Integral As Net Change

(1)

Slide 7- 1

(2)

Chapter 7

(3)

7.1

(4)

Quick Review

 













2

-2 2

Find all values of (if any) at which the function changes sign on the given interval.

1. cos 2 on 0,1

2. 5 6 on -5,5

3. on 0, 1

4. on -5,5

4

x

x x

e x x

 

 

(5)

Slide 7- 5

Quick Review Solutions

 













2 -2 2

Find all values of (if any) at which the function changes sign on the given interval.

1. cos 2 on 0,1

2. 5 6 on -5,5

3. on 0,

4

2, 3 always positiv

1

4. on -5,5

4 e 2, -x x x x x e x x     

(6)

What you’ll learn about



Linear Motion Revisited



General Strategy



Consumption Over Time



Net Change from Data



Work

…and why

(7)

Slide 7- 7

Example

Linear Motion Revisited

( ) 10 - 2 is the velocity in m/sec of a particle moving along

the -axis when 0 9. Use analytic methods to:

Determine when the particle is moving to the right, to the left, and stopped.

Find

v t t

x t



 

(a)

(b) the particle's displacement for the given time interval.

If (0) 3, what is the particle's final position? Find the total distance traveled by the particle.

s 

(c) (d)

When ( ) 0, the particle is moving right. This occurs when

0 5. When ( ) 0, the particle is stopped. This occurs

when 5. When ( ) 0, the particle is moving left. This

occurs when 5 9.

v t

t v t

t 

  

 

 

(8)

Example

Linear Motion Revisited





9 0 9 0 9 0 2 10

displacement ( )

10 - 2

90 -81 9

During the first 9 seconds o

t t v t dt

t dt            (b)

f motion, the particle moves 9 m to the right.

It starts at (0) 3, so its position at 9 is

New position initial position displacement 3 9.

s  t 

    (c)









9 9 0 0 5 9 0 5

Total distance ( ) 10 - 2

10 2 2 -10

v t dt t dt

t dt t dt

 

 

  

(9)

Slide 7- 9

Strategy for Modeling with Integrals

1. Approximate what you want to find as a Riemann sum of values of a continuous function multiplied by interval lengths. If ( ) is the function and [ , ] the interval, and you partition the interval

f x a b

into subintervals of length , the approximating

sums will have the form ( ) with a point in the th

subinterval.

2. Write a definite integral, here ( ) , to express the limit

of these sum

k k

b a

x

f c x c k

f x dx 



 

s as the norms of the partitions go to zero.

(10)

Example

Potato Consumption

From 1970 to 1980, the rate of potato consumption in a particular country was ( ) 2.2 1.1 millions of bushels per year, with being years since the beginning of 1970. How many bushels were consumed

t

C t t

 

from the beginning of 1972 to the end of 1975?

 

We seek the cummulative effect of the consumption rate for 2 6.

Step 1: Reimann sum: We partition 2,6 into subintervals of length and let be a time in the th subinterval. The amount consumed d

k

t

t t k

 



 

(11)

Slide 7- 11

Example

Potato Consumption





6 6

2 2

Step 2: Definite integral: The amount consumed from 2 to

6 is the limit of these sums as the norms of the partitions

go to zero ( ) 2.2 1.1t million bushels.

t t

C t dt dt

 

 

 

Step 3: Evaluate: Evaluating numerically, we obtain

(12)

Work

When a body moves a distance along a straight line as a result of the action of a force of constant magnitude

in the direction of motion, the done by the force is

.

The equation

d

F

W Fd



work

is the for work.

The units of work are force distance. In the metric system, the

unit is the or . In the U.S. customary system,

the most common unit is the



constant - force formula

newton - meter joule

(13)

Slide 7- 13

Hooke’s Law

for springs says that the force it takes to stretch or compress a spring units from its natural length is a

constant times . In symbols: , where , measured in

force units per lengt

x

x F  kx k

Hooke's Law

(14)

Example

A Bit of Work

It takes a force of 6N to stretch a spring 2 m beyond its natural length. How much work is done in stretching the spring 5 m from its natural length?

