4033-properties of the definite integral (5.3)
DESCRIPTION
4033-Properties of the Definite Integral (5.3). AB Calculus. Properties of Definite Integrals. Think rectangles Distance. A) B) C) D). f (x). a to a nowhere. a dx b. rectangle. Opposite direction. Constant multiplier. Properties of Definite Integrals. Think rectangles - PowerPoint PPT PresentationTRANSCRIPT
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
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
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 .