Aplikasi Integral

5/27/2018 Aplikasi Integral

1/85

CHAPTER SIX:

APPLICATIONS OF THEINTEGRALReleased by Krsna Dhenu

February 03, 2002

Edited on October 7, 2003

Hare Krsna Hare Krsna Krsna Krsna Hare Hare

Hare Rama Hare Rama Rama Rama Hare Hare

Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of Math and

Scien

ce)

KRSNA CALCULUS PRESENTS:


2/85

HARI BOL! WELCOME BACK!

Jaya Sril a Prabhupada! Jaya Sri Sri Gaura Nitai! Jaya Sri Sri RadhaVijnanesvara!

I hope you are spiritually and materially healthy. This chapter is where we start todepart the AB calculus students and College Calculus I.

If you take AB Calculus or Calculus I, then I urge you to please check out my veryvery final slide show. Thank you and much love.

If you are a BC student, however, this is your middle point of the course. ForCalculus II students, welcome to Krsna Calculus . I will enjoy having you as mystudents.

Calculus II entrants: please check out chapter 4 and 5 extensively.

Everyone: This chapter deals highly with other fancy things you could do with theintegral. First, finding general area, and techniques will be discussed. Then the bigpart of the chapter on volume of a solid of revolution. This may seem hard, but it isreally not.

Other applications include arc length AND surface area.

When we start talking about cylindrical shells, AB and Calculus I students arecompleted with this course and are urge to go and take a review slide show forthem.


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AREA

In chapter 4, we introduced integration in

terms of area between the curve and the x-axis.

This chapter, we will discuss area between two

curves and methods.

Also, we will talk about integral and derivative

properties in terms of their graphs.


4/85

AREA BETWEEN TWO CURVES

The area between two curves is really simple if you really lookat it. Lets consider f(x)=6-x2. Also consider g(x)=x.

Find the area between these two curves.

Here is a graphical view of the scenario. (Magenta region)

ALWAYS DRAW A PICTURE TO SEE HOW THEREGION WILL LOOK LIKE!!!!!!!


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STEP 1: FIND THE POINTS OF

INTERSECTION

In order to see where the

region begins andends, we must find thex limits to see where we

can actually computearea.

Solve for the x limits

The two limits of

integration are x=2 and

x=-3

3

2320

606

2

2

x

xxx

xxxx


6/85

STEP 2

Compute the area of both functions.

Subtract the areas.

6

125

2

5

3

55

2

5

2

92

2

1

3

559

3

28

3

166

2

3

22

3

2

3

3

2

32

xxdx

xxdxx

THIS IS OUR AREA BETWEEN THE TWO CURVES


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GEOMETRICALLY

This is a shaded region style problem. Let the maroon be the regionbetween the parabola and the x axis. with the limits of -3 and 2. Let thelight blue be the region between the line and the x axis with the samelimits. Let the purple represent overlapping regions.

We must subtract the two areas of the regions (that is Af(x)-Ag(x)) to get

rid of the overlapping areas (i.e. purple region).


8/85

FORMULA FOR FINDING AREA

BETWEEN TWO CURVES

If f(x)>g(x), then

b

adxxgxfA )()(

Top Function Bottom Function

b

a bottomtop dxyyA


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If f(x)>g(x) top vs. bottom

function???

Very simple..

Since the

parabola, f(x), ison top of the

line, g(x),

therefore,

f(x)>g(x).

TOP

BOTTOM


10/85


We just tackled one of the three types of area

between curves problems. They are

1) Area between two y(x) functions. (We just

did)

2) Area between a y(x) and x(y)

3) Area between two x(y) functions.


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AREA BY SUBDIVIDING

Sometimes, it is necessary to split the regionup into two parts before integrating. Thissituation often occurs when you have y as a

function of x and x as a function of y. Consider this example: Find the region

bounded by y=x and x=-y2+8.

First thing you do is to draw a picture to seehow the region will look like. Then solve fortheir intersecting points.


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DRAW A PICTURE!!!

It helps to draw a picture to get an understanding on

what is going on. The red line is y=x and the blue

parabola is x=-y2+8.


