Post on 29-Jan-2016
transcript
How tall am I?
• My head is 232 feet above sea level.
• My feet are 226 feet above sea level.
• My head is 2 feet above sea level.
• My feet are (-4) feet ‘above’ sea level.
How many feet tall am I?
How many feet tall am I?
7.0
0.1
What is the area of the green rectangle?
What is the area of the green rectangle?
What is the area of the green rectangle?
7.0
0.1
What is the area
below f(x), above g(x) and
between x=a and x=b ?
Set up n rectangles of width x
And height is top – bottom or f(x) – g(x)
* *
1
lim ( ( ) ( ))n
i i in i
Area x xf xg
The area of one rectangle is height times width
* *
1
lim ( ( ) ( ))n
i i in i
Area x xf xg
By the definition of the definite integral
Area = * *
1
( ( ) ( )) (m ) ( )libn
i i in i a
f g f xx x x dxg x
Example 1
• Find the area over y=(2x-2)2 and under y=5• Between x=0 and x=2• Just add up all of the red rectangles• As they slide from x=0 to x=2• The top function is . . . • Y=5• And the bottom function is . . .• Y=(2x-2)2
By the definition of the definite integral
Area = * *
1
( ( ) ( )) (m ) ( )libn
i i in i a
f g f xx x x dxg x
Example 1• Find the area over y=(2x-2)2 and under y=5• Between x=0 and x=2
22
0
5 (2 2)x dA xrea
( ) ( )x b
x a
top x bott dxomAr a xe
2
0
( ) ( )x
x
top x bott dxomAr a xe
Example 1• Find the area over y=(2x-2)2 and under y=5• Between x=0 and x=2
=5x-1/2 (2x-2)3/3
=10 - 0.5 8/3-[0 - 0.5 (-8/3)]
=10 - 4/3 - 4/3 = 10 – 8/3 = 22/3 square feet
20|
22
0
5 (2 2)x dA xrea 2 2
2
0 0
15 (2 2) [2]
2dx x dx
2 22
0 0
5 (2 2)dx x dx
Example 2
• Set the two functons equal to each other• Solve for x x2 = x3 or 0 = x3 - x2
• By factoring 0 = x2 ( x – 1 ) • so x2 =0 or x–1=0• Next we add up all of the red rectangles• From 0 to 1• Area =
Area =
= 1/12 square feet
* *
1
( ( ) ( )) (m ) ( )libn
i i in i a
f g f xx x x dxg x
0
21
3x
x
x x dx
3 4
10
1 1|
3 4 3 4
x x
The area over y=2 and under y=x2+3 between x=-1 and x=1
A. [
B. [
C. [
1 2
1( 3) 2x dx
1 2
1( 3) 2x dx
1 2
12 ( 3)x dx
The area over y=2 and under y=x2+3 between x=-1 and x=1
A. [
B. [
C. [
1 2
1( 3) 2x dx
1 2
1( 3) 2x dx
1 2
12 ( 3)x dx
.
A. [
B. [
C. [
112 |x
2 211
( 1)|
2
x
311( ) |
3
xx
1 12 2
1 1( 3) 2 ( 1)x dx x dx
.
A. [
B. [
C. [
112 |x
2 211
( 1)|
2
x
311( ) |
3
xx
1 12 2
1 1( 3) 2 ( 1)x dx x dx
]3
11( ) |
3
xx
]
2.667
0.1
311( ) |
3
xx
Area = * *
1
( ( ) ( )) (m ) ( )libn
i i in i a
f g f xx x x dxg x
/12
0
2sin(3 ) sec(3 ) tan(3 )x
x
dxxAr xea x
.
• Set the two functons equal to each other
2
sin 32sin(3 )
cos 3
xx
x
( ) ( )x b
x a
top x bott dxomAr a xe
.