By Hooke's Law, ( ) . We know that (2) 6 2 , so

3 N/m and ( ) 3 for this spring.

F x kx F k

k F x x

  

 

Step 1: Riemann sums: We partition the interval into subintervals on each of which is so nearly constant that we can apply the constant-force formula for work. If is any point in the

th subinter

k

F

x

k val, the value of throughout the interval is

approximately ( ) 3 . The work done across the interval is

approximately 5 , where is the length of the interval. The

k k

F

F x x

x x x



(15)

Slide 7- 15

Example

A Bit of Work

5

5 5

0 0

0 2 3 2

Steps 2 and 3: Integrate: The limit of these sums as the norm of the partitions go to zero is

75

( ) 3 N m.

2

x

F x dx xdx 

(16)

7.2

(17)

Slide 7- 17

Quick Review

 





2

1. Find the area between the -axis and the graph of the given function over the given interval.

a. cos over ,

2 2

b. over 0,1

c. 3 4 1 over -1,1

2. Find the - and - coordinate

x

y x

y e

y x x

x y      _ _       2 3

s of all points where the graphs of the given functions intersect.

a. and 2

b. and 1

c. cos and

x

y x y x

y e y x

y x y x

  

(18)

Quick Review Solutions

 





2

1. Find the area between the -axis and the graph of the given function over the given interval.

a. cos over ,

2 2

b. over 0,1

c. 3 4 1 over -1,1

2. Find the

2 -1 4 - x x y x y e

y x x

e x      _ _       2

and - coordinates of all points where the graphs of the given functions intersect.

a. and 2

b. and 1

(-1,1) and (2,4) (0,1)

x

y

y x y x

y e y x

  

(19)

Slide 7- 19

What you’ll learn about



Area Between Curves



Area Enclosed by Intersecting Curves



Boundaries with Changing Functions



Integrating with Respect to

y



Saving Time with Geometric Formulas

…and why

(20)

Area Between Curves

 





 

Partition the region into vertical strips of equal width .

Each rectangle has area for some in its

respective subinterval. Approximate the area of each region with the Riemann sum

k k k

k

x

f c g c x c

f c g c



 



 



_k



x.

(21)

Slide 7- 21

Area Between Curves













If and g are continuous with ( ) ( ) throughout , ,

then the area between the curves ( ) and ( ) from

to is the integral of - from to ,

b ( ) - ( )

a

f f x g x a b

y f x y g x

a b f g a b

A _ f x g x dx



 

(22)

Example

Applying the Definition

Find the area of the region between cos and sin

from 0 to .

3

y x y x

x x 

 









3 0

3 0 sin cos

Note that cos is above sin in the given interval.

cos - sin

3 1

units squared.

2

x x

y x y x

A  x x dx







 

(23)

Slide 7- 23

Example

Area of an Enclosed Region

2

Find the area of the region enclosed by the parabola

1 and

1. y

x

y

x





 





2

Graph the curves to view the region.

The limits of integration are found by solving the equation 1 1. The solutions are -1 and 2.

Since the line lies above the parabola on -1,2 , the area integra

x   x x  x 









2 2 2 1 2 -1 3 2 2 3 2

nd is 1 1 .

2 9 units squared. 2 x x x x x

A  x x dx

(24)

Integrating with Respect to y

If the boundaries of a region are more easily described by

(25)

Slide 7- 25

Example

Integrating with Respect to y

2

Find the area of the region bounded by the curves 1

and 1.

x y y x    





 



 











2 1 2 0 1 2 0 1 0 2 3 2 3

Solve for in terms of : 1 and 1. Find the points

of intersection: -1,0 and 0,1 .