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THE INTERSECTING POINTS

ARE (-8,-4) and (4,2)

2

4

4

8

048

0324

432

844

14

8

8

41

21

2

2

2

2

2

22

y

x

y

x

xx

xx

xx

xx

xy

yx

xyxy


14/85

INTEGRATING

It is very important to

let all the functions be

in the form y(x).

Therefore, solve for y. Using the formula given

previously for finding

the area between curves,

apply that formula using-8 and 4 as your limits.

376

3

80

3

4

83

2

4

1

82

1

8

2

1

4

8

2/32

4

8

xx

dxxx

xy

xy


15/85

WHAT AREA DID WE JUST

DO???

Notice how much area we did accumulate using the integral.

Also consider how much area we have left to take into

account.

The purple is the area we jus did. (76/3) The green is the areawe have left to do.


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THE GREEN AREA

Since we know that the area of the green is the

area of that sideways parabola, and since we

also know that the area above the x-axis from

[4,8] equal to the area below the x-axis. So in

effect, we can double the area of that to take

into account the top and the bottom of the x-

axis.


17/85

CALCULATION OF AREA

Notice the 2 outside theintegral. We have todouble the area.

Fundamental theorem.

Add the two areas.

THE TOTAL AREAOF THE REGION IS36!!

363

32

3

76

3

3208

3

4

82

8

4

2/3

8

4

x

dxx


18/85

INTEGRATING A FUNCTION

WITH RESPECT TO Y.

At times, we may have to curves that there is

no top or bottom y, rather there are leftand right curves. By definition, these curves

would not be functions of x, since they would

fail to honor the vertical line test (the test that

determines if a relation is actually a function).

But these functions, on the hand, seem simplerif they were in terms of y.


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INTEGRATING WITH RESPECT

TO Y.

To calculate an area, using the right being x=f(y) and

the left one being x=g(y) function, between limits of

y (c and d)then

d

cdyygyfA )()(

d

c leftright dyxxA

OR


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EXAMPLE

Lets do the previous problem using this method of integrating with respectto y.

The graphs are y=x and x=-y2+8.

First, always always always!! Look at a picture of this! This will help you!

The red is the line. The blue is the parabola. The purple is the region.

LEFT

RIGHT


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SOLVE FOR INTERSECTING

POINTS

2

4

4

8

048

0324

432

84414

8

8

41

21

2

2

2

2

2

22

y

x

y

x

xx

xx

xx

xx

xy

yx

xyxy

SAME SLIDE AS BEFORE,

HOWEVER, CONSIDER THAT

THE Y VALUES ARE NOW

MORE RELEVANT, SINCE WEARE INTEGRATING WITH

RESPECT TO Y!!!!

THIS MEANS THE INTEGRAL

WILL HAVE THE LIMITS

BETWEEN y=-4 AND y=2!!!


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SETTING UP THE INTEGRAL

First solve everything for x before putting the f(y)

and g(y) inside. You will get x=2y AND x=8-y2.

363

108

3

80

3

28

83

1

82

28

)()(

2

4

23

2

4

2

2

4

2

yyy

dyyyA

ydyyA

dyygyfAd

c

Simpler, eh???


23/85


In reality, there is no set formula to finding areabetween curves. Its important to know what to doand how you go about doing it. For example, younoticed that integrating a function in terms of y wassimpler than taking the region, cutting it, and findingareas that way.

But sometimes, you will see that integrating withrespect to y can be a hassle. Sometimes, integratingwith x would be better.

But in either case, the general formula stays the same:top/rightbottom/left. (for x and y respectively)


24/85

EXAMINING THE INTEGRAL

A calculus course is not only about, Do you know how tocalculate the integral. However, this course also asks you,Do you know how to find the integral qualitatively?

Consider this graph of g(x). The function f(x) is its derivative.

Therefore f(x) = g(x).


