .
At t = 2 its velocity is 5 and its position is 10.
Find the function, , that models the particle’s motion.
( ) 2a t
( )x t
Initial Condition Problems: EX 7.
B) Initial Condition Problems:
EX 7:
If no Initial Conditions are given:
Find if/// ( ) 1f x ( )f x
Last Update:
• 12/17/10
• Assignment– Xerox