1 1 1 6 y y

x y x y x y

A y y dy

(26)

Example

Using Geometry

Find the area of the region enclosed by the graphs of 1,

1 and the -axis.

y x

y x x

   









   





3 -1 3 1 3 2 2 1 3

Find the area under the curve 1 over the interval -1,3

and subtract the area of the triangle: 1

1 2 2

2 2 10 units squared 3 x y x

A x dx

(27)

7.3

(28)

Quick Review

Give a formula for the area of the plane region in terms of the single variable .

1. a square with side length . 2. a semicircle of radius . 3. a semicircle of diameter .

4. an equilateral triangle

x

x x

x

with sides of length . 5. an isosceles triangle with two sides of length 2 and one of length .

x

(29)

Slide 7- 29

Quick Review Solutions

2 2

2

Give a formula for the area of the plane region in terms of the single variable .

1. a square with si

/ 2

de length . 2. a semicircle of radius .

3. a semicircle of diameter . / 8

4. an eq

x

x x

x

x x

x

 









2

uilateral triangle with sides of length .

5. an isosceles triangle with two sides of length 2 and one of length .

3 / 4

15 / 4

x

x x

(30)

What you’ll learn about



Volumes As an Integral



Square Cross Sections



Circular Cross Sections



Cylindrical Shells



Other Cross Sections

…and why

(31)

Slide 7- 31

Volume of a Solid

The definition of a solid of known integrable cross section

area ( ) from to is the integral of from to ,

( ) .

b a

A x x a x b A a b

V _ A x dx

 

(32)

How to Find Volumes by the Method of

Slicing

1. Sketch the solid and a typical cross section. 2. Find a formula for ( ).

3. Find the limits of integration.

4. Integrate ( ) to find the volume.

A x

(33)

Slide 7- 33

Example

Square Cross Sections

A pyramid 3 m high has congruent triangular sides and a square base that is 3 m on each side. Each cross section of the pyramid parallel to the base is a square. Find the volume of the pyramid.

1. Sketch: Draw the pyramid with its vertex at the origin and its

altitude along the interval 0 3.

Sketch a typical cross section at a point between 0 and 3.

x

(34)

Example

Square Cross Sections

2

2. Find a formula for ( ): The cross section at is a square meters on a side, so ( ) .

A x x

x A x  x

3. Find the limits of integration: The square goes from

0 to 3.

x  x 

3

3 3 2 3

0 0

0 3 3

4. Integrate to find the volume:

( ) x 9 m

V _ A x dx _ x dx _ 

(35)

Slide 7- 35

Example

A Solid of Revolution





The region between the graph of ( ) 2 cos and the -axis

over the interval -2,2 is revolved about the -axis to generate a solid. Find the volume of the solid.

f x x x x

x  





2

Revolving the region about the -axis generates a vase-shaped solid. The cross section at a typical point

is circular, with radius equal to ( ).

Its area is ( ) ( ) .

The volume of the solid is

x

x f x

A x f x

V   





2 -2 2 ( )

NINT( 2 cos , , 2, 2)

52.43 units cubed.

A x dx

x x x

 

  

(36)

Example

Finding Volumes Using

Cylindrical Shells

The region bounded by the curve , the -axis, and the

line 4 is revolved about the -axis to generate a solid. Find

the volume of the solid.

y x x

x x

 

1. Sketch the region and draw a line segment across it parallel to the axis of revolution. Label the segment's length and distance from the axis of revolution. The width of the segment is the

(37)

Slide 7- 37

Example

Finding Volumes Using

Cylindrical Shells

2. Identify the limits of integration: runs from 0 to 2.y



 



  



2 0

2 2

0

2 shell radius shell height

2 4

8

V dy

y y dy

   

 

 

(38)

Example

Finding Volumes Using

Cylindrical Shells

2

The region bounded by the curve 4 - , , and 0

is revolved about the -axis to form a solid. Use cylindrical shells to find the volume of the solid.

y x y x x

y

  

2

1. Sketch the region and draw a line segment across it parallel

to the -axis. The segment's length is 4 - . The distance

of the segment from the axis of revolution is .

y x x

(39)

