Calculus July 19-21Calculus July 19-21

Review of Week 1Review of Week 1

Your thoughts, questions, musings, Your thoughts, questions, musings, etc...etc...

A problem that was around long before the A problem that was around long before the invention of calculus is to find the area of a invention of calculus is to find the area of a general plane region (with curved sides).general plane region (with curved sides).

And a method of solution that goes all And a method of solution that goes all the way back to Archimedes is to the way back to Archimedes is to divide the region up into lots of little divide the region up into lots of little regions, so that you can find the area regions, so that you can find the area of almost all of the little regions, and of almost all of the little regions, and so that the total area of the ones you so that the total area of the ones you can't measure is very small. can't measure is very small.

AmebaAmeba

By Newton's time, people realized that it would be By Newton's time, people realized that it would be sufficient to handle regions that had three straight sufficient to handle regions that had three straight sides and one curved side (or two or one straight side sides and one curved side (or two or one straight side -- the important thing is that all but one side is -- the important thing is that all but one side is straight). Essentially all regions can be divided up into straight). Essentially all regions can be divided up into such regions. such regions.

These all-but-one-side-straight regions look like These all-but-one-side-straight regions look like areas under the graphs of functions. And there is a areas under the graphs of functions. And there is a standard strategy for calculating (at least standard strategy for calculating (at least approximately) such areas. For instance, to calculate approximately) such areas. For instance, to calculate the area between the graph of y = 4x - xthe area between the graph of y = 4x - x22 and the x and the x axis, we draw it and subdivide it as follows: axis, we draw it and subdivide it as follows:

Since the green pieces are all Since the green pieces are all rectangles, their areas are easy to rectangles, their areas are easy to calculate. The blue parts under the calculate. The blue parts under the curve are relatively small, so if we curve are relatively small, so if we add up the areas of the rectangles, we add up the areas of the rectangles, we won't be far from the area under the won't be far from the area under the curve. For the record, the total area curve. For the record, the total area of all the green rectangles is: of all the green rectangles is: 246246

2525

whereas the actual area under the curve is:whereas the actual area under the curve is:

Also for the record, Also for the record, 246/25 = 9.84246/25 = 9.84 while while 32/332/3 is about is about 10.666710.6667. .

4

0

2

3

324 dxxx

We can improve the approximation by dividing into more We can improve the approximation by dividing into more rectangles: rectangles:

Now there are 60 boxes instead of 20, and their total area is: Now there are 60 boxes instead of 20, and their total area is: which is about 10.397. Getting better. We can in fact which is about 10.397. Getting better. We can in fact

take the limit as the number of rectangles goes to infinity, which take the limit as the number of rectangles goes to infinity, which will give the same value as the integral. will give the same value as the integral. This was Newton's and This was Newton's and Leibniz's great discovery -- derivatives and integrals are related Leibniz's great discovery -- derivatives and integrals are related and they are related to the area problem.and they are related to the area problem.

Area 60 boxesArea 60 boxes

70187018675675

Limits of Riemann sumsLimits of Riemann sumsA kind of limit that comes up occasionally A kind of limit that comes up occasionally

is an integral described as the limit of a is an integral described as the limit of a Riemann sum. One way to recognize Riemann sum. One way to recognize these is that they are generally these is that they are generally

expressed as expressed as , where , where thethe

““something” depends on something” depends on nn as well as on as well as on ii..

n

in

something1

lim

Green graphGreen graphAgain, recall that one way to look at integrals is Again, recall that one way to look at integrals is

as areas under graphs, and we approximate as areas under graphs, and we approximate these areas as sums of areas of rectangles. these areas as sums of areas of rectangles.

