O p tim ization P rob lem s - H om ew ork · D im ensions are 16.402 in. w ide by 8.401 in. tall ... 8!)64!A;5K!&+!)64!,*041 ;'K!;)!;,A440!&8!M!844)[,49 ... ic O p tim ization P rob

www.MasterMathMentor.com - AB Solutions - 111 - Stu Schwartz

Optimization Problems - Homework

1. Find two numbers whose sum is 10 for which the sum of their squares is a minimum.

S x x

S x xxx y

= + !( )= ! +

! =

= =

2 2

2

10

2 20 1004 20 0

5 5,

2. Find nonnegative numbers x and y whose sum is 75 and for which the value of xy 2 is as large as possible.

P y y

P y yy y

= !( )= !

= !

75

750 150 3

2

2 3

2

3 50 050 25

y yy x

!( ) == =,

3. A ball is thrown straight up in the air from ground level. Its height after t seconds is given bys t t t( ) = ! +16 502 . When does the ball reach it maximum heightG What is its maximum heightG

0 32 5032 50

2516

39 063

= ! +

=

= ( ) =

tt

t s t sec, ft.

4. A farmer has 2,000 feet of fencing to enclose a pasture area. The field will be in the shape of a rectangle andwill be placed against a river where there is no fencing needed. What is the largest area field that can be createdand what are its dimensionsG

River

x

y y

P x yA xyA y y

A y yy

yy x

= + =

=

= !( )= !

= !

=

= =

2 2000

2000 2

2000 20 2000 44 2000

500 1000

2

ft, ft,Area = 500,000 ft 2


5. A fisheries biologist is stocking fish in a lake. She knows that when there are n fish per unit of water, theaverage weight of each fish will be W n n( ) = !500 2 , measured in grams. What is the value of n that willmaximize the total fish weight after one season. Hint: Total Weight = number of fish • average weight of a fish.

W n n

W n nn

n

= !( )= !

= !

=

500 2

500 20 500 4

125

2

fish

6. The size of a population of bacteria introduced to a food grows according to the formula P t tt

( ) =+

600060 2

where t is measured in weeks. Determine when the bacteria will reach its maximum size. What is the

maximum size of the populationG

060 6000 6000 2

60

0 360000 600060

60

2

2 2

2

2

=+( )( ) ! ( )

+( )= !

=

=

t t t

t

tt

t - Week 8Size = 387 bacteria

7. The U.S. Postal Service will accept a box for domestic shipping only if the sum of the length and the girth(distance around) does not exceed 108 inches. Find the dimensions of the largest volume box with a squareend that can be sent.

girth = distance around this square

length

y

y

x

V xyV y y

V y yy yy y

y x

=

= !( )= !

= !

= !( )= =

2

2

2 3

2

108 4

108 40 216 120 12 18

18 in, 36 in

8. Blood pressure in a patient will drop by an amount D x( ) where D x x x( ) = !( )0 025 302. where x is the amountof drug inTected in cm3. Find the dosage that provides the greatest drop in blood pressure. What is the drop inblood pressureG

D x x x

D x xx x

x

x

( ) = !( )= !

= !

= !( )=

0 025 30

75 0250 1 5 0750 075 20

20

2

2 3

2

.

. .. .

.

, cm Drop = 100 pts3


9. A wire 24 inches long is cut into two pieces. One piece is to be shaped into a square and the other piece into acircle. Where should the wire be cut to maximize the total area enclosed by the square and circleG

24 inches

cut the wire at point x

S x x x

x x x x

x x x

= + "! +

"

#

$%

&

'(

= +!"

) = " + !

" + = ) =+ "

=

2 2

21648 5764

08

2 484

0 4 96

4 96 964

13 442.

This is a minimum - max occurs when it is all circle Let x be the point where the cut is made. Assume the square is on the left and the circle on the right. Complete the chart.

x 4 8 12 20 xArea square 1 4 9 16 x

4

2#

$%

&

'(

Area circle 1004"

2564"

1444"

164"

244

2!( )"

x

Total area 32.1 24.3 20.5 20.0

10. A designer of custom windows wishes to build a Norman Window with a total outside perimeter of 60 feet.How should the window be designed to maximize the area of the window. A Norman Window contains arectangle bordered above by a semicircle.

Dimensions are 16.402 in. wide by 8.401 in. tall

xx

y

A xy x

A x x x x

x x

x

= + "

=! !"#

$%

&

'(+ "

= ! !"

=" +

=

2 12

2 60 22

12

0 60 460

48 401

2

2

.

2 2 6060 2

2

x y x

y x x+ + " =

=! !"

11. Alaina wants to get to the bus stop as quickly as possible. The bus stop is across a grassy park, 2,000 feet eastand 600 feet north of her starting position. Alaina can walk along the edge of the park on the sidewalk at aspeed of 6 feet/sec. She can also travel through the grass in the park, but only at a rate of 4 ft/sec (dogs arewalked here, so she must move with care). What path will get her to the bus stop the fastest.

Alaina

600feet

2,000 feetBus Stop

PARK

x T x x

xx

x xx xx

=+

+!

=+

!

= +

= +

=

2

2

2

2 2

3600004

20006

04 360000

16

6 4 3600009 4 1440000

536 656. or 1463.344 ft from bus stop


12. On the same side of a straight river are two towns, and the townspeople want to build a pumping station, S,that supplies water to them. The pumping station is to be at the river’s edge with pipes extending straight tothe two towns. The distances are shown in the figure below. Where should the pumping station be located tominimize the total length of pipeG

Town 1

Town 2

1 mile

4 miles

S

4 milesx D x x x

xx

xx x

xx

xx x

xx

x xx x

x x x x x x xx x

x x

= + + ! + +

=+

+!

! +

!+

=!

! +

+=

! +

! +! + ! + = ! +

+ ! =

!( ) +

2 2

2 2

2 2

2

2

2

2

4 3 2 4 3 2

2

1 16 8 16

01

48 32

14

8 32

18 168 32

8 17 8 16 8 3215 8 16 0

5 4 3 44 0 45( ) = ) =x miles

Easierx x

x x

x

14

44 4

45

=!

= !

=

13. A physical fitness room consists of a rectangular region with a semicircle on each end. If the perimeter of theroom is to be a 200-meter running track, find the dimensions that will make the area of the rectangular regionas large as possible.

y

x

Total distance around track = 200 m

A xy x x y

A xx

x y x

x y

= = + " =

=!( )"

= =!"

=!"

="

200

m

m

4 2 200200 2

50 200 2

0 200 2 1002

14. Below is the graph of y x= !1 2. Find the point on this curve which is closest to the origin. (Remember, you needa primary equation. What is it that you wish to minimize?)

D x y

D x x

D x x x

x x

x xx

x y

= +

= + !( )= + ! +

=!( )

! +

=

= =#

$%

&

'(

2 2

2 2 2

2 4 2

2

4 2

2

1

2 1

02 1

12 1

12

12

12

12

, ,


Economic Optimization Problems - Classwork

Example 1) A trucking company has determined that the cost per hour to operate a single truck is given by C s s s( ) = ! +0 0001 0 01 1122. . where s is the speed that the truck travels. At what speed is the total cost per hour a minimumG What is the hourly cost to operate the truckG

0 0002 010002 01= !