Slide 7- 39

Example

Finding Volumes Using

Cylindrical Shells

2

2. Identify the limits of integration: The -coordinate of the

point of intersection of the curves 4 - and in the first

quadrant is about 1.562. So runs from 0 to 1.562.

x

y x y x

x

 



 



  



1.562 0

1.562 2

0

2 shell radius shell height

2 4

13.327

V dx

x x x dx

  

 

  

(40)

Example

Other Cross Sections

A solid is made so that its base is the shape of the region between

the -axis and one arch of the curve 2sin . Each cross section

cut perpendicular to the -axis is a semicircle whose diameter runs

x y x

x



(41)

Slide 7- 41

Example

Other Cross Sections





















2 2 0 2 2 3 2sin

The semicircle at each point has radius sin

2 1

and area ( ) sin .

2

So, sin

2

NINT sin , , 0,

2 1.570796327 2 in . 4 x x x

A x x

V x dx

(42)

Quick Quiz

Sections 7.1-7.3

-1

You may use a graphing calculator to solve the following problems.

1. The base of a solid is the region in the first quadrant bounded by

the -axis, the graph of x y  sin , and the vertical line x x 1. For this

solid, each cross section perpendicular to the -axis is a square. What is its volume?

(A) 0.117 (B) 0.285 (C) 0.467 (D) 0.571 (E) 1.571

(43)

Slide 7- 43

Quick Quiz

Sections 7.1-7.3

-1

You may use a graphing calculator to solve the following problems.

1. The base of a solid is the region in the first quadrant bounded by

the -axis, the graph of x y  sin , and the vertical line x x 1. For this

solid, each cross section perpendicular to the -axis is a square. What is its volume?

(A) 0.117 (B) 0.285

(D) 0.571

(C) 0.46

(E)

7

1.571

(44)

Quick Quiz

Sections 7.1-7.3

2

2. Let be the region in the first quadrant bounded by the

graph of 3 - and the -axis. A solid is generated when

is revolved about the vertical line -1. Set up, but do not

integrate, the def

R

y x x x

R x  



 





 



  











3 2 0 3 2 1 3 2 0 3 2 0 3 2 0

inite integral that gives the volume of this solid.

(A) 2 1 3

(B) 2 1 3

(C) 2 3

(D) 2 3

(E) 3

x x x dx

x x dx x x dx

(45)

Slide 7- 45

Quick Quiz

Sections 7.1-7.3

2

2. Let be the region in the first quadrant bounded by the

graph of 3 - and the -axis. A solid is generated when

is revolved about the vertical line -1. Set up, but do not

integrate, the def

R

y x x x

R x  



 





 



  











3 3 2 1 3 2 0 3 2 0 3 2 0 2 0

inite integral that gives the volume of this solid.

(B) 2 1 3

(C) 2 3

(D) 2 3

(E) 3

(A) 2 x 1 3x x

x x x dx

x x x d

d

x x x dx x x dx

(46)

Quick Quiz

Sections 7.1-7.3

0.2

3. A developing country consumes oil at a rate given by

( ) 20 million barrels per year, where is time measured

in years, for 0 10. Which of the following expressions

gives the amount of oi

t

r t e t

t 

 

10 0 10 0

l consumed by the country during

the time interval 0 10?

(A) (10)

(B) (10) - (0)

(C) '( )

(D) ( )

(E) 10 (10)

t r

r r

r t dt r t dt

r  

 

(47)

Slide 7- 47

Quick Quiz

Sections 7.1-7.3

0.2

3. A developing country consumes oil at a rate given by

( ) 20 million barrels per year, where is time measured

in years, for 0 10. Which of the following expressions

gives the amount of oi

t

r t e t

t    10 0 10 0

l consumed by the country during

the time interval 0 10?

