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ANTIDERIVATIVES AND INDEFINITE INTEGRATION. AB 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) - PowerPoint PPT Presentation
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ANTIDERIVATIVES AND INDEFINITE INTEGRATION
AB 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
2 𝑥+3
2 𝑥+3
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.
𝑎𝑑𝑑𝑜𝑛𝑒𝑡𝑜 h𝑡 𝑒𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑛𝑒𝑤𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡
Family of Graphs +C
The Family of Functions whose derivative is given.
All same equations with different y intercept
𝑑𝑦𝑑𝑥
=cos (𝑥)𝑦=− sin (𝑥 )+𝑐
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
small8-8.1
=sumSumming a bunch of little changes
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!
The Family of Functions whose derivative is given.
𝑦 ′=𝑥3
𝑦=? 𝑦= 𝑥4
4
𝑦 ′=𝑥− 2
𝑦= 𝑥−1
−1
𝑦 ′=𝑥12
𝑦=𝑥
32
32
¿23𝑥
32
𝑦 ′=𝑥− 2
3
𝑦=𝑥
13
13
=3 𝑥13
Notation:
Differential Equation
Differential Form
( REM: A Quantity of change)Increment of change
Antiderivative or Indefinite Integral
Total (Net) change
3 4dy
xdx
y
(3𝑥+4 ) 𝑑𝑥
(3𝑥+4 ) 𝑑𝑥
𝑑𝑦=¿¿
General Solution
A) Indefinite Integration and the Antiderivative are the same thing. General Solution _________________________________________________________ ILL:
( ) ( )f x dx F x c
(3𝑥+4 ) 𝑑𝑥 3𝑥𝑑𝑥+ 4𝑑𝑥3 (𝑥
2
2 )+4 𝑥+𝑐
General Solution: EX 1.
General Solution: The Family of Functions ( ) ( )f x dx F x c
3
1dx
xEX 1:
𝑥−3 𝑑𝑥𝑥− 2
−2+𝑐
General Solution: EX 2.
General Solution: The Family of Functions ( ) ( )f x dx F x c
(2sin )x dxEX 2:
2 (−cos 𝑥 )+𝑐
−2¿
General Solution: EX 3.
General Solution: The Family of Functions ( ) ( )f x dx F x c / 1( )f x
xEX 3:
Careful !!!!!
¿ 𝑥−1
𝑥−1𝑑𝑥𝑥0
0
𝑙𝑛|𝑥|+𝑐
Verify the statement by showing the derivative of the right side equals the integral of the left side.
(− 9
𝑥4 )𝑑𝑥
(𝑡2 −sin (𝑡 ) )𝑑𝑡=¿¿ 𝑡3
3+cos (𝑡 )+𝑐
¿3
𝑥3 +𝑐
¿𝑥3 (0 )− 3(3 𝑥2)
(𝑥3 )2 ¿ −9 𝑥2
𝑥6
¿−9
𝑥4 +𝑐
General Solution
A) Indefinite Integration and the Antiderivative are the same thing. General Solution _________________________________________________________ ILL:
( ) ( )f x dx F x c 𝑥
43
43
=34𝑥
43 +𝑐
𝑥− 1𝑥−12 =
𝑥− 3
2+ 2
2
−12
=− 2𝑥− 1
2 +𝑐
12𝑥− 3=
12 ( 𝑥
− 2
−2 )=−14𝑥− 2
3√𝑥 𝑑𝑥 1𝑥√𝑥
𝑑𝑥
1
2 𝑥3 𝑑𝑥
Special Considerations
43x dx3𝑥5
5+𝑐
2( 3)x dx
(𝑥2 −6 𝑥+9 )𝑑𝑥𝑥2𝑑𝑥+− 6𝑥𝑑𝑥+ 9𝑑𝑥𝑥3
3+(−6 ) 𝑥
2
2+9 𝑥
𝑥3
3− 3 𝑥2+9𝑥+𝑐
𝑥2−2 𝑥+1𝑥−1
(𝑥−1)(𝑥−1)(𝑥− 1)
(𝑥− 1 )𝑑𝑥
𝑥𝑑𝑥−1𝑑𝑥𝑥2
2−𝑥+𝑐
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
𝑑𝑦= (2𝑥− 1 )𝑑𝑥
𝑑𝑦= (2 𝑥=1 )𝑑𝑥𝑦=2( 𝑥
2
2 )−𝑥+𝑐
𝑦=𝑥2−𝑥+𝑐1= (12 ) −1+c
1+c
1=𝑐𝑦=𝑥2−𝑥+1
through the point (1,1)
𝑦=𝑥2−𝑥+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
𝑦= 12𝑥𝑑𝑥
𝑦=12 (𝑥
2
2 )+𝑐𝑦=
14𝑥2+𝑐
−1=14 ( 1
2 )2
+𝑐
−1=1
16+𝑐
−1716
=𝑐
𝑦=14𝑥2−
1716
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
𝑓 (𝑥 )= cos (𝑥 )𝑑𝑥¿ sin 𝑥+𝑐
1=sin( 𝜋3 )+𝑐1=√3
2+𝑐
1 − √32
=𝑐
𝑓 (𝑥 )=sin (𝑥 )+(1 − √32 )
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 t𝑥 (𝑡)=𝑣(𝑡)
𝑥′ ′ (𝑡 )=𝑎(𝑡)
𝑣 (𝑡 )=𝑎 (𝑡 )𝑑𝑡𝑣 (𝑡 )= 2𝑑𝑡𝑣 (𝑡 )=2𝑡+𝑐5=2 (2 )+𝑐
1=𝑐
𝑣 (𝑡 )=2𝑡+1 𝑥 (𝑡 )=𝑣 (𝑡 )𝑑𝑡𝑥 (𝑡 )= (2 𝑡+1 )𝑑𝑡𝑥 (𝑡 )=2( 𝑡
2
2 )+𝑡+𝑐10=22+2+𝑐
4=𝑐𝑥 (𝑡 )=𝑡 2+𝑡+4
Initial Condition Problems: EX 7.
B) Initial Condition Problems:
EX 7:
If no Initial Conditions are given:
Find if/// ( ) 1f x ( )f x
𝑓 ′ ′ (𝑥 )=1𝑑𝑥𝑓 ′ ′ (𝑥 )=𝑥+𝑐
𝑓 ′ (𝑥 )= (𝑥+𝑐 )𝑑𝑥𝑓 ′ (𝑥 )= 𝑥2
2+𝑐1𝑥+𝑐2
𝑓 (𝑥 )= 𝑥2
2+𝑐1𝑥+𝑐2
𝑓 (𝑥 )=12 ( 𝑥
3
3 )+𝑐1(𝑥2
2 )+𝑐2 𝑥+𝑐3
Last Update:
• 12/17/10
• Assignment– Xerox