=

. .. .

ss

s = =50 mph, Cost $111.75

Example 2) A nursery wants to add a 1,000-square-foot rectangular area to its greenhouse to sell seedlings. For aesthetic reasons, they have decided to border the area on three sides by cedar siding at a cost of $10 per foot. The remaining side is to be a wall with a brick mosaic that costs $25 per foot. What should the dimensions of the sides be so that the cost of the proTect will be minimizedG

Cedar$10 per foot

Mosiac - $25 per foot

width

length

C x x

C xx

xx

x

x= +( ) +

= +

= !

=

=

10 25

35 20000

0 35 20000

35 2000023 905

2000

2

2

.Width = 23.905 ft, Length = 41.833 ft, Cost = $1,673.32

Width Length Cedar Cost Mosaic Cost Total Cost10 100 2100 250 2350100 10 1200 2500 37001000 1 10020 25000 35020x 1000

x 10 2000x x+( ) 25x 10 252000x xx+( ) +

Example 3) A real estate company owns 100 apartments in New òrk City. At $1,000 per month, each apartment can be rented. However, for each $50 increase, there will be two additional vacancies. How much should the real estate company charge for rent to maximize its revenuesG

$50increases

Rent Apts. rented Revenue

0 1000 100 100,0001 1050 98 102,9002 1100 96 105,6003 1150 94 108,1004 1200 92 110,400

50 3500 0 0 x 1000 50+ x 100 2! x 1000 50 100 2+( ) !( )x x

R x x

R x x

= +( ) !( )= + !

1000 50 100 2

10000 3000 100 2

0 3000 20015

= !

=

xx Rent = $1,750


Example 4) A closed box with a square base is to have a volume of 1,800 in3. The material for the top and bottom of the box costs $3 per square inch while the material for the sides cost $1 per square inch. Find the dimensions of the box that will lead to the minimum total cost. What is the minimum total costG

xxy

y

xx

C x xy x y

C x xx

yx

xx

x

x

= + =

= +#

$%

&

'( =

= !

=

=

6 4 1800

6 4 1800 1800

0 12 7200

12 7200

600

2 2

22

2

3

3

Base = 8.434 in, Height = 25.303 in, Cost = $1,280.48Example 5. A telephone wire is to be laid from the telephone company to an island 7 miles off shore at a cost of $200,000 per mile along the shoreline and $300,000 per mile under the sea. How should the wire be laid at the least expensive cost if the distance along the shoreline is 12 miles. What is that costG

7miles

TelephoneCompany

x12 miles

Island C x xx

xx

xx xxx

= + + !( )

=+

!

=+

+ =

=

=

3 49 2 12

0 349

2

2 349

4 196 95 196

6 261

2

2

2

2 2

2

. Miles, Cost = $3,965,247.58

x water cost land cost total cost0 2.4 Million 2.1 Million 4.5 Million

12 4.17 Million 0 4.17 Millionx 3 492x + 2 12 !( )x 3 49 2 122x x+ + !( )

Example 6) A small television company estimates that the cost (in dollars) of producing x units of a certain

product is given by 800 04 00002 2+ +. .x x . Find the production level that minimizes the

average cost per unit.

Units Cost Average Cost100 904.20 9.04

1,000 960 .965,000 1600 .32

10,000 3300 .33

C x x

AVGx

x

= + +

= + +

800 04 00002800 04 00002

2. .

. . 0 800 00002

00002 800

2

2

=!

+

=xx

.

. x * 6325 units


Economic Optimization Problems - Homework1. The profit for Ace Advertising Co. is P s s= + !230 20 1

22 where s is the amount (in hundreds of dollars) spent

on advertising. What amount of advertising gives the maximum profitG

P s s

ss

= + !

! =

=

230 20 12

20 020

2

Amount is $2,000Profit - $25,000

2. North American Van Lines calculates charges for delivery according to the following rules.

Fuel cost bv 2

120 per hour Driver cost b $30 per hour

Find the speed v that a truck should travel in order to minimze costs for a trip of 120 miles. Hint: remember thatrate • time = distance. Make a chart of possible speeds v and total costs.

C

tt t

Ct

t

=#

$%

&

'( +

= +

1120

120 30

120 30

2

0 120 30

30 1202

2

2

=!

+

=

= )

tt

t hrs speed = 60 mph,cost = $120

3. Normally a pear tree will produce 30 bushels of pears per tree when 20 (or fewer) pear trees are planted peracre. However, for each additional pear tree planted above 20 trees per acre, the yield per tree will fall by onebushel per tree (whyG). How many trees ahould be planted per acre to maximize the total yieldG Hint: Make achart like the Apartment Housing sample problem.

xB x x

B x x

=

= !( ) +( )= + !

number of pear trees above 2030 20

600 10 2

0 10 22 10 525

= !

= ) =

xx x trees

4. Midas Muffler charges $28 to replace a muffler. At this rate, the company replaces 75,000 mufflers per weeknationally. For each additional dollar that the company charges, it tends to lose 1,000 customers a week. Foreach dollar the company subtracts from the $28, the company gains 1,000 per week. How much should Midascharge to change a muffler in order to maximize their revenueG What would that revenue beG Hint: Make achart like the Apartment Housing sample problem.

xR x x

R x xx

=

= !( ) +( )= + !

= !

number of dollars charged above $2875000 1000 28

2100000 47000 10000 47000 2000

2

2000 4700023 5x

x=

= .Charge $51.50, Revenue = $2,652,250


5. A concert promoter knows that 5,000 people will attend an event with tickets set at $50. For each dollar less inticket price, an additional 1,000 tickets will be sold. What should the price of a ticket be in order to maximizethe total receipts. Hint: Make a chart like the Apartment Housing sample problem.

xR x x

R x xx

=

= +( ) !( )= + !

= !

number of dollars less than $505000 1000 50

250000 45000 10000 45000 2000

2

2000 4500022 5x

x=

= .Charge $28.50 Revenue = $756,250

6. A travel agent is offering charter holidays in the Bahamas for college students. For groups of size up to 100, thefare is $1,000 per student. For larger groups, the fare per person decreases by $5 for each additional person inexcess of 100. Find the size of the group that will maximize the travel agent’s revenues. Hint: Make a chart likethe Apartment Housing sample problem.

xF x x

F x xx

x x

=

= +( ) !( )= + !

= !

= ) =

number of additional people above 100

Size =150 at $250 a student. Revenue is $112,500.

100 1000 5

10000 500 50 500 1010 500 50

2

7. A real estate office handles 50 apartment units. When the rent is $540 per month, all units are occupied.However, on the average, for each $30 increase in rent, one unit becomes vacant. Each occupied unit requiresan average of $36 per month for service and repairs. What rent should be charged to realize the most profitG

xR x x x

R x x xx

x x

=

= !( ) +( ) ! !( )= + ! ! +

= ! +

= ) = *

=

number of vacant units

Rent $1050 a student. 33 apts rented. Revenue is $33,462

50 540 30 36 50

27000 960 30 1800 360 960 60 3660 996 16 6 17

2

.


