1

if 11

nn xx dx C n

n

x xe dx e C ln | |

dxx C

x

Current circulation is 3000 per month. Growth rate is C’(t)= . Find C(t).

• C(t) =

• C(0) = c = C(t) =

'( )C t dt 1

24 5t dt 4 5 t

3

2

4 532

tt c

3

210

4 30003

t t

3

210

43

t t c

3000

1( )( ) '( ) if 1

1

nn g x

g x g x dx C nn

( ) ( )'( )g x g xe g x dx e C '( )

ln | ( ) |( )

g xdx g x C

g x

Fundamental Theorem of Calculus

If f is continuous on [a, b] and F is any antiderivative of f, then

( )b

af x dx ( ) ( )F b F a

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?

7.0

0.1

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

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)]

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

[5x- 0.5 (2x-2)3/3] [10 - 0.5 (8/3)] - [0 - 0.5 (-8/3)]=7.33333

0.1

20|

1 8 1 8[10 ] [0 ]

2 3 2 3

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

( ( ) ( )) (m ) ( )libn

i i in i a

f g f xx x x dxg x

0

21

3x

x

x x dx

.

A. .

B. .

C. ..

0

21

3x

x

x x dx

4 3

10

1 1|

4 3 4 3

x x

3 410

1 1|

3 4 3 4

x x

2 102 3 | 2 3x x

Area =

= 1/12

( ) ( )b

a

f x g x dx

0

21

3x

x

x x dx

3 4

10|3 4

x x 1 1

3 4

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

3 11( ) |x x

311( ) |

3

xx

1 12 2

1 1( 3) 2 ( 1)x dx x dx

]

2.667

0.1

311( ) |

3

xx

Find the enclosed area.

• y = x = -2 x = 2

• Area =

=

2

2

4

x

x 0 2

2 0top bottom top bottomy y dx y y dx

0

22

20

4

xdx

x

2

20

20

4

xdx

x

Find the enclosed area.

Area =

= -ln + ln

= - ln(4) + ln(8) + ln(8) – ln(4)

= ln(8/4) + ln (8/4) = 2ln(2)= ln22

= ln(4) = 1.386 sq. ft.

0

22

20

4

xdx

x

2

20

20

4

xdx

x

2 0

2( 4) |x 2 20( 4) |x

Find the enclosed area.

Area = ln(4) = 1.386 sq. ft.

= ln(8) - ln(8) = 02

22

2

4

xdx

x

2 22ln( 4) |x

Example 4• Find the area bounded by y2 = 2x + 6

and y = x – 1.• Solve for x = 0.5 y2 – 3 and we see a

parabola opening to the right with the vertex at (-3, 0).

• Replacing y by (x-1) gives (x-1)2=2x + 6• x2 - 2x + 1 = 2x + 6• x2 - 4x – 5 = (x - 5)(x + 1) = 0• x = 5 or x = -1

Example 4• Find the area bounded by y2 = 2x + 6

and y = x – 1.• vertex at (-3, 0).• x = 5 or x = -1• Area =• = •

1

32 6 [ 2 6]x x dx

5

12 6 [ 1]x x dx

1

32 62x dx

5 5

1 1

12 62 1

2x dx x dx

3 322 2

1 53 1

[2 6] 1 [2 6]| |

3 32 22 2

x x xx

.

16/3 – 0 + 1/3 (64) - 12.5 + 5 –[8/3-1/2-1]

72/3 – 12 + 6 = 24 – 6 = 18

3 322 2

1 53 1

[2 6] 1 [2 6]| |

3 32 22 2

x x xx

• X2 - 2x + 1 = 2x + 6• X2 - 4x – 5 = (x - 5)(x + 1) = 0• X = 5 or x = -1• y = x – 1 so y = 4 or y = -2.

4

2 right leftx 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