This is a picture of This is a picture of

the“right endpoint”the“right endpoint”

approximation to theapproximation to the

integral of aintegral of a

function.function.

approximatingapproximating

b

a

n

in

aab

nab

nabi

nabi

nab

dxx

a

iax

i

nba

x

nabi

help willexampleAn .)(f integral theisinfinity approaches

n as sum thisoflimit theand , thus

is rectangles theof areas theof sum The . f

is rectangleth theof area theso and ,

at is rectangleth theof sideright The . width has

rectangleeach then ,rectangles using to from interval

over the )f( of integral theingapproximat are weIf

1

)f()(

)(

)(

)(

Example...Example... ?lim isWhat 1

4

3

n

in n

i

solutionsolution

. toequal is sum theof

limit theThus .)( have llthen we'

,f(x) If 0.a choose should wefactor,

other in the appearsi/n And 1.a-b need we

1/n,a)/n-(bFor b. and aout figure toneed

weNow .))(( assign summation the

under expression therewritecan weSo

a)/n.-(bour for 1/n a need weFirst,

1

0 413

31(f)(

3

31

)(

dxx

x

ni

nn

aab

ni

n

nabi

Area between two curves:Area between two curves:

A standard kind of problem is to find the area A standard kind of problem is to find the area above one curve and below another (or to the left above one curve and below another (or to the left of one curve and to the right of another). This is of one curve and to the right of another). This is easy using integrals. easy using integrals.

Note that the "Note that the "area between a curve and area between a curve and the axisthe axis" " is a special case of this problem where one of the is a special case of this problem where one of the curves simply has the equation curves simply has the equation y = 0y = 0 (or perhaps (or perhaps x=0x=0 ) )

1. Graph the equations if possible 1. Graph the equations if possible

2. Find points of intersection of the curves to 2. Find points of intersection of the curves to determine limits of integration, if none are determine limits of integration, if none are given given

3. Integrate the top curve's function minus the 3. Integrate the top curve's function minus the bottom curve's (or right curve minus left bottom curve's (or right curve minus left curve). curve).

Example:Example:Find the area between the graphs of y=sin(x) and y=x(Find the area between the graphs of y=sin(x) and y=x(-x)-x)

26

)sin()(3

0

dxxxx

It’s easy to see that the curves It’s easy to see that the curves intersect on the x-axis, and theintersect on the x-axis, and thevalues of x are 0 and values of x are 0 and ..

The parabola is on top, so we integrate:The parabola is on top, so we integrate:

And this is the area between the two curves.And this is the area between the two curves.

An Area Question:An Area Question:

Find the area of the region bounded by the Find the area of the region bounded by the curves y=4xcurves y=4x22 and y=x and y=x22+3. +3.

A. 1/2 A. 1/2

B. 1 B. 1

C. 3/2 C. 3/2

D. 2 D. 2

E. 5/2 E. 5/2

F. 3 F. 3

G.7/2 G.7/2

H. 4H. 4

Position, velocity, and Position, velocity, and acceleration:acceleration:

Since velocity is the Since velocity is the derivative of position and derivative of position and acceleration is the derivative acceleration is the derivative of velocity, of velocity,

Velocity is the integral of Velocity is the integral of accelerationacceleration, and , and position is position is the integral of velocitythe integral of velocity. .

(Of course, you must know (Of course, you must know starting values of position starting values of position and/or velocityand/or velocity to determine to determine the constant of integration.) the constant of integration.)

Example...Example...

An object moves in a force field so that its An object moves in a force field so that its

acceleration at time t is a(t) = acceleration at time t is a(t) = t -t+12 (meterst -t+12 (meters

per second squared). Assuming the object isper second squared). Assuming the object is

moving at a speed of 5 meters per second at timemoving at a speed of 5 meters per second at time

t=0, determine how far it travels in the first 10t=0, determine how far it travels in the first 10

seconds. seconds.

22

First we determine the velocity, by integrating the acceleration. First we determine the velocity, by integrating the acceleration. Because v(0) = 5, we can write the velocity v(t) as 5 + a Because v(0) = 5, we can write the velocity v(t) as 5 + a definite definite integral, as follows: integral, as follows:

The distance the object moves in the first 10 seconds is the total The distance the object moves in the first 10 seconds is the total change in position. In other words, it is the integral of dx as t change in position. In other words, it is the integral of dx as t goes from 0 to 10. But dx = v(t) dt. So we can write: goes from 0 to 10. But dx = v(t) dt. So we can write:

(distance traveled between t=0 and t=10) =(distance traveled between t=0 and t=10) =

= = = 3950/3 = 1316.666... meters . = 3950/3 = 1316.666... meters .