8. A power station is on one side of a river that is .5 mile wide, and a factory is 6 miles downstream on the otherside. It costs $6,000 per mile to run power lines overland and $8,000 per mile to run them underwater. Findthe most economical path to lay transmission lines from the station to the factory.

Power Station

Factory6 miles

0.5miles

C x xx

xx

x

x x

x x

x

= + + !( )

=+

!

=+

+ =

= ) =

=

8 25 6 6

0 825

6

3 425

9 94

16

7 94

28 9

567

2

2

2

2 2

2 2

.

.

.

. miles from power station

9. A rectangular area is to be fenced in using two types of fencing. The front and back uses fencing costing $5 afoot while the sides uses fencing costing $4 a foot. If the area of the rectangle must contain 500 square feet,what should be the dimensions of the rectangle in order to keep the cost a minimumG

C x y

C xx

xx

x

= ( ) + ( )

= +

= !

=

=

5 2 4 2

10 4000

0 10 4000

10 400020

2

2

xy

yx

=

=

500500

Front/back = 20 ft, Sides = 25 ft, Cost = $400

10. The same rectangular area is to be built, but now the builder has only $800 to spend. What is the largest area that can be fenced in using the same two types of fencing mentioned above.

A xy

A x x

A x x

x

x

=

= !#

$%

&

'(

= !

= !

=

100 54

100 54

0 100 52

40

2 10 8 800

100 54

x y

y x+ =

= !

Front/back = 40 ft, Sides = 50 ft, Area = 2,000 ft 2


Indefinite Integration - Classwork

Take a piece of notebook paper and cover up the paragraph under the chart below. Below, there are 5 terms. Writewhat you feel are the inverses of each of these terms.

term 5 1/3 boy dog hot dogits inverse nonsense nonsense nonsense nonsense nonsense

As ridiculous as the last problem is (how can you have an inverse of a hot dogG) so are they all ridiculous. Theproblem is that all of these terms aboves are nouns. Inverses refer not Tust to an opposite but to an oppositeprocess. We take inverses of verbs, not nouns. Write the inverses of these processes.

process sit down get dressed take a book home get wet go to sleepits inverse stand up get undressed take it to school dry off wake up

In mathematics, you have learned about operations on functions f . The inverse operation is denoted as f !1 (notto be confused with x!1, the reciprocal of x (remember x is a noun and f is a process which is a verb.) When everyou perform an operation and immediately perform its inverse, you will end up exactly where you started with. Wesay that f f x x! ( )[ ] =1 and also f f x x! ( )[ ] =1 . Below write some mathematical functions and their inverses.

Function Inverse Function Inverse

addition subtraction log 10x

multiplication division trig inverse trig

square square root raising to n power taking nth root

So, obviously since differentiation of functions is a process, we must have an inverse of that process. We call thatprocess antidifferentiation. For instance, we know that the derivative of y x x= 3 23 is . So it makes sense to saythat an antiderivative of 3 2x x is 3.

However, it is important to say an antiderivative of 3 2x rather than the antiderivative of 3 2x for the simple reasonthat the derivative of y x x= 3 23 is , but so is the derivative of y x= +3 2 , y x= !3 5., and y x= + "3 6 . There are

an infinite number of functions whose derivative is 3 2x . So when we go backwards to the antiderivative it isimpossible to determine which function it came from. So to cover our bets, we say that the antiderivative of 3 2xis x C3 + , where C represents a constant. We call C the constant of integration. It is important to attach the cCafter every antiderivative. What you are doing is saying that the antiderivative is a family of functions rather thanone specific function.

The process of taking antiderivatives is called integration, specifically indefinite integration because of the constant ofintegration C. So we do not have to write the word antiderivative again, we use a symbol to represent anantiderivative. That symbol is called an integral sign which is written like this: !. The way we write an integral is:

f x dx F x C( ) = ( ) ++ . The dx tells you what the important variable is when you are integrating

Tust as you need to know what the important variable is when you differentiate dydt

dydx

as opposed to #

$%

&

'(.


So, since ddx

x4 4( ) = , we will say that 4 4 dx x C+ = + and

and since ddx

x x x2 3 1 2 3+ !( ) = + , we will say that 2 3 32x dx x x C+ ( ) = + ++eust as we have derivative rules, we have a corresponding rule for integrals. Here are some basic integration rules.

Differentiation formula Integration formuladdx

C[ ] = 0 0 dx C+ =

ddx

kx k[ ] = k dx kx C + = +

ddx

kf x kf x( )[ ] = ,( )a constant can be “factored out” kf x dx k f x dx C( ) = ( ) ++ + - factor out constant

ddx

f x g x f x g x( ) ± ( )[ ] = ,( ) ± ,( ) f x g x dx f x dx g x dx C( ) ± ( )[ ] = ( ) + ( ) ++ + +

the derivative of a sum is the sum of derivatives integral of a sum is the sum of the integralsddx

x nxn n[ ] = !1 - the power rule x dx xn

Cnn

+ =+

++1

1 - the power rule reversed

Examples - find the integral of each of the following:

1) 7dx+ = +7x C 2) x dx5+ = +x C

6

63) x dx12+ = +

x C13

13

4) x x dx4 2!( )+ b x x C

5 3

5 3! + 5) t t dt3 1+ +( )+ b

t t t C

4 2

4 2+ + + 6) 3 3x dx + b

34

4x C+

7) 2 7 82x x dx! !( )+ 8) 34

53 2

5 2x x x dx+ !#

$%

&

'(+ 9) " +

"

#

$%

&

'(+ x dx1

2

372

83 2x x x C! ! +

x x x C

62

2

859 4

+ ! + "

+"+

x x C2

2

10) 1

2xdx + b

! +

1x

C 11) 4 5

3 4x xdx!

#

$%

&

'(+ b

!+ +

2 532 3x x

C 12) x dx + b 2

3

32x C+

13) 2 43 4y y dy!( )+ 14) 1 2

3

xx dx!

#

$%

&

'(+ 15) x dx" + "( )+

32

165

43

54y y C! + 2 3

51

25

3x x C! + x x C"+

" ++ " +

1

1

In taking integrals, you may have to be clever. There are only a certain set of rules and if an integration problemdoesn’t fit one of the rules, you may have to change the expression so that it does. I call this a “bag of tricks.”

Trick 1 - multiply out, then integrate Trick 2 - Split into individual fractions, then integrate

16) 2 3x dx!( )+ 2 17) x x

xdx

2

43 1+ ++ 18)

2 5 3 2x xx

dx!( ) +( )+

4 12 9

43

6 9

2

32

x x dx

x x x C

! +( )

! + +

+

x x x dx

x x xC

! ! !+ +( )!

! ! +

+ 2 3 4

2 3

3

1 32

1

6 11 10

125

223

20

32

12

12

52

32

12

x x x dx

x x x C

! !( )! ! +

!+


We took derivatives of trig functions earlier in the year. So, natually, we should be able to go backwards and takeintegrals involving trig functions.