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PROPERTIES OF THE GRAPH

Lets talk about limits. Notice at x=1, the limit from the left side is a little more than5, but from the left side, its 0. Therefore, the limit at x=1 does not exist.

g(x) is not differentiable at some point between -6 and -7 and at -2, since they causevertical tangents. Also x=-2 is not differentiable, since it is a corner. x=-1 is also acorner, thus it is not differentiable either. x=1 is not differentiable, since it is adiscontinuity. Lastly, x=5 cannot be differentiated since it is the end point of thegraph.

g(x) is positive at approximately (-6.5,-4.3), (-1,1), and (2.75,5). g(x) is negative at (-4.3,-2) and (1,2.75). g(x) is zero with maxima at x=-4.3 and minima at x=2.75. f(x) is the area between the curve and the x-axis. Thus, the curve of f(x) is the

antiderivative. Since g(x) is the slope for f(x), this means that f(x) is increasingslowly, and maintains an equilibrium from x=-2 to x=-1. From x=-1, the the

function increases like a parabola, but due to the discontinuity of g(x), a cusp wasformed so that the slope is negative throughout.

REMEMBER!! THE SIGN OF f(x) OR g(x) IS THE SIGN OF THE SLOPE!!!!


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RED = f(x) BLACK = g(x)


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INTEGRAL MEANS AREA

ACCUMULATION

Look at the left graph f(x)=1/x and the right graph g(x)=ln x.

Obviously, g(x)=f(x). Notice how the 1/x decreases veryquickly, but nevertheless, adds more area. Although not at a

fast RATE, but it still ACCUMULATES area!


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VOLUME OF A SOLID OF

REVOLUTION

FOR AP CALCULUS AB STUDENTS THIS IS

YOUR FINAL TOPIC

FOR CALCULUS I: YOU ARE COMPLETED FOR

THE COURSE. YOU ARE WELCOMED TOSTAY, BUT YOU CAN LEAVE IF YOU WISH.

TAKE CARE! BEST WISHES! HARE KRSNA!

FOR CALCULUS II STUDENTS: HARI BOL!

WELCOME! THIS IS YOUR FIRST TOPIC FOR

THIS CLASS!!!!


29/85

VOLUME

Volume and 2-D calculus?? Possible?

Yeah, I guess.

Imagine if you were to take a region between

f(x), the limits of integration, and the x-axis,and you were to swing that region about the x-axis.

Take your hand and try that. Put your hand onthe region, and turn your hand such that yourotate about the x-axis.


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SEE THE BEFORE AND AFTER

PICTURES!!!


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CLARIFICATION

What we did was we took the region under theline from x=0 to x=7 and rotated that region360 about the x-axis.

You will notice that a solid is formed. Thatsolid looks like a cone in this example.

My computer does not really have great 3-d

effects so please forgive my drawings. How do we find the volume of this cone using

calculus?


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TORICELLIS LAW

An Italian mathematician suggested that you can find the

volume by take adding the areas of the cross sections.

He also proved that the volume for two geometric objects,

despite slant, would still be equal. Take a look at these twocylinders. Although it looks very obscure on the right, it will

still have the same volume as the cylinder on the left.


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THEREFORE:

If A(x) is the cross section for such a volume, and you

could get a Reimann sum saying that the width of the

solid multiplied by the area of the cross section

n

i xxAxV 1 )()(


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IF THE WIDTH

If the width was infinitesimally small as dx, then the

area would represented as an integral.

b

adxxAxV )()(


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CUTTING UP THE SOLID

Imagine if we were to cut up the solid to infinity

slices! That would that the thickness of the slice

would be virtually 0. But lets take out one of these

slices

dx


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THE SLICE

The cross section of such volume is a circle.

The radius of the circle would merely be f(x)

value, since the center of the circle is the x-

axis. Look at the 2-D and the 3-D illustration.

r=f(x)

x axis

The radius of the cirlce

is f(x).


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AREA OF A CIRCLE?

Lets say that the line was the function f(x)= x

Remember that the integral of the area of the cross

sections from a to b is the volume? Also rememberthat f(x) is the radius?

If the circle is the cross-section for the solid, thenwhat is the area of a circle?

Cmon! Dont tell me you are in calculus and dontknow the area of a circle! A=pr2! No! pies are round!

If the radius and the )( 22 xfr


38/85

If the radius and the

function are equal, so are

their squares

f(x)=x Apply the squared radius

into the area of circle

expression

Volume-as-an-integral

expression

Limits of the original

problem were x=0 and x=7. Evaluation of the integral.