Solution...Solution...

ttt

ddatvt t

1223

5125)(5)(0 0

232

dttv10

0

)(

dtttt

1223

52310

0

Surfaces of revolution:Surfaces of revolution:VolumeVolume

A A "surface of revolution""surface of revolution" is formed when a curve is is formed when a curve is revolved around a line (usually the x or y axis). The revolved around a line (usually the x or y axis). The curve sweeps out a surfacecurve sweeps out a surface

Interesting problems that can be solved by integration are to find the volume enclosed inside such a surface or to find its surface area.

VolumesVolumes

You might already be familiar with finding volumes of You might already be familiar with finding volumes of

revolution. revolution.

Once a surface is formed by rotating around the x-axis, youOnce a surface is formed by rotating around the x-axis, you

can sweep out the volume it encloses with disks perpendicularcan sweep out the volume it encloses with disks perpendicular

to the x axis.to the x axis.

Here is the surface formed...Here is the surface formed...Here is the surface formed by revolving Here is the surface formed by revolving around the around the

xx axis for axis for xx between 0 and 2, showing one of the disks that sweep between 0 and 2, showing one of the disks that sweep

out the volume: out the volume:

xy

enclosed inside the surface, we need toenclosed inside the surface, we need to

add up the volumes of all the disks.add up the volumes of all the disks.

The disks are (approximately) cylinders turned sideways, and The disks are (approximately) cylinders turned sideways, and the disk centered at (the disk centered at (xx,0) has radius,0) has radius and width (or and width (or height) height) dxdx. The volume of the disk is thus. The volume of the disk is thus

To find the total volume of the solid we have to integrate thisTo find the total volume of the solid we have to integrate this

quantity for x from 0 to 2. We getquantity for x from 0 to 2. We get

To calculate the volumeTo calculate the volume

, or

x

dxx2

dxx

22

0

dxxV cubic unitscubic units

In general, if the piece of the graph of the function of In general, if the piece of the graph of the function of y y = = ff ( (xx) ) between between x x = = a a and and x x = = b b is revolved around the is revolved around the xx axis, the axis, the volume inside the resulting solid of revolution is calculated as: volume inside the resulting solid of revolution is calculated as:

The same sort of formula applies if we rotate the region The same sort of formula applies if we rotate the region between the between the yy-axis and a curve around the -axis and a curve around the yy-axis (just change -axis (just change all the all the xx's to 's to yy's). 's).

A formula for volume:A formula for volume:

b

a

dxxfV 2)(

The same region The same region rotated around rotated around

the y axisthe y axis

A different kind of problem is to rotate the regionA different kind of problem is to rotate the region

between a curve and the between a curve and the xx axis around the axis around the yy axis (or axis (or

vice versa). For instance, let's look at the same region vice versa). For instance, let's look at the same region

(between (between yy=0 and =0 and yy= = for for xx between 0 and 2), but between 0 and 2), but

rotated around the rotated around the yy axis instead: axis instead:

x

The generating curveThe generating curve

Here is the surface being swept Here is the surface being swept out by the generating curve:out by the generating curve:

WashersWashers

We could sweep out this volume with “washers” We could sweep out this volume with “washers” with inner radius with inner radius yy22 and outer radius 2 as and outer radius 2 as yy goes goes from 0 tofrom 0 to

Each washer is (approximately) a cylinder with a hole in Each washer is (approximately) a cylinder with a hole in the middle. The volume of such a washer is then the the middle. The volume of such a washer is then the volume of the big cylinder minus the volume of the hole. volume of the big cylinder minus the volume of the hole.