(A) (10)

(B) (10) - (0)

(C) '( )

(E) (D) 1 (10) ( 0 ) t r r r

r t dt

r

r t dt



 

(48)

7.4

(49)

Slide 7- 49

Quick Review

 





2 2 2 2 3

1. Simplify the function.

a. 4 4 on 1,5

b. 1 cot on 1, 2

1

c. 1 on 4,12

4

2. Identify all values of for which the function fails to be differentiable.

a. ( ) -1

b. ( ) 3 c. ( )

x x x x

x

f x x

f x       _  _     2

1 2x x

(50)

Quick Review Solutions

 





2 ₂ 2 2

1. Simplify the function.

a. 4 4 on 1,5

b. 1 cot on 1, 2

1

c. 1 on 4,12

4

2. Identify all values of for which the function fails t

2 csc

o be differentiable.

a. ( )

4 4 x x x x x x f x x x x x      _        2 3 -1

b. ( ) 3

1 0

(51)

Slide 7- 51

What you’ll learn about



A Sine Wave



Length of a Smooth Curve



Vertical Tangents, Corners, and Cusps

…and why

(52)

Example

The Length of a Sine Wave

What is the length of the curve y  sin from x x  0 to x  2 ?





Partition 0,2 into intervals so short that the pieces of curve

lying directly directly above the intervals are nearly straight. Each arc is nearly the same as the line segment joining its

(53)

Slide 7- 53

Example

The Length of a Sine Wave

2 2 k k 2 2 2 2 2 2 2

The sum x over the entire partition approximates

the length of the curve. Rewrite as a Riemann

sum:

1

k k k k

k k k

k k k k y x y x y

x y x

x y x x                         _ _    





2

Rewrite the last square root as a function evaluated at some in the th subinterval. Use the Mean-Value Theorem for

differentiable functions to obtain the sum: 1 sin' .

Take the limit as the n

k

k k

c k

c x

  





2





2

2 2

0 0

orms of the subdivisions go to zero:

1 sin' 1 cos 7.64.

L  x dx  x dx

 

(54)

Arc Length: Length of a Smooth Curve









 

2 b

a

2

If a smooth curve begins at , and ends at , , ,

, then the length (arc length) of the curve is

1 if is a smooth function of on , ;

1 if is a smooth

a c b d a b

c d

dy

L dx y x a b

dx dx

L dy x

dy 

 

 

 _{  }

 

 

 _{  }

 





d

c function of on ,y c d .

(55)

Slide 7- 55

Example

Applying the Definition

2

Find the length of the curve y  x for 0  x 1.

 

 

2 1

0

2 , this is continous on 0,1 . Therefore,

1 2 using NINT

1.479

dy

x dx

L x

L  

 

(56)

Example

A Vertical Tangent

1 3

Find the length of the curve y  x between (-1,-1) and (1,1).

2 3

1

is not defined at = 0. There is a vertical tangent 3

at (0,0). Change to as a function of in order to make the tangent at the origin horizontal and the derivative equal to zero instea

dy

x

dx _x

x y







1

3 2

3

2

1 2

1

d of undefined.

Solve for . and 3

1 3 3.096 using NINT.

dx

y x x x y y

dy

L  y dy

  

(57)

Slide 7- 57

Example

Getting Around a Corner

Find the length of the curve y  x 1 from x  -2 to x 1.

There is a corner at -1 so neither nor can exist.

To find the length, split the curve at -1 to write the function

1 if -1,

without absolute values: 1

1 if 1.

Then,

-dy dx

x

dx dy

x

x x

x

x x

L



  



   _ _{ }



 -1





1





-2 x -1 dx 1 x 1 dx 2.5

(58)

7.5

(59)

Slide 7- 59

Quick Review





1 -1 3 0 3 4 0 2 3 1 ₃

Find the definite integral by antiderivatives and using NINT.

1.

2. cos 3.

3 4.

1

Find, but do not evaluate, the definite integral that is the limit as the norm

x

e dx xdx x x dx

x dx x         (a) (b)

 













2 s of the partitions go to zero of the Reimann sums on the closed interval 0,7 .

