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