2

For the washer centered at the point (0, For the washer centered at the point (0, yy), the ), the

radius of the outside cylinder is always equal radius of the outside cylinder is always equal to 2, and the radius of the hole is equal to the to 2, and the radius of the hole is equal to the corresponding corresponding xx (which, since (which, since , is equal , is equal

to to yy22 ). And the height of the washer is equal to ). And the height of the washer is equal to

dydy. So the volume of the washer is. So the volume of the washer is

The volume of the The volume of the washers...washers...

xy

dyydyydV 4222 42

Therefore the volume of the Therefore the volume of the entire solid isentire solid is

cubic unitscubic units

5

2164

2

0

4 dyyV

Cylindrical shellsCylindrical shells

Another way to sweep out this volume is with Another way to sweep out this volume is with "cylindrical shells". "cylindrical shells".

The volume of a cylindrical The volume of a cylindrical shellshell

Each cylindrical shell, if you cut it along a vertical line, can beEach cylindrical shell, if you cut it along a vertical line, can be

laid out as a rectangular box, with lengthlaid out as a rectangular box, with length , with width, with width

and with thickness and with thickness dxdx. The volume of the cylindrical. The volume of the cylindrical

shell that goes through the point (shell that goes through the point (xx,0) is thus,0) is thus

So, we can calculate the volume of the entire solid to be:So, we can calculate the volume of the entire solid to be:

cubic units, which agrees with the answer we got the other way.cubic units, which agrees with the answer we got the other way.

x2

xy

dxxxdV 2

5

2162

2

0

2/3 dxxV

Another family of volume Another family of volume problems involves volumes of problems involves volumes of three-dimensional objects three-dimensional objects whose cross-sections in some whose cross-sections in some direction all have the same direction all have the same shape. shape.

For exampleFor example: Calculate the : Calculate the volume of the solid S if the base volume of the solid S if the base of S is the triangular region of S is the triangular region with vertices (0,0), (2,0) and with vertices (0,0), (2,0) and (0,1) and cross sections (0,1) and cross sections perpendicular to the x-axis are perpendicular to the x-axis are semicircles. semicircles.

First, we have to visualize the First, we have to visualize the solid. Here is the base triangle, solid. Here is the base triangle, with a few vertical lines drawnwith a few vertical lines drawn

Other volumes Other volumes with known with known

cross sectionscross sections

on it (perpendicular to theon it (perpendicular to thex-axis). These will be diameters of the semicircles in the solid.x-axis). These will be diameters of the semicircles in the solid.

3-D3-D

Now, we'll make the three-dimensional plot that has Now, we'll make the three-dimensional plot that has this triangle as the base and the semi-circular cross this triangle as the base and the semi-circular cross sections. sections.

3-D BOX3-D BOX

From that point of view you can see some of the base From that point of view you can see some of the base as well as the cross section. We'll sweep out the volume as well as the cross section. We'll sweep out the volume with slices perpendicular to the x-axis, each will look with slices perpendicular to the x-axis, each will look like half a disk: like half a disk:

Since the line connecting the two points (0,1) and (2,0) has Since the line connecting the two points (0,1) and (2,0) has equation equation yy = 1 - = 1 - xx/2, the centers of the half-disks are at the /2, the centers of the half-disks are at the points (points (xx, 1/2 - , 1/2 - xx/4), and their radii are likewise 1/2 - /4), and their radii are likewise 1/2 - xx/4. /4. Therefore the little bit of volume at Therefore the little bit of volume at xx is half the volume of a is half the volume of a cylinder of radius 1/2 - cylinder of radius 1/2 - xx/4 and height /4 and height dxdx, namely , namely

Therefore, the volume of the solid S is:Therefore, the volume of the solid S is:

The volume of that The volume of that object:object:

dxxdV 2

41

21

2

12

2

0

2

41

21

2

dxxV

Note that we could also have calculated the volume by noticing Note that we could also have calculated the volume by noticing that the solid S is half of a (skewed) cone of height 2 with base that the solid S is half of a (skewed) cone of height 2 with base radius = 1/2. radius = 1/2.

Using the formula Using the formula for a cone, we arrive for a cone, we arrive

at the same answer, at the same answer, cubic units. cubic units.