2
sin 32sin(3 )
cos 3
xx
x
2
2
2sin3 cos 3 sin 3 0
sin3 [2cos 3 1] 0
x x x
x x
( ) ( )x b
x a
top x bott dxomAr a xe
3 0 34
x or x
12
0
( ) ( )
x
x
top x bottoAr a dxe m x
Area =
A. True
B. False
/12
0
2sin(3 ) sec(3 ) tan(3 )x
x
x x x dx
Area =
A. True
B. False
/12
0
2sin(3 ) sec(3 ) tan(3 )x
x
x x x dx
Review
sin cosxdx x c 1
33
sin 3 sin 3xdx x dx cos1
33x c
.
/12
0
2 1sin(3 )[ ]3
3 3
x
x
x
/12
0
2sin(3 ) sec(3 ) tan(3 )x
x
x x x dx
( ) ( )x b
x a
top x bott dxomAr a xe
/12
0
sec(3 ) tan(3 )[ ]3x x dx
/12 /12
0 0
2 1sin(3 )[3] sec(3 ) tan(3 )[3]
3 3
x
x
x x x dx
/120
2 1cos(3 ) sec(3 ) |
3 3x x
2 2 2 2 1 2 21
3(2) 3 3 3 3
Example 4• Find the area bounded by y2 = 2x + 6
and y = x – 1.• 2x = y2 – 6 = 2y + 2
y2 - 6 = 2y + 2 Solve for y.
A. y = -2 or y = -4
B. y = - 2 or y = 4
y2 - 6 = 2y + 2 Solve for y.
A. y = -2 or y = -4
B. y = - 2 or y = 4
• x = 0.5y2 - 3• x = y + 1• y2 – 6 = 2y + 2• y2 - 2y – 8 = (y – 4)(y + 2 ) = 0• y = 4 or y = -2
4
2
y
right leftyx x dy
• x = y + 1.• x = 0.5 y2 – 3 .
4
2 right leftx x dy
4 2
2( 1) (0.5 3)y y dy
4 2
20.5 4y y dy
3
6
y
2
2
y 4
24 |y 64
6
8 16 ( ) 8
62 8
32 824 6
3 6
18
First two places y=cosxand y=2 sinx cosx cross?
]A. x= /3, B. x=/3, /2
C. x=/6, /2
First two places y=cosxand y=2 sinx cosx cross?
]A. x= /3, B. x=/3, /2
C. x=/6, /2
What is the area between them?
A. .
B. .
C. .
2
6
. sin(2 ) cos( )A x x dx
2
6
. cos( ) sin(2 )B x x dx
6
2
. cos( ) sin(2 )C x x dx
What is the area between them?
A. .
B. .
C. .
2
6
. sin(2 ) cos( )A x x dx
2
6
. cos( ) sin(2 )B x x dx
6
2
. cos( ) sin(2 )C x x dx
What is the area between them?
A. .
B. .
C. .
2
6
. sin(2 ) cos( )A x x dx
2
6
. cos( ) sin(2 )B x x dx
6
2
. cos( ) sin(2 )C x x dx
]
A. ]
B. ]
C. ]
2
6
sin(2 ) cos( )x x dx
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
C x x
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
B x x
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
A x x
]
A. ]
B. ]
C. ]
2
6
sin(2 ) cos( )x x dx
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
C x x
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
B x x
/ 2
/ 6
1 |. cos(2 ) sin( )2 |
A x x
]
A. ]
B. ]
C. ]
2
6
2sin cos cos( )x x x dx
/22
/6
|. sin sin( )
|C x x
/22
/6
|. cos ( ) sin( )
|B x x
/22
/6
|. sin sin( )
|A x x
]
A. ]
B. ]
C. ]
2
6
2sin cos cos( )x x x dx
/22
/6
|. sin sin( )
|C x x
/22
/6
|. cos ( ) sin( )
|B x x
/22
/6
|. sin sin( )
|A x x
]/ 2
/ 6
1 |cos(2 ) sin( )2 |
x x
]
0.25
0.1
/ 2
/ 6
1 |cos(2 ) sin( )2 |
x x