Differentiation formula Integration formuladdx

x xsin cos[ ] = cos sinx dx x C + = +

ddx

x xcos sin[ ] = ! sin cosx dx x C + = ! +

ddx

x xtan sec[ ] = 2 sec tan2 x dx x C + = +

ddx

x x xcsc csc cot[ ] = ! csc cot cscx x dx x C + = ! +

ddx

x x xsec sec tan[ ] = sec tan secx x dx x C + = +

ddx

x xcot csc[ ] = ! 2 csc cot2 dx x C+ = ! +

Differentiation and integration using the sine and cosine functions occur in calculus all the time and studentsalways get confused with signs. A good way to remember is to follow this chart.

Differentiation Integration

sin

cos

sin

cos

sin

-

-

!

-

!

-

When you differentiate, you follow the chart down.

When you integrate, you follow the chart up.

Examples - find the integral of each of the following:

20) 4sin x dx + 21) !+ 2

3cos x dx 22)

52cos x

dx +

! +4cos x C !

+23

sin x C 5

5

2sec

tan

x dx

x C

++

23) 4 9cos sinx x dx!( )+ 24) !#

$%

&

'(+ sin

cosxx

dx2 25) . . .2 22!( )+ csc d

4 9sin cosx x C+ + !

! +

+ sec tan

sec

x x dx

x C

. . .

..

2 2

3

2

32

!( )

+ +

+ csc

cot

d

C


If you were given the statement that dydx

x= 4 , we can cross multiply to get dy x dx= 4 . We can now

integrate each side of the equation to get dy x dx+ += 4 . From there, we can solve for y.

y C x C y x C+ = + ) = +12

222 2

The original statement dydx

x= 4 is called a differential equation (DEh). In a differential equation, you are given a

statement about the derivative of y :dydx

. òur goal is to solve for y . We have done so with the exception of the

cC, the constant of integration. So we have a general solution of the DEh. But suppose we were told that if x b 0,then y b 5. From there we can solve for C and we will thus have the specific solution of the DEh. Let’s do so.

y C C y x= + = = = +0 5 5 52 so and thus

Example 18) Solve the differential equation. Example 19) Solve the differential equation.

,( ) = ! ( ) =f x x f3 1 2 3, ,( ) = ! + ( ) = !f x x x f2 2 2 3 1,

f x x dx

f x x x C

f C C

f x x x

( ) = !( )

( ) = ! +

( ) = ! + = ) = !

( ) = ! !

+ 3 1

32

2 6 2 3 1

3 1

2

2

f x x x dx

f x x x x C

f C C

f x x x x

( ) = ! +( )

( ) = ! + +

( ) = ! + + = ! ) = !

( ) = ! + !

+ 2

32

32

2 2

32

3 9 9 6 1 7

32 7

Example 20) Solve the differential equation. Example 21) Solve the differential equation.

,,( ) = ,( ) = !( ) =f x f f2 4 1 1 2, , ,,( ) = , !( ) = ( ) = !f x x f f2 5 30 2 1, ,

,( ) =,( ) = +

,( ) = + =

= !

,( ) = !

+f x dx

f x x C

f CCf x x

2

24 8 1

72 7

1

1

1

f x x dx

f x x x C

f CCf x x x

( ) = !( )

( ) = ! +

!( ) = + + =

= !

( ) = ! !

+ 2 7

71 1 7 2

67 6

22

2

22

,( ) =

,( ) = +

, !( ) = + =

=

,( ) = +

+f x x dx

f x x C

f CCf x x

2

5 25 305

5

21

1

12

f x x dx

f x x x C

f C

C

f x x x

( ) = +( )

( ) = + +

( ) = + + = !

= !

( ) = + !

+ 2

3

2

2

2

3

5

35

2 83

10 1

413

35 41

3

Example 22) Given that the graph of f x( ) passes through the point (1, 6) and that the slope of its tangent line

at x f x x, ( )( ) + is 2 1, find f 6( ) .

,( ) = +

( ) = +( )

( ) = + +

+f x x

f x x dx

f x x x C

2 1

2 12

f CCf x x x

1 1 1 64

42

( ) = + + =

=

( ) = + +

f x x6 4 462( ) = + + =


Indefinite Integration - Homework

1. !+ 9 dx 2. !+ 5 x dx 3. 6 2+( )+ x dx

! +9x C ! +52

2x C 6 2x x C+ +

4. x dx7 + 5. x x x dx4 3 2+ !( )+ 6. 3 43 2x x dx!( )+

x C

8

8+

x x x C5 4 3

5 4 3+ ! +

34

43

4 3x x C! +

7. 23

52

12

5x x dx! +#

$%

&

'(+ 8.

34xdx

#

$%

&

'(+ 9. 2 1 7

5 3! +#

$%

&

'(+ x xdx

x x x C

6 2

954 2

! + + !

+13x

C 2 14

724 2x

x xC+ ! +

10. 5 x dx + 11. 5 5 x dx( )+ 12. xx

dx3

43

4

1!

#

$%

&

'(+

103

32x C+

256

65x C+

47

474

14x x C! +

13. 3 23 x dx + 14. x dx!( )+ 5 2 15. 4 3 2x dx!( )+ 3

95

53x C+

x x dx

x x x C

2

32

10 25

35 25

! +( )

! + +

+

4 27 54 36 8

4 274

18 18 8

27 72 72 32

3 2

43 3

4 3 2

x x x dx

x x x x C

x x x x C

! + !( )

! + ! +#

$%

&

'(

! + ! +

+

16. x x

xdx

3 4 1! !+ 2 3 17. t t dt2 23+( )+ 18.

3 2 2xx

dx!( )+

12

2 12

12

2 14

2 3

2

! !#

$%

&

'(

+ + +

+ x xdx

xx x

C

t t t dt

t t t C

4 3 2

5 43

6 9

532

3

+ +( )

+ + +

+

9 12 4

185

8 8

32

12

12

52 3

212

x x x dx

x x x C

! +#

$%

&

'(

! + +

!

+


19. 3

5cos x dx + 20. 1 6!( )+ cos x dx 21.

12x

x dx!#

$%

&

'(+ sin

35

sin x C+ x x C! +6sin !

+ +1x

x Ccos

22. sec cos2 1t t dt+ +( )+ 23. sin cos2 2x x dx+( )+ 24. sinsinxxdx

1 2!+

tan sint t t C+ + + 1 dx

x C++

sincostan sec

sec

xxdx

x x dx

x C

2

+

++

Solve the following differential equations.25. ,,( ) = ,( ) = ( ) = !f x f f2 1 4 2 2, , 26. ,,( ) = ,( ) = ! ( ) =f x x f f2 2 1 3 1, ,

,( ) =,( ) = +

,( ) = + =

=

,( ) = +

+f x dx

f x x C

f CCf x x

2

21 2 4

22 2

1

1

1

f x x dx

f x x x C

f CCf x x x

( ) = +( )

( ) = + +

( ) = + + = !

= !

( ) = + !

+ 2 2

22 4 4 2

102 10

22

2

22

,( ) =

,( ) = +

,( ) = + = !

= !

,( ) = !