Finally...Finally...

hrV 231

12

Methods of integrationMethods of integrationBefore we get too involved with applications of the integral, we Before we get too involved with applications of the integral, we have to make sure we're good at calculating antiderivatives. have to make sure we're good at calculating antiderivatives. There are four basic tricks that you have to learn (and hundreds There are four basic tricks that you have to learn (and hundreds of of ad hocad hoc ones that only work in special situations): ones that only work in special situations):

1. Integration by substitution (chain rule in reverse)1. Integration by substitution (chain rule in reverse)

2. Trigonometric substitutions (using trig identities to your 2. Trigonometric substitutions (using trig identities to your advantage) advantage)

3. Partial fractions (an algebraic trick that is good for more than 3. Partial fractions (an algebraic trick that is good for more than doing integrals)doing integrals)

4. Integration by parts (the product rule in reverse) 4. Integration by parts (the product rule in reverse)

We'll did #1 last week, and we’ll do the others this week. LOTS of We'll did #1 last week, and we’ll do the others this week. LOTS of practice is needed to master these!practice is needed to master these!

Integration by partsIntegration by partsThis is another way to try and integrate This is another way to try and integrate productsproducts. .

It is in fact the opposite of the product rule for derivatives:It is in fact the opposite of the product rule for derivatives:

Product rule for derivativesProduct rule for derivatives: d: d((uvuv) = ) = u dvu dv + + v duv du . .

Integration by partsIntegration by parts: :

Use this when you have a product under the integral sign, and it Use this when you have a product under the integral sign, and it appears that integrating one factor and differentiating the other appears that integrating one factor and differentiating the other will make the resulting integral easier. will make the resulting integral easier.

duvuvdvu

If we If we differentiate differentiate x2 , we'll get , we'll get 22xx which seems which seems simplersimpler. . And And integratingintegrating ex doesn't change anything. doesn't change anything.

Let Let u = x and and dv = ex dx. Then . Then du = 2x dx and and v = ex. .

The integration by parts formula then gives us: The integration by parts formula then gives us:

Example:Example:-- There's -- There's no other rule for thisno other rule for this, ,

so we try so we try partsparts. . dxex x 2

dxexexdxex xxx 222

Continuing:Continuing:

We can use parts again on this integralWe can use parts again on this integral

(with (with u = 2x and and dv = ex dx) to get: ) to get:

dxexexdxex xxx 222

Ceexex

dxeexexdxex

xxx

xxxx

22

22

2

22

You try one...You try one...

……contrast this withcontrast this with

dxxx 2sin

dxxx 2sin

XOXXOXXOXXOX

There is a method called the "tic-tac-toe" There is a method called the "tic-tac-toe" method for integration by parts. method for integration by parts.

Different strategies for choosing Different strategies for choosing u and and dv are are called for to integrate called for to integrate x ln x or or x tan-1 x or or something like something like ex sin xxx

A problem for you...A problem for you...

Find the value ofFind the value of

00

x x cos(cos(xx) ) dxdx

A) A)

B) 2B) 2

C) 2C) 2

D) 0D) 0

E) -2E) -2

F) 1F) 1

G) 1/2G) 1/2

H) H) /2/2

And another:And another:

EvaluateEvaluate

11

00

x x ln ln xx dxdx

A) -1/4A) -1/4

B) -1/2B) -1/2

C) 0C) 0

D) 1/4D) 1/4

E) 1/2E) 1/2

Trigonomeric substitutionsTrigonomeric substitutions

These are just substitutions like the ordinary These are just substitutions like the ordinary substitution method, but some of them are a little substitution method, but some of them are a little

surprising. There are two kinds:surprising. There are two kinds:

To integrate products of powers of sine and cosineTo integrate products of powers of sine and cosine

(like sin (x) cos (x) ) . You need the identities: (like sin (x) cos (x) ) . You need the identities:

sin (x) + cos (x) = 1 (everybody knows that one!),sin (x) + cos (x) = 1 (everybody knows that one!),

and the double angle formulas: cos (x) = and the double angle formulas: cos (x) =

and sin (x) = and sin (x) =

3 6

2 2

2 1+cos(2x)2

1-cos(2x)2

2

The first,The first, 1111

The trick is as follows...The trick is as follows...