5. 2 1 sin

6. 2 2

3

7. 5

4

k k

k

x x x

x x x

(60)

Quick Review Solutions





 

1 -1 3 0 3 4 0 2 3 1 ₃ 1 2.

Find the definite integral by antiderivatives and using NINT.

1.

2. cos

350 3 3. 3 0.866 2 441 44.1 10 ln x 4.

x 1 14

x

e dx

xdx

x x dx

x e d e          (a

a. b.

) (b)

a. b.

 





7





0

Find, but do not evaluate, the definite integral that is the limit as the norms of the partitions go to zero of the Reimann sums on the closed interval 0,7 .

5. 2 1 sin

2.639 2 -6. 1 s in

k k x x

x x  x dx

 

  





7





2 2

2 x 2 x x 2 x 2 x dx

(61)

Slide 7- 61

What you’ll learn about



Work Revisited



Fluid Force and Fluid Pressure



Normal Probabilities

…and why

(62)

Example

Finding the Work Done by a

Force

 

Find the work done by the force ( ) sin newtons along the

1

-axis from 0 meters to meter.

2

F x x

x x x

   

 





1 2 0 1 2 0 1 cos sin 1 0 1 1 x

W x dx

(63)

Slide 7- 63

Example

Work Done Lifting

A leaky bucket weighs 15 newtons (N) empty. It is lifted from the ground at a constant rate to a point 10 m above the ground by a rope weighing 0.5 N/m. The bucket starts with 70 N of water, but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done

lifting the bucket alone; lifting the water alone; lifting the rope alone;

li (a) (b) (c)

(d) fting the bucket, water, and rope together.



 



Since the bucket's weight is constant, you must exert a force of 15 N through the entire 10-meter interval.

W= 15 N  10 m 150 N m=150 J

(64)

Example

Work Done Lifting

The force needed to lift the water is equal to the water's weight, which decreases steadily from 70 N to 0 N over the 10-m lift. When the bucket is meters off the ground, it

origina weighs: ( )

x F x 

(b)

 





10 10 0 0 2 7 70 2

l weight proportion left at

of water elevation

10

( ) 70 70 7 N.

10 The work done is

70 - 7 x x 350 J.

x x

F x x

W _ x dx _  _

(65)

Slide 7- 65

Example

Work Done Lifting

The force needed to lift the rope also varies, starting at (0.5)(10)=5 N when the bucket is on the ground and 0 N when the bucket is is at the top. The rate of decrease is constant. At elevation

(c)









10 0

10 0 2 5 -0.25

meters, the (10 - ) meters of rope

still there to lift weigh ( ) 0.5 10 - N.

0.5 10

=25 J.

x x

F x x

W _ x dx

 

 

 



(66)

Fluid Force and Fluid Pressure

In any liquid, the (force per unit area) at

depth is , where is the weight-density (weight

per unit volume) of the liquid.

p

h p wh w

(67)

Slide 7- 67

Example

The Great Molasses Flood if

1919

At 1:00 pm on January 15, 1919, a 90-ft-high, 90-foot-diameter cylindrical metal tank in which the Puritan Distilling Company stored molasses at the corner of Foster and Commercial streets in Boston’s North End exploded. Molasses flooded the streets 30 feet deep, trapping pedestrians and horses, knocking down buildings and oozing into homes. It was eventually tracked all over town and even made its way into the suburbs via trolley cars and people’s shoes. It took weeks to clean up.

(a) Given that the tank was full of molasses weighing 100 lb/ft3_{, what was the}

total force exerted by the molasses on the bottom of the tank at the time it ruptured?

(68)

Example

The Great Molasses Flood if

1919





 

3 2

2

At the bottom of the tank, the molasses exerted a constant

lb lb

pressure of 100 90 ft 9000 .

ft ft

The area of the base was 45 and the total force on the base

lb

was 9000 2025 ft

ft

p wh



 

  _ _ 

 

 

 

 

(a)

(69)

Slide 7- 69

Example

The Great Molasses Flood if

1919

2

Partition the band from depth 89 ft to depth 90 ft into narrower

bands of width and choose a depth in each one. The pressure

at this depth is 100 lb/ft . The force against each na

k

k k

y y

y p wh y



 

(b)



 



rrow

band is approximately pressure area 100 90

9000 lb.