+f x x dx

f x x C

f CCf x x

2

2 4 15

5

21

1

12

f x x dx

f x x x C

f CC

f x x x

( ) = !( )

( ) = ! +

( ) = ! + =

=

( ) = ! +

+ 2

3

2

2

23

5

35

3 9 15 17

35 7

27. ,,( ) = ,( ) = ( ) =f xx

f f1 4 2 0 132

, , 28. ,,( ) = , "( ) = "( ) = !f x x f fcos , ,2 1

,( ) =

,( ) = ! +

,( ) = ! + =

=

,( ) = ! +

!

!

!

+f x x dx

f x x C

f CC

f x x

32

12

1

1

11

2

22 1 2

3

2 3

f x x dx

f x x x C

f CC

f x x x

( ) = ! +#

$%

&

'(

( ) = ! + +

( ) = + =

=

( ) = ! + +

!

+ 2 3

4 30 0 1

1

4 3 1

12

12

2

2

212

,( ) =,( ) = +

, "( ) = + =

=

,( ) = +

+f x x dx

f x x C

f CCf x x

cos

sin

sin

1

1

1

0 22

2

f x x dx

f x x x C

f CCf x x x

( ) = +( )

( ) = ! + +

"( ) = + " + = !

= ! ! "

( ) = ! + ! ! "

+ sin

cos

cos

2

21 2 12 2

2 2 2

2

2

2


u-Substitution - Classwork

When you take derivatives of more complex expressions, you frequently have to use the chain rule to differentiate.The integration equivalent of the chain rule is called u-substitution. u-substitution allows you integrate expressionswhich do not appear integratable.

1) x x dx2 51!( )+ Set up a u b x 2 1! Find

dudx

b 2x . Solve for du b 2x dx

12

2 12 5x x dx!( )+ òu need to manufacture your du in the original expression. So you will have to

multiply by 2 on the inside and thus multiply by 12

on the outside.

12

12 6

112

56 2

u du u xC+ =

#

$%

&

'( =

+( )+ Now change everything to u. Now integrate in terms of u.

Finally, change back to the variable x and add C.

2) 3 2 4x dx!( )+ 3) 5 2x dx!+

13 153 2

15

45

5

u du u

xC

+ =

!( )+

u xdu dx= !

=

3 23

15

215

2 5 215

12

32

32

u du u

xC

+ =

!( )+

u xdu dx= !

=

5 25

4) 4 6 12

3x dx!( )+ 5) x x dx2 2!+

46

46

35

2 6 15

23

53

53

u du u

xC

+ = /

!( )+

u xdu dx= !

=

6 16

12

12

23

23

12

32

232

u du u

xC

+ = /

!( )+

u xdu x dx= !

=

2 22

6) x x dx2 31 4!+ 7) xx

dx2 123 !

+

! =

!/

! !( )+

+112

112

23

1 418

12

32

332

u du u

xC

u xdu x dx= !

= !

1 412

3

2

14

14

32

3 2 18

13

23

223

u du u

xC

!

+ = /

!( )+

u xdu x dx= !

=

2 14

2

8) x x dx1

23

29

2+#$% &

'(+ 9) x x x dx+( ) + !+ 2 4 32

23

23 10

2

15

910

32

10

u du u

xC

+ = /

+#$% &

'(

+

u x

du x dx

= +

=

32

12

232

12

12

23

4 33

12

32

232

u du u

x xC

+ = /

+ !( )+

u x xdu x dx= + !

= +( )

2 4 32 2


10) x x dx+( ) !+ 2 4 11) xx

dx!

!+ 5

6

u u du

u u du

u u

x x C

+( )

+#

$%

&

'(

+ /

!( ) + !( ) +

+

+

6

6

25

6 23

25

4 4 4

12

32

12

52

32

52

32

u x x udu dx= ! = +

=

4 4

u u du

u u du

u u

x x C

+( )

+#

$%

&

'(

+

!( ) + !( ) +

!

!

+

+

1

23

2

23

6 2 6

12

12

12

32

12

32

12

u x x udu dx= ! = +

=

6 6

12) xx

dx2

1++ 13) cos4x dx+

u u u du

u u u

x x x C

32

12

12

52

32

12

52

32

12

2

25

43

2

25

1 43

1 2 1

! +#

$%

&

'(

! +

+( ) ! +( ) + +( ) +

!

+

u x x udu dx= + = !

=

1 1

14

14

44

cos sin

sin

u du u

x C

+ = /

+

u xdu dx=

=

44

14) 3 1 3sin !( )+ x dx 15) sin cos3 x x dx+

33

1 3!

= ! !( )

!( ) +

+ sin cos

cos

u du u

x C

u xdu dx= !

= !

1 33

u du u

x C

3 + =

+

4

44

4sin

u xdu x dx=

=

sincos

16) tan sec10 10x x dx+ 17) tan sec2 2x x dx+

110

1010

tan sec sec

sec

u u du u

x C

+ =

+

u xdu dx=

=

1010

u du u

x C

2 + =

+

3

33

3tan

u xdu x dx=

=

tansec2

18) sin cosx x dx+ 19) cos

sinxxdx

1!+

! =

!

! ( )+

+ u du u

xC

12

32

32

23

23

cos

u xdu x dx=

= !

cossin ! = !

! ! +

!

+ u du u

x C

12

122

2 1

sin

u xdu x dx= !

= !

1 sincos


u-Substitution - Homework1. x dx!+ 2 2. 2 3 11x dx+( )+

u du u x du dx

ux

C

1 2

3 2

32

2

23

2 23

+ = ! =

=!( )

+

, 12

2 3 2

12 12

2 324

11

12 12

u du u x du dx

u xC

+ = ! =

#

$%

&

'( =

+( )+

,

3. 5 1x dx!+ 4. 6 13 x dx++

15

5 1 5

15

23

2 5 115

1 2

3 2

32

u du u x du dx

ux

C

+ = ! =

#

$%&

'(#

$%&

'( =

!( )+

,

16

6 1 6

16

34

6 18

1 3

4 3

43

u du u x du dx

ux

C

+ = + =

#

$%&

'( =

+( )+

,

5. 5 3 42

3!( )+ x dx 6. dxx8 1 3!( )

+

!= ! = !

! ( )#

$%&

'( =

! !( )+

+14

5 3 4 4

14

5 35

3 3 44

2 3

5 3

53

u du u x du dx

ux

C

,

18

8 1 8

18 2

116 8 1

3

2

2

u du u x du dx

ux

C

!

!

+ = ! =

!

#

$%

&

'( =

!

!( )+

,

7. x x dx2 62+( )+ 8. 6 3 12 3x x dx!+

12

2 2

12 7

214

6 2

7 2 7

u du u x du x dx

u xC

+ = + =

#

$%

&

'( =

+( )+

,

19

6 3 1 9

6 19

23

4 3 19

1 2 3 2

3 23

32

u du u x du x dx

ux

C

+ = ! =

#

$%&

'(#

$%&

'( =

!( )+

,

9. 1 1 13

2+#

$%

&

'(#

$%

&

'(+ x xdx 10. x x dx

13

43 9+ +#

$% &

'(

8

! = + =!

