4033-Properties of the Definite Integral (5.3)

AB Calculus

Properties of Definite Integrals

 

    

A)

 

B)

 

C)  

D)

A b h

D r t

• Think rectangles• Distance

( ) 0a

af x dx

1 ( )b

adx b a ( ) ( )

b a

a bf x dx f x dx

a dx b

f (x)

( ) ( )b b

a akf x dx k f x dx

a to a

nowhere

rectan

gle

Oppos

ite

direc

tion

Constant

multiplier

𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥=2𝑎−𝑏

𝑎

𝑏

( 𝑓 (𝑥 )+5¿)𝑑𝑥¿

𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥+¿𝑎

𝑏

5𝑑𝑥¿

2𝑎−𝑏+5𝑎−5𝑏

Properties of Definite Integrals  

A b h

D r t

• Think rectangles• Distance

a c b

( ( ) ( )) ( ) ( )b b b

a a af x g x dx f x dx g x dx E)

NOTE: Same Interval

(2). IMPORTANT: Finding Area between curves.

(1). Shows the method to work Definite Integrals – like Σ

𝑓 (𝑥 )−𝑔(𝑥

)

𝑓subtract

Properties of Definite Integrals

 

  

A b h

D r t

• Think rectangles• Distance

a c b

( ) ( ) ( )b c b

a a cf x dx f x dx f x dx

F) If c is between a and b , then:

 

Placement of c important: upper bound of 1st, lower bound of 2nd.

REM: The Definite Integral is a number, so may solve the above like an equation.

( ) ( ) ( )b c b

a a cf x dx f x dx f x dx

Examples:

Show all the steps to integrate.

3 2

1(2 3 5)x x dx

Step 1:Break into parts

1

3

2𝑥2𝑑𝑥+1

3

3 𝑥𝑑𝑥−1

3

5𝑑𝑥

21

3

𝑥2𝑑𝑥+31

3

𝑥𝑑𝑥−51

3

𝑑𝑥

FTC FTC rectangle

Remove constant multiplier

Examples:

GIVEN: 5

0( ) 10f x dx

7

5( ) 3f x dx

5

0( ) 4g x dx

1)

7

0( )f x dx 2)

0

5( )f x dx

3)7

54 ( )f x dx

5

3( ) 2g x dx

−0

5

𝑓 (𝑥 )𝑑𝑥=−10

0

5

𝑓 (𝑥 )𝑑𝑥+5

7

𝑓 (𝑥 )𝑑𝑥=10+3=13

45

7

𝑓 (𝑥 )𝑑𝑥=4 (3 )=12

Examples: (cont.)

GIVEN: 5

0( ) 10f x dx

7

5( ) 3f x dx

5

0( ) 4g x dx

4)

5)

5

3( ) 2g x dx

3

3( )g x dx

3

0( )g x dx

0

0

5

𝑔 (𝑥 )𝑑𝑥−3

5

𝑔 (𝑥 )𝑑𝑥=−4−2=−6

Properties of Definite Integrals

 

Distance  

A b h

D r t

* Think rectangles

(min)( ) ( ) (max)( )c

af b a f x dx f b a

G) If f(min) is the minimum value of f(x) and f(max) is the maximum value of f(x) on the closed interval [a,b], then

 

a c b

Example:

1 2

0sin( )x dx

Show that the integral cannot possibly equal 2.

Show that the value of lies between 2 and 3 1

08x dx

sin (1)2

sin (0 )2

0<0

1

sin (𝑥 )2≤1∴𝑐𝑎𝑛𝑛𝑜𝑡=2

√0+8=2√2√1+8=32√2≤

0

1

√𝑥+8𝑑𝑥≤3

∴𝑚𝑢𝑠𝑡 𝑙𝑖𝑒𝑏𝑒𝑡𝑤𝑒𝑒𝑛2𝑎𝑛𝑑3

AVERAGE VALUE THEOREM (for Integrals)

Remember the Mean Value Theorem for Derivatives.

( ) ( )( ) ( )

F b F aF c f c

b a

And the Fundamental Theorem of Calculus

( ) ( ) ( )b

af x dx F b F a

Then:

( )( )

b

af x dx

f cb a

1( )

b

a

f x dxb a

a c b

1

4

𝑥2𝑑𝑥1

4−1

( )( )

b

af x dx

f cb a

1( )

b

a

f x dxb a

𝑥3

3 |41643−

13=

633 ¿21

( 13 )21=7

AVERAGE VALUE THEOREM (for Integrals)

( )( )

b

af x dx

f cb a

f (c)f (c) is the average of the function under consideration

i.e. On the velocity graph f (c)is the average velocity (value).

c is where that average occurs.

AVERAGE VALUE THEOREM (for Integrals)

( )( )

b

af x dx

f cb a

f (c)f (c) is the average of the function under consideration

NOTICE: f (c) is the height of a rectangle with the exact area of the region under the curve.

( ) ( )b

af c b a f x dx

Method:

Find the average value of the function

on [ 2,4].

2( ) 2 1f x x x

14−2

2

4

(𝑥2+2 𝑥+1 )𝑑𝑥

12 ( 𝑥

3

3+2

𝑥2

2+𝑥)|42

12 ( 64

3+16+4)− 1

2 ( 83+4+2)

12 ( 56

3+14)=1

2 ( 56+423 )

12 ( 98

3 )=986

= 493

Example 2:

A car accelerates for three seconds. Its velocity in meters

per second is modeled by on

t = [ 1, 4].

Find the average velocity.

2( ) 3 2v t t t

14−1

1

4

( 3 𝑡2−2 𝑡 )𝑑𝑡

13 (3

1

4

𝑡 2𝑑𝑡−21

4

𝑡𝑑𝑡 )13 (3 𝑡

3

3 )− 13 (2

𝑡 2

2 )|41𝑡3

3−𝑡 2

3 |41

( 643−

163 )−( 1

3−

13 )

483−

03=

483

=16

Last Update:

• 01/27/11

• Assignment: Worksheet

Example 3 (AP):At different altitudes in the earth’s atmosphere, sound travels at different speeds The speed of sound s(x) (in meters per second) can be modeled by:

4 341, 0 11.5

295 11.5 22

3278.5 22 32

4( )3

254.5 32 5023

404.5 50 802

x x

x

x xs x

x x

x x

Where x is the altitude in kilometers. Find the average speed of sound over the interval [ 0,80 ].

SHOW ALL PROPERTY STEPS .