Add the forces against all bands in the partition and take the limit

k k

y y

y y

 

  

 

90 90

89 89

as the norms go to zero. The force against the bottom foot of tank wall is:

9000 9000 2,530,553 lb.

(70)

Probability Density Function (pdf)





-A is a function ( ) with domain

all reals such that ( ) 0 for all and ( ) 1.

Then the probability associated with an interval , is ( ) .

b a

f x

f x x f x dx

a b f x dx

 



 

(71)

Slide 7- 71

Normal Probability Density Function (pdf)

 2  ₂ 2

The for

a population with mean and standard deviation is 1

( ) .

2

x

f x e  

 

 

 



(72)

The 68-95-99.7 Rule for Normal

Distributions

Given a normal curve,

68% of the area will lie within of the mean , 95% of the area will lie within 2 of the mean , 99.7% of the area will lie within 3 of the mean .

 

(73)

Slide 7- 73

Example

A Telephone Help Line

Suppose a telephone help line takes a mean of

1.5 minutes to answer a call. If the standard

deviation is 0.2, then 68% of the calls are

(74)

Example

Weights of Coffee Cans

Suppose that coffee cans marked as "8 ounces" of coffee have

a mean weight of 8.2 ounces and a standard deviation of 0.3 ounces. What percentage of all such cans can be expected to weigh

between 8 a (a)

nd 9 ounces?

What percentage would we expect to weigh less than 8 ounces? What is the probability that a can weighs exactly 8 ounces?

(b) (c)

 2

8.2 / 0.18

Assume a normal pdf will model these probabilities. 1

( )

0.3 2

Find the area under the curve of ( ) from 8 to 9: NINT( ( ), ,8,9) 0.744

x

f x e

f x f x x



 





(75)

Slide 7- 75

Example

Weights of Coffee Cans

The graph of ( ) approaches the -axis as an asymptote.

When 5, the graph of ( ) is very close to the -axis.

NINT( ( ), ,5,8) 0.252.

Expect about 25.2% of the boxes to weigh less than 8 ounc

f x x

x f x x

f x x 



(b)

es.

(76)

Quick Quiz

Sections 7.4 and 7.5

1 6

0

You should solve the following problems without using a graphing calculator.

1. The length of a curve from 0 to 1 is given by 1 16 .

If the curve contains the point (1,4), which of the followi

x  x  _  x dx

4 4

6 6 7

ng could be an equation for this curve?

(A) 3

(B) 1

(C) 1 16

(D) 1 16

y x

x

 

(77)

Slide 7- 77

Quick Quiz

Sections 7.4 and 7.5

1 6

0

You should solve the following problems without using a graphing calculator.

1. The length of a curve from 0 to 1 is given by 1 16 .

If the curve contains the point (1,4), which of the followi

x  x  _  x dx

4

6 6

4

7

ng could be an equation for this curve?

(B) 1

(C) 1 16

(D) 1 1

(A) 3

6

(E)

7

y

y x

x

y x

x

 

(78)

Quick Quiz

Sections 7.4 and 7.5

4 3

2 6 4 0

2 6 0

2 4

0

2 6 4 0

2 3 2 0

2. Which of the following gives the length of the path described 1

by the parametric equations and , where 0 2?

4

(A) 9

(B) 1

(C) 1 9

(D) 9

(E) 3

x t y t t

t t dt

t dt

(79)

Slide 7- 79

Quick Quiz

Sections 7.4 and 7.5

2 6

4 3

2 6 4 0

2 6 0

2 4

0

2 3 2

4 0

0

2. Which of the following gives the length of the path described 1

by the parametric equations and , where 0 2?