!#

$%

&

'( =

! +#

$%

&

'(+

+ u du ux

dux

dx

u x C

32

4

4

1 1 1

4

1 1

4

,

34

9 43

34 9

9

12

8 43

13

94

3

u du u x du x dx

u xC

+ = + =

#

$%

&

'( =

+#$% &

'(+

,

9

11. 23

4 35

!+ x dx 12. 3 15 10 42x x x dx+( ) + ++

23

53

4 35

35

23

53

23

20 4 35

27

1 2

3 2

32

!#

$%

&

'( = ! =

!

!#

$%

&

'(#

$%&

'( =

! !#

$%

&

'(+

+ u du u x du dx

ux

C

,

32

5 10 4 2 5

32

23

10 4

1 2 2

3 2 232

x u du u x x du x dx

u x x C

+( ) = + + = +( )

#

$%&

'( = + +( ) +

+ ,


13. x x dx+( ) !+ 2 2 14. xx

dx2

4!+

u u du u x du dx x u

u u du

u ux x

C

+ +( ) = ! = = +

+( )

+#

$%&

'( =

!( )+

!( )+

++

2 2 2 2

4

25

4 23

2 25

8 23

1 2

3 2 1 2

5 2 3 2

52

32

,

u u du u x du dx x u

u u u du

u u u du

u u u

x xx

+( ) = ! = = +

+ +( )+ +( )

+#

$%&

'( + ( )

!( )+

!( )+ !

!

!

!

+++

4 4 4

8 16

8 16

25

8 23

16 2

2 45

16 43

32 4

2 1 2

2 1 2

3 2 1 2 1 2

5 2 3 2 1 2

5 2 3 2

,

(( ) +1 2 C

15. sin5x dx+ 16. cos x dx2+

15

5 5

15

55

sin ,

cos cos

u du u x du dx

u x C

+ = =

!( ) = ! +

22

12

2 22

cos ,

sin sin

u du u x du dx

u x C

+ = =

( ) = +

17. 13

82sec x dx+ 18. sin cos4 4x x dx+

13

18

8 8

13

18

824

2#

$%&

'( = =

#

$%&

'(( ) = +

+ sec ,

tan tan

u du u x du dx

u x C

14

4 4 4

14

14 2

48

48

2 2 2

sin cos sin , cos

sin cos

u u du u x du x dx

u du u x C x C

or

+

+

= =

=#

$%

&

'( = +

!+

19. cos sin3 x x dx+ 20. tan secx x dx2+

! = = !

!#

$%

&

'( =

!+

+ u du u x du x dx

u x C

3 cos , sin

cos4 4

4 4

u du u x du x dx

u x C

+ = =

= +

tan , sec

tan

2

2 2

2 2

21. cos sin6 6x x dx+ 22. sincosxx

dx4 3!( )

+

!= = !

! #

$%&

'(( ) =

! ( )( )+

+16

6 6

16

23

69

32

u du u x du x dx

ux

C

1 2

3 2

cos , sin

cos

u du u x du x dx

ux

C

-3

-2

+ = ! =

!=

!

!( )+

cos , sin

cos

4

21

2 4 2


Sigma Notation - Classwork

We will switch gears for a section and learn a completely different type of problem. It will be apparent within afew sections why we are seemingly learning something unrelated to integration.

Suppose you were asked to find the sum of the first 5 terms of the following sequence:

1 c 2 c 4 c j b kkkkkkkk How did you arrive at the answerG kkkkkkkkkkkkkkkkkkkkkkkkkkkk

The problem with writing such addition problems with the ellipsis (j) , is that the rule for each term is notapparent. We introduce notation called sigma notation for such problems using the Greek letter sigma ".

The sum of n terms a a a an1 2 3+ + + +... is written as

aii

n

=

01

where i is the index of summation and ai is the ith

term of the sum. Note that sigma notation does not help you to calculate the sum, only to write the sum.

Examples - Find the following sums.

1) 3 241

8

i=0 = 2. i

i=0 =

1

6

21 3.

jj

2

1

7

140=

0 = 4. kk

3

2

3

27=!

0 =

Since

aii

n

=

01

represents a summation of numbers, we can apply basic properties of addition and multiplication.

ka k aii

n

ii

n

= =

0 0=1 1

(meaning you can factor out k ) a b a bi ii

n

ii

n

ii

n

±[ ] = ±= = =

0 0 01 1 1

(write one sum as 2 sums)

Find the following sums (calculators allowed)

5) 8 2241

7

ii=0 = 6) 5 2 65

1

5

ii

![ ] ==

0 7) i ii

+( ) +( )[ ] ==

0 2 3 1601

5

8) i

i 312

1

8

=

0 =

9) iii

!=

=

0 2 11101

6

10) ii

2

2

8

1 34 114! ==

0 . 11) !( ) ==

0 1 01

100i

i

12) !( ) ==

0 1 552

0

10i

i

i

Technology: òu can use your TI-84 to generate and add terms of these sequences. To create a sequence you willuse the SEh command found in 2nd LIST OPS. The format of this command is Seq(formula in x, x, starting x,

ending x). For instance, problem 5 above 81

7

ii=0 would be Seq(8l,l,1,7). This will generate the sequence. Now to

add the terms, you use the sum command found in 2nd LIST MATH. Use Sum(Ans). òu can do this in onefell swoop: Sum(Seq(8l,l,1,7)). òu may only sum up to 999 terms.


As quick as you can, find the sum

ii=0 =

1

15

120 .

Suppose you were asked to find the sum

ii=0

1

100

. Would you add all 100 termsG There must be an easier way.

ii

= + + + + + + + + +=

0 1 2 3 50 51 98 99 1001

100

... ...

Instead of adding your terms in sequential order,add the first plus the last, 2nd and next to last, etc.Each gives a sum of 101. Altogether you have 101added 50 times or 50(101) = 5050.

There are formulas you can use to add many terms. While it is not necessary that you memorize the formulas, youwill find them extremely useful for difficult summations.

c cni

n

=

0 =1

in n

i

n

=

0 =+( )

1

12

in n n

i

n2

1

1 2 16=

0 =+( ) +( ) i

n n

i

n3

1

2 214=

0 =+( )

Examples - Find the following sums.

13) 3 3

45 462

31051

45

ii=0 =

( )14)

ii

2

1

30 30 31 616

9455=

0 =( )( )

15)

2 1

250 51 101

650

85 800

2

1

50

ii

!( )

( )( )!

=

0

,

16)

i ii

2

1

60

4 2

60 61 1216

460 61

2120

66 610

! +( )

( )( )!

( )+

=

0

,

17)

3 5

938 39 77

630

38 392

25 38

149 891

2

1

38

ii

!( )

( )( )!

( )+ ( )

=

0

,

18)

i i ii

3 2

1

20

2 220 214

20 21 416

20 112

41 440

! +( )

( )( )!

( )( )+

( )=

0

,


Sigma Notation - Homework

For each problem, determine the sum by generating each term and calculate using the calculator.

1) 3 2 511

6

kk

!( ) ==

0 2) 2 1201

60

j=0 =

3) kk

2

1

10

1 375!( ) ==

0 4) kk

!( ) ==

0 1 2852

1

10

5) i ii

+( ) !( ) ==

0 1 2 3 801

5

6) i

ii += =

=

0 11479280

5 2821

7

.