The volume is found

12

243

12

)(

4)(

)()(

4

4

12

1

)(

7

0

3

7

0

2

2

22

2

2

xxV

dxxxV

dxxAxV

xA

xr

xr

xfr

b

a


39/85

RECALL

We took a region under f(x)=x from x=0 tox=7 and rotated it around the x- axis to form asolid. We took this solid and cut it up to

infinite slices. We took one of those slices andexamined that the cross-section is a circle(most commonly called a disk). We found thearea formula for the disk. A=pr2. Since r=f(x),

we put it into the formula. The volume formulasays that we take the integral from 0 to 7, forthe [f(x)]2. That result yields into the volume.


40/85

DISK METHOD FORMULA

If the cross sections arecircles, and if region are is

being rotated around the x-axis, then the formula forfinding volume is:

If the region between g(y)and x -axis is being rotatedabout the y-axis, with diskcross section, then the area

formula is this. Dont forget to put the pin

there.

b

adxxfxV

2)()(

d

c

dyygyV 2

)()(


41/85

DISK METHOD

Remember how we had instances that we were

forced to integrate with respect to y?

If that is the case, solve for y, use the y-values

for the limits of integration, and use the same

formula.


42/85

WASHER METHOD

Find the volume of the region between y2=x,

and y=x3, which intersect at (0,0) and (1,1),

revolved about the xaxis.

First, DRAW A PICTURE


43/85

PICTURES


44/85

THE SOLID

Notice that the solid now has a hole inside it(very light blue). The edges of this solid

(darker light blue) (is now created by two

functions.

Since we are revolving this solid around the x-

axis, we must solve everything in terms of x.


45/85

SHELLS FORMULA

If we solve for x, then we will get x. (square root)and x3.

If you look at the cross section, you will getsomething similar but somewhat different from adisk. In order to take into account, the hole in the

bottom, we have to include a hole inside the disk.Therefore, the slice will look like this.

f(x) TOP FUNCTIONg(x) BOTTOM FUNCTION


46/85

THE AREA OF THE ORANGE

If you wanted to find the area of the orange portion,

then find the area of the entire circle and subtract the

area of the inner cirlce. A typical shaded region

problem!Capital R means outer radius, and lowercase r means inner radius.

22 rRA


47/85

VOLUME

Having infinite doughnuts that are very thin, youcan use the integral to find the volume. This is called

WASHERS FORMULA.

b

adxrRxV 22)(

d

cdyrRyV 22)(


48/85

R and r

If you also consider it, the outer radius is thetop function while the inner radius is thebottom function, since the bottom function

borders the hole in the center of thedoughnut. Lets call this doughnut a washer from now

on!!!!

I dont wanna make ya guys hungry yet! (15 mins. Later) Now Im hungry!!!!


49/85

FINDING THE VOLUME

Wahsers formula

DONT FORGET THEp!

Original limits and top

and bottom functions

are included. The final volume

evaluated.

145

71

21

72)(

)(

)(

)(

1

0

72

1

0

6

1

0

23

2

22

xxxV

dxxxxV

dxxxxV

dxrRxVb

a


50/85

INTEGRATING w/ RESPECT TO Y.

Lets take the previous problem and rotate itabout the y-axis.

That means, we will have R as right function

and r as the left function

Therefore: x=y2and x=y.

ALWAYS DRAW A BEFORE/AFTER

PICTURE!!!


51/85

PICTURES


52/85

VOLUME rotated about y axis

5

2

5

1

5

3

55

3)(

)(

)(

)(

1

0

53/5

1

0

43/2

1

0

2223/1

22

xxyV

dyyyyV

dyyyyV

dyrRyVd

c


53/85

LINE ROTATIONS

Suppose we take the previous example and

rotated it about the following lines

A) x=-1

B) y=-1

Find both volumes.


54/85

DRAW A PICTURE

Draw both before/after diagrams for the x=-1

rotation and y=-1 rotation.

Please forgive the cut-offs in my drawings.

PICTURES f t ti b t


55/85

PICTURES for rotation about

x=-1

PICTURES f t ti b t


56/85

PICTURES for rotation about

y=-1


57/85

THE SHIFT!!!

Its not a movie. LOL

Look at the reference line (reference line is theline you are rotating the region about.)

Actually draw out the radii. For the x=-1rotation, you will need to add 1 to bothfunctions to take into account.

General Rule: If rotation around a line x=k,then f(x)-k is the big radius and g(x)-k is thesmall radius.