4

(A) 9

(B) 1

(C) 1 9

(E)

(D) 9

3

x t y t t

t t dt

(80)

Quick Quiz

Sections 7.4 and 7.5

3. The base of a solid is a circle of radius 2 inches. Each cross section perpendicular to a certain diameter is a square with one side lying in the circle. The volume of the solid in cubic inches is

(A) 4 (B) 4

32 (C)

3 32 (D)

3 (E) 8



(81)

Slide 7- 81

Quick Quiz

Sections 7.4 and 7.5

3. The base of a solid is a circle of radius 2 inches. Each cross section perpendicular to a certain diameter is a square with one side lying in the circle. The volume of the solid in cubic inches i

32

s (

(C) 3

A) 4 (B) 4

32 (D)

3 (E) 8



(82)

Chapter Test

2 3

1. A toy car slides down a ramp and coasts to a stop after

5 sec. Its velocity from 0 to 5 is modeled by

( ) 0.2 ft/sec. How far does it travel? Set up an

integral and evaluate it to answ

t t

v t t t

 

 

2

3 2

er the question.

2. Find the area of the region enclosed by 1 and

3 - .

3. Find the area of the region enclosed by and

. 1

y x

y x x

x y

x

  

 



(83)

Slide 7- 83

Chapter Test

4. You drove an 800-gallon tank truck from the base of Mt. Washington to the summit and discovered on arrival that the tank was only half full. You had started out with a full tank of water, had climbed at a steady rate, and had taken 50 minutes to accomplish the 4750-ft elevation change. Assuming that the water leaked out at a steady rate, how much work was spent in carrying the water to the summit? Water weighs 8 lb/gal. (Do not count the work done getting you and your truck to the top.)

(84)

Chapter Test

(85)

Slide 7- 85

Chapter Test

7. A solid lies between planes perpendicular to the -axis

at 0 and at 6. The cross sections between the planes

are squares whose bases run from the -axis up to the

curve 6. Find the volume

x

x x

x

x y

 

(86)

Chapter Test

(87)

Slide 7- 87

Chapter Test Solutions

2 3

1. A toy car slides down a ramp and coasts to a stop after

5 sec. Its velocity from 0 to 5 is modeled by

( ) 0.2 ft/sec. How far does it travel? Set up an

integral and evaluate it to answ

t t

v t t t

 

 

2

3 2

er the question.

2. Find the area of the region enclosed by 1 and

3 - .

3. Find the area of the region enclosed by

10.417 ft and . 9 2 1.2 1 956 y x y x

y x x

(88)

Chapter Test Solutions

4. You drove an 800-gallon tank truck from the base of Mt. Washington to the summit and discovered on arrival that the tank was only half full. You had started out with a full tank of water, had climbed at a steady rate, and had taken 50 minutes to accomplish the 4750-ft elevation change. Assuming that the water leaked out at a steady rate, how much work was spent in carrying the water to the summit

22,800,000 ft-l

? Water weighs 8 lb/gal. (Do not count the work done getting you and your truck to the top.)

5. If a force of 80 N is required to hold a spring 0.3 m beyond its unstressed length, h

b

(89)

Slide 7- 89

Chapter Test Solutions

6. The vertical triangular plate shown below is the end plate of a feeding trough full of hog slop, weighing 80 pounds per cubic foot. What is the force against t

426.67 lbs

he plate?



7. A solid lies between planes perpendicular to the -axis

at 0 and at 6. The cross sections between the planes

are squares whose bases run from the -axis up to the

curve 6. Find the volume

x

x x

x

x y

 

  of the solid.

14.4

(90)

Chapter Test Solutions

8. A researcher measures the lengths of 3-year-old yellow perch in a fish hatchery and finds that they have a mean length of 17.2 cm with a standard deviation of 3.4 cm. What proportion of 3-year-old yellow perch raised under similar conditions can be expected to reach a length of 20

0.2

cm

051

or mor

(2

e?

0.5%)