7) 4

22 8392

1

6

ii +=

=

0 . 8) !( ) ==

0 1 3043

1

8i

i

i

9) 2 23 8893

9

ii=0 = . 10) i i

i

3 2

1

5

1 135! +( )( ) ==

0

Use your formulas and calculators to calculate the values of the following.

11) 5 7651

17

ii=0 = 12) i

i

2

1

20

2870=

0 =

13) ii

2

1

20

1 2850!( ) ==

0 14) ii

+( ) ==

0 4 85252

1

25

15) i ii

3

1

30

216 690+( ) ==

0 , 16) i ii

3 2

1

30

206 770!( ) ==

0 ,

17) i i ii

3 2

1

10

2 5 3 3 550+ ! +( ) ==

0 , 18) 2 431 7953

5

30

i ii

!( ) ==

0 ,


Area Under Curve - Classwork

One of the basic problems of calculus is to find the slope of the tangent line (i.e. the derivative) at any point on thecurve. The other basic problem is to find the area under the curve, that is the area between the curve and the x-axis between any two values of x.

Below you are given a curve y f x= ( ) . Estimate what you think the area is. Then on the next three graphs, draw2 rectangles, 4 rectangles, and 8 rectangles, total the areas of each and sum them for another estimate of the totalarea under the curve.

Estimate of area b 30 2 Rectangles: 6 c 32 b 38

4 Rectangles: 3 c 3.5 c 7 c 16 b 29.5 8 Rectangles: 1.5 c 1.5 c 1.5 c 2 c 2.3 c 3.5 c 5.0 c 8 b 25.3

As you can see, as you use more rectangles, your estimate gets better but the amount of work you have to doincreases. This is the idea we bring as we start our study of area. Now let’s get more exact. Let’s do the samething with another curve, but this time you will be given the function: f x x( ) = +2 1. Our problem is to estimatethe area under the curve between x b 0 and x b 4 using “right” rectangles.


A1

A2

Estimate the area 30Area A1 A2Area

* + = +

* ( ) + ( ) =b h b h

b f b f1 1 2 2

1 22 4 42

A1A2

A3

A4

Area A1+ A2 + A3 + A4AreaArea

*

* + + +

* ( ) + ( ) + ( ) + ( ) =b h b h b h b hb f b f b f b f

1 1 2 2 3 3 4 4

1 2 3 41 2 3 4 34

Area A1 A2 A3 A4 A5 A6 A7 A8AreaArea

* + + + + + + +

* + + + + + + +

* ( ) + ( ) + ( ) + ( ) +( ) + ( ) + (

b h b h b h b h b h b h b h b hb f b f b f b f

b f b f b f

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8

1 2 3 4

5 6 7

5 1 1 5 22 5 3 3 5. .

. . )) + ( ) =b f8 4 29 5.

As you go through this process, several things should be apparent:n Drawing the function is not really necessary.n The more rectangles you create, the more work you have to do. It is Tust a lot of arithmetic.n In each case, the base is same allowing you to factor it out. For instance in the last case above,

b f b f b f b f b f b f b f b f

b f f f f f f f f1 2 3 4 5 6 7 85 1 1 5 2 2 5 3 3 5 4

5 1 1 5 2 2 5 3 3 5 4

. . . .

. . . .( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) =

( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( )[ ]

n The more rectangles you create, the more accurate the area should be. So it should be apparent that

True Area =12

=

0limn i

i

n

A1

Examples) Find the area under the following functions using the indicated number of rectangles:

1) f x x( ) = +3 1 1 5 on [ , ] 2. f x x( ) = +2 2 53 on [ , ] 3. f x x x( ) = !2 3 2 4 6- on [ , ]

a) 4 = 46 b) 8 = 43 a) 3 b 59 b) 6 b 53 375. a) 8 - 18 438.


Riemann Sums - Classwork

As a common example, this worksheet will use this problem. Find the area under the function f x( ) given in thepicture below from x b 1 to x b 5. What we are looking for is the picture on the right. We will look at 4techniques: right rectangles, left rectangles, midpoint rectangles and trapezoids.

First some common statements. We will use 4 rectangles or trapezoids in this worksheet but you are expected tolearn the technique for any number of rectangles or trapezoids. Obviously, if we wish 4 rectangles and the valuesof x run from 1 to 5, the base of each rectangle is 1. Here are the 4 pictures of what we are looking for.

Right rectangles Left rectangles

The height of the rectangle is on the right side. The height of the rectangle is on the left side. This will underestimate the area (this case only). This will overrestimate the area (this case only).

Midpoint rectangles Trapezoids

The height of the rectangle is in the middle. The vertical lines represent the bases of the trapezoids This ends up both over and underestimating the area. The result is a very good approximation to the area.


When we divide our picture into 4 rectangles, we have to find the base of each rectangle. In this case, since wewere interested in doing this from x b 1 to x b 5, and there were 4 rectangles, we lucked out because the base ofeach rectangle is 1. That won’t always happen. In general, let’s call the base b b.

Now let us define x x x x xn0 1 2 3, , ..., as the places on the x-axis where we will build our heights, where n representsthe number of rectangles (or trapezoids). In this case,

x x x x x0 1 2 3 41 2 3 4 5= = = = =, , , , .

In the case of left rectangles, the area will be: In the case of right rectangles, the area will be:

A bh bh bh bhA b h h h h* + + +

* + + +( )0 1 2 3

0 1 2 3

A bh bh bh bhA b h h h h* + + +

* + + +( )1 2 3 4

1 2 3 4

but since h f xi i= ( ) , we can say

A b f x f x f x f x* ( ) + ( ) + ( ) + ( )( )0 1 2 3 A b f x f x f x f x* ( ) + ( ) + ( ) + ( )( )1 2 3 4

so, in the specific case above

A f f f f* ( ) + ( ) + ( ) + ( )( )1 1 2 3 4 A f f f f* ( ) + ( ) + ( ) + ( )( )1 2 3 4 5 so in general:

A b f xii

n

* ( )=

!

00

1

A b f xii

n

* ( )=

01

These are called Riemann Sums.

In the case of midpoint rectangles, you have to find the midpoint between your x x x x xn0 1 2 3, , ...,The midpoint between any two x values is their sum divided by 2, so you will use:

A b fx x

fx x

fx x

fx x

*+( )#

$%

&

'(+

+( )#

$%

&

'(+

+( )#

$%

&

'(+

+( )#

$%

&

'(

3

45

6

78 0 1 1 2 2 3 3 4

2 2 2 2

In our case, A f f f f*+( )#

$%

&

'(+

+( )#

$%

&

'(+

+( )#

$%

&

'(+

+( )#

$%

&

'(

3

45

6

781

1 22

2 32

3 42

4 52

or A f f f f* ( ) + ( ) + ( ) + ( )( )1 1 5 2 5 3 5 4 5. . . .