58/85

THE RADII

Therefore, the radii is 1+y2and 1+y1/3.

Since by this time, you are getting the

fundamental theorem of calculus down pretty

well (or at least you should be), I will stop

showing the plugging in of numbers. I will

show the integral, substitution, actual

integration and answer.


59/85

INTEGRATION

30

37

15

13

20

23)(

53

2

5

3

4

6)(

2121)(

11)(

)(

1

0

533/53/4

1

0

423/23/1

1

0

2223/1

22

yV

yyyyyyyV

dyyyyyyV

dyyyyV

dyrRyV dc


60/85

ROTATING about y=-1

Remember! If you have y=k as line of rotation,

then you will f(x)-k and g(x)-k

Therefore, you will have f(x)=x1/2+1 and

g(x)=x3+1

Using the washers formula, lets find thevolume.


61/85

VOLUME

3

7

1

2

1

2

1

3

4)(

72234)(

2121)(

11)(

)(

1

0

742

2/3

1

0

63

1

0

232

22

xV

xxxxxxxV

dxxxxxxV

dxxxxV

dxrRxVb

a


62/85

KRSNACULULUS STATUS

END OF AP CALCULUS AB

END OF CALCULUS I

MIDDLE POINT OF AP CALCULUS BC.

START OF CALCULUS II.

** AP CALCULUS AB AND CALCULUS I,

PLEASE GO TO THE SUMMARYSECTION.**


63/85

VOLUME BY SHELLS

If we wanted revolve a region to form a solid,we could use really thin cylinders, find thearea for all of these cylinders and thus, get the

volume. It is merely the sum of the surface area of each

cylinder. The formula of the surface area of acylinder is 2prh. If we make the thickness as

thick as dx, then the shells formula would likethis.


64/85

SHELLS FORMULA

RydyxV

RxdxyV

2)(

2)(

rh

For rotation over the y -axis

PROPOSED SLIDES.


65/85

TEMPORARY UNDER

CONSTRUCTION

ARC LENGTH AND SURFACE


66/85

ARC LENGTH AND SURFACE

AREA

This topic requires very strong algebra skills.

We will introduce arc length and surface area

of a solid of revolution, since their formulas

are very much similar.

First, we must discuss ds.

WHAT IS ds?


67/85

PYTHAGOREAN THEOREM

Remember the Pythagorean theorem?

x2+y2=r2

Lets call they hypotenuse as s instead ofr.

y

x

s


68/85

SIMILARLY

Works with differentials!

ds2=dx2+dy2

Same concept applies

You could rewrite it as ds2=[1+(dy/dx)2]dx2

dy

dx

ds


69/85

FINDING DISTANCE

y

x

x

x

y

ys

s

s


70/85

DISTANCE

Notice if you make right triangles with equal x, thehypotenuse will be really close to the graph line. Thehypotenuse would be s.

The sum of s will give you the distance or the lengthof the line.

If we took infinitesimal right triangles such that xwould become the width, dx. Therefore, s will

become ds.

The sum of these very minute lines or dots (hehe)would become the length of the line.

REIMANN SUM AND INTEGRAL


71/85

REIMANN SUM AND INTEGRAL

FOR ARC LENGTH

Definition for ds.

Reimann sum to find the

distance of the line

THE ARC LEGNTH

FORMULA with respect to

x.

THE ARC LENGTH

FORMULA with respect to

y.

d

c

d

c

b

a

b

a

n

i

n

i

dydy

dxds

dxdx

dyds

xyxs

dxdx

dyds

2

2

11

2

1

1

1


72/85

COMPUTATING ARC LENGTH

Due to the radical sign in the ds expression,

integration becomes really difficult.

In the later chapters, we will discuss

integration of such integrals.

We will only do very simple functions that we

can integrate with the knowledge we have

now.


73/85

FROM THIS FORMULA

You can get surface

area.

d

csurface

b

asurface

dydy

dxxyA

dxdx

dyyxA

2

2

1)(

1)(


74/85

SAMPLE PROBLEM

Lets say you are given the great honor to makegarland for the Deities. Of course, you dont wantthem to be so short that the garlands become anecklace. But then again, you dont want to makethem as long as they would touch the floor! In other

words, you gotta make em just right!