For trapezoids, remember that area A b b= / / +( )12 1 2height . That is when the trapezoid looks like this:

b1

b2

hSince our traezoids are on their sides, we will say

A h h= / / +( )1

2 1 2base

So, the total area A b f x f x f x f x f x f x f x f x* ( ) + ( )( ) + ( ) + ( )( ) + ( ) + ( )( ) + ( ) + ( )( )[ ]12 0 1 1 2 2 3 3 4

or, in our case A b f x f x f x f x f x b f f f f f* ( ) + ( ) + ( ) + ( ) + ( )[ ] = ( ) + ( ) + ( ) + ( ) + ( )[ ]12

2 2 2 12

1 2 2 2 3 2 4 50 1 2 3 4

In general, the trapezoidal rule: A b f x f x f x f x f x f xn n* ( ) + ( ) + ( ) + ( ) + + ( ) + ( )[ ]!

12

2 2 2 20 1 2 3 1 ...


Let’s try one: Let f x x( ) = !2 3. We want to find the area under the curve using 8 rectangles/trapezoids fromx b 2 to x b 6. First, let’s draw it. Note that the curve is completely above the axis. If it dips below, the methodchanges slightly.

The drawing of the curve is helpful, but not necessary.

Since there are 8 rectangles, and we are finding the area between x b 2 and x b 6, the base is .5

Let’s complete the chart:i xi f xi( )0 2.0 1.001 2.5 3.252 3.0 6.003 3.5 9.254 4.0 13.005 4.5 17.256 5.0 22.007 5.5 27.258 6.0 33.00

So, the right rectangle formula gives . . . . . .5 3 25 6 9 25 13 17 25 22 27 25 33 65 5+ + + + + + +( ) = the left rectangle formula gives . . . . . .5 1 3 25 6 9 25 13 17 25 22 27 25 49 5+ + + + + + +( ) =

the trapezoid formula gives . . . . . .52

1 2 3 25 2 6 2 9 25 2 13 2 17 25 2 22 2 27 25 33 57 5+ / + / + / + / + / + / + / +( ) =Note that the chart will not give you the midpoint formula. Let’s do it here:

3 25 9 25 17 25 27 25 57. . . .+ + + =

The calculator can generate this chart. Let’s use right rectangles. Go to STAT EDIT and clear out L1 and L2.

Place your xi in L1. It will look like this: Now L2 contains f xi( ) . Since your function is in `1, use

Now, you want to sum your L2 list and multiply it by your base which is .5. So go to your home screen and use:

òu will find the SUM command in your LIST MATH menu.

How do you adTust this for left rectanglesG replace the last data point 6 with the first data point 2


Interpretation of Area

1. A car comes to a stop 5 seconds after the driver slams on the brakes. While the brakes are on, the followingvelocities are recorded. Estimate the total distance the car took to stop.

Time since brakes applied (sec) 0 1 2 3 4 5Velocity (ft/sec) 88 60 40 25 10 0

d * + + + + +( ) =12

88 120 80 50 20 0 179 ft

2. òu Tump out of an airplane. Before your parachute opens, you fall faster and faster. òur accelerationdecreases as you fall because of air resistance. The table below gives your acceleration a (in m/sec2) after tseconds. Estimate the velocity after 5 seconds.

t 0 1 2 3 4 5a 9.81 8.03 6.53 5.38 4.41 3.61

v * + + + + +( ) =12

9 81 16 06 13 06 10 78 8 82 3 61 31 06. . . . . . . ft/sec

3. Cedarbrook golf course is constructing a new green. To estimate the area A of the green, the caretaker drawsparallel lines 10 feet apart and then measures the width of the green along that line. Determine how manysquare feet of grass sod that must be purchased to cover the green if

a) The caretaker is lazy and uses midpoint rectangles to calculate the area.b) The caretaker uses left rectangles to calculate the area.c) The caretaker uses right rectangles to calculate the area.d) The caretaker uses trapezoids to calculate the area.

Width in feet 0 28 50 62 60 55 51 30 3

a) midpoints:

A A* * + + +( ) =20 28 62 55 30 3500 ftNote that you cannot assume that halfway between the first two givenwiths, the width will be 14. You can only use given data.

2

b) left: A * + + + + + +( ) =10 28 50 62 60 55 51 30 3 360, ft 2

c) right A * + + + + + + +( ) =10 28 50 62 60 55 51 30 3 3 390, ft 2

d) trapezoids:

A * + + + + + + +( ) =102

56 100 124 120 110 102 60 3 3 375, ft

Note that the trapezoid answer is the average of the left and right Riemannsums. This will always be true if the base is constant.

2


Riemann Sums - Homework

For each problem, approximate the area under the given function using the specified number of rectangles/trapezoids. òu are to do all 4 methods to approximate the areas.

o Function Interval Number Left

Rectangles

Right

Rectangles

Midpoint

Rectangles

Trapezoids

1 f x x x( ) = ! +2 3 4 p1,4q 6 9.125 12.125 10.438 10.625

2 f x x( ) = p2,6q 8 7.650 8.168 7.914 7.909

3 f x x( ) = 2 p0,1q 5 1.345 1.545 1.442 1.445

4 f x x( ) = sin p0,#q 8 1.974 1.974 2.013 1.974

5. Roger decides to run a marathon. Roger’s friend eeff rides behind him on a bicycle and clocks his pace every 15minutes. Roger starts out strong, but after an hour and a half he is so exhausted that he has to stop. The dataeeff collected is summarized below. Assuming that Roger’s speed is always decreasing, estimate the distancethat Roger ran in a) the first half hour and b) the entire race. (Trapezoids)

Time spent running (min) 0 15 30 45 60 75 90Speed (mph) 12 11 10 10 8 7 0

a. d * + +( ) =. .252

12 22 10 5 5 miles b. d * + + + + + +( ) =.252

12 22 20 20 16 14 0 13 miles

6. Coal gas is produced at a gasworks. Pollutants in the air are removed by scrubbers, which become less and lessefficient as time goes on. Measurements are made at the start of each month (although some months wereneglected) showing the rate at which pollutants in the gas are as follows. Use trapezoids to estimate the totalnumber of tons of coal removed over 9 months.

Time (months) 0 1 3 4 6 7 9Rate pollutants are escaping(tons/month)

5 7 8 10 13 16 20

T * +( ) + +( ) + +( ) + +( ) + +( ) + +( ) =. . . .5 5 7 1 7 8 5 8 10 1 10 13 5 13 16 1 16 20 103 5 tons

7. For 0 $ t $ 1, a bug is crawling at a velocity v, determined by the formula vt

=+1

1 , where t is in hours and v is

in meters/hr. Find the distance that the bug crawls during this hour using 10 minute increments.

d v v v v v v v*#

$%&

'( ( ) +

#

$%&

'(+

#

$%&

'(+

#

$%&

'(+

#

$%&

'(+

#

$%&

'(+ ( )

3

45

6

78=

12

16

0 2 16

2 26

2 36

2 46

2 56

1 0 695. meters

8. An obTect has zero initial velocity and a constant acceleration of 32 ft sec2 . Complete the chart to find the velocity at these specified times. Then determine the distance traveled in 4 seconds. v t= 32

t sec( ) 0 .5 1 1.5 2 2.5 3 3.5 4

v ft sec( ) 0 16 32 48 64 80 96 112 128

d *#

$%&

'( + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) +[ ] =1

212

0 2 16 2 32 2 48 2 64 2 80 2 96 2 112 128 256 feet

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