Make this a bonafide problem as much as possible.If we were to break the loop and lie it along thetilak graph (y= (2/3)x3/2), youll see that the graphstarts from x=3 to x = 15. Find the length of thisgarland.


75/85

SAMPLE PROBLEM

3

11256

3

2864

3

2

3

2

1

1

1

1

16

4

2/3

16

4

15

3

15

3

15

3

2

2/3

15

3

215

3

u

duuLdx

dxdu

xu

dxxLdx

xdxdy

xdx

dy

xy

dxdx

dyLdx

You guessed it! Arc lengthformula!

You differentiate the functionand square the derivative as

shown. Put the formula to some

action!!

Oops! Dont forget u-substitution. If u=1+x, then

du/dx = 1, therefore du = dx. Using the fundamental

theorem of calculus along withthe u-substitution, evaluate theintegral!


76/85

SURFACE AREA PROBLEM

Remember in the Mahabharata, the five Pandavasneeded to be in disguise during their thirteenth yearof exile! Arjuna is gonna be Brhanalla, and he needs a

pakhwaj drum.

You are his personal assistant who makes these drum.Its almost cylindrical but lets say that is 100%cylindrical.

Theoretically, the pakhawaj resembles the solidformed by rotating the region over the x-axis. The

region is defined as being bound between the liney=3, x-axis, y axis and x=10.

Find the surface area of this pakhawaj.


77/85

GRAPH, BEFORE, AFTER

Always know what region

you are talking about.

Draw a graph!! It helps!

Eh.. I know.. Not the best

pic but you get the ideaof how it looks like when

rotated!


78/85

FINDING THE SURFACE AREA

3003033

3

13

0

3

1)(

10

0

10

0

10

0

10

0

2

xdx

dxA

dxA

dxdy

y

dxdx

dyyxA

surface

surface

b

asurface

Remember! Formula first!Dont forget! We rotatedabout the x-axis, so we use

the A(x) formula.

The function is y=3. dy/dx =0.

The integral becomes a

piece of cake

Simply apply fundamentaltheorem of calculus.

Dont forget. Area = unitssqaured!


79/85

END OF CHAPTER SIX

Considering this the end of Calculus I (AP

Calculus AB) students and the beginning of

Calculus II (mid-AP Calc BC), everyone

should keep the following in mind. Dont forget the most important formulas:

d

b

a bottomtop

dA

dxyyA Area between 2

curves


80/85

d

csurface

b

asurface

d

c

d

c

b

a

b

a

d

c

b

a

d

c

b

a

c leftright

dydy

dx

xyA

dxdx

dyyxA

dydy

dxds

dxdx

dyds

RydyxV

RxdxyV

dyrRyV

dxrRxV

dyygyV

dxxfxV

dyxxA

2

2

2

2

22

22

2

2

1)(

1)(

1

1

2)(

2)(

)(

)(

)()(

)()(

curves (all)

Disk method (AB II)

Washers method (AB II)

Cylindrical Shells (BC II)

Arc Length (BC II)

Surface Area (BC II)

S


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SUMMARY

When doing these problems with area, volume,

arc length, or surface area, always draw a

picture of the all graphs and conditions

involved. Then draw a before/after scenario.

Write the equations necessary to solve the

problem.

Check with a calculator if available

CONCLUSION


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CONCLUSION

We are about to reach a point in calculus wheresimply memorizing equations and plugging numbers

will not be useful all on its own.

Calculus requires critical thinking. The secret behindmath is planning. How do you plan to solve a

problem? With method would work the best and

efficiently? These questions will come up soon.

Never forget the important rules of integration. Thenext few chapters are strictly devoted to integration.

END OF CHAPTER SIX


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END OF CHAPTER SIX

Sri Krsna Caitanya Prabhu Nityananda

Sri Advaita Gadadhara Srivasadi Gaura Bhakta

Vrnda

Hare Krsna Hare Krsna Krsna Krsna Hare Hare

Hare Rama Hare Rama Rama Rama Hare Hare

CREDITS


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CREDITS

Mr. J. Trapani

Mr. G. Chomiak

Dr. A. Moslow

Mr. D. Ireland

Calculus and Early Transcendental Functions

5thEd.

Single-variable Calculus (SUNY Buffalo)


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Aplikasi Integral