(a) If both the power of sine and the power of cosine (a) If both the power of sine and the power of cosine are are eveneven, then use the double angle formulas to , then use the double angle formulas to divide both in half. Keep doing this until at least one divide both in half. Keep doing this until at least one of the powers is odd. of the powers is odd.

(b) Once at least one of the powers is (b) Once at least one of the powers is oddodd, say it's the , say it's the power of cosine, then let u = sin( -- ) -- then use the power of cosine, then let u = sin( -- ) -- then use the Pythagorean identity to convert all but one power of Pythagorean identity to convert all but one power of cosine to sines, and the last power of cosine will be cosine to sines, and the last power of cosine will be the du in a substitution. the du in a substitution.

Let’s do one...Let’s do one... dxxx 44 cossin

Both the powers are even, so use Both the powers are even, so use the double angle formula trick:the double angle formula trick:

xx

xxxx

2cos2cos21

2cos12cos1cossin42

161

22

16144

In the last two terms, use the identities In the last two terms, use the identities again to get what’s on the next slide.again to get what’s on the next slide.

Trig identity city!Trig identity city!

xx

xx

xxxx

8cos4cos43

4cos4cos21

2

4cos14cos112cos2cos21

1281

2641

2

16142

161

……and (finally!) this is something we can and (finally!) this is something we can integrate!integrate!

dxxxdxxx 8cos4cos43cossin 128144

……and at last:and at last:

dxxxdxxx 8cos4cos43cossin 128144

Cxxx

Cxxx

8sin4sin

8sin4sin3

10241

1281

1283

81

1281

A) 2/15A) 2/15

B) 4/15B) 4/15

C) 2/5C) 2/5

D) 8/15D) 8/15

E) 2/3E) 2/3

F) 4/5F) 4/5

G) 14/15G) 14/15

Here’s one with an odd power for you Here’s one with an odd power for you to do:to do:

2

0

32 cossin

dxxx

Try this one:Try this one:

00

cos x dx =cos x dx =44

A) 2A) 2

B) B)

C) C) - 1/2 - 1/2

D) 2 D) 2

E) 3E) 3

The second,The second,

-- and more important, kind of trig substitution -- and more important, kind of trig substitution happens when there is a sum or difference of happens when there is a sum or difference of squares in the integrand. Usually one of the squares in the integrand. Usually one of the squares is a constant and the other involves squares is a constant and the other involves the variable. the variable.

If some other substitution doesn't suggest If some other substitution doesn't suggest itself firstitself first, then try the Pythagorean trig , then try the Pythagorean trig identity that has the same pattern of signs as identity that has the same pattern of signs as the one in the problem.the one in the problem.

,, , or, or

EitherEither

22 22

22 sin1cos 22 tan1sec 1sectan 22

Example:Example:

dxx

x 2

2

5

If you think about it, the If you think about it, the substitution substitution uu = 5 - = 5 - xx22 won’t won’t work, because of the extra work, because of the extra factor of factor of x x in the numerator.in the numerator.

But 5 - But 5 - xx22 is vaguely reminiscent of 1 - sinis vaguely reminiscent of 1 - sin22, so let, so let

22 sin5x

ThenThen cos5x cos5dx

222 cos5sin555 xandand

We make all the substitutions and get:We make all the substitutions and get:

d

ddxx

x

2

2

2

2

sin5

cos5cos5

sin5

5

This is a trig integral of the other sort . And we can This is a trig integral of the other sort . And we can use the double-angle formula to get:use the double-angle formula to get:

Cdd

2sin2

2cos15sin5 4

5252

Now we have to get back to x’sNow we have to get back to x’s

So far,So far, Cx

x

2sin5

45

25

2

2

wherewhere sin5x

thereforetherefore

5

arcsinx

5sin

x and soand so

Also,Also, cossin22sin

x5

25 x

And from the triangle,And from the triangle,

5

5cos

2x

The conclusion is:The conclusion is:

Cxxx

Cxxx

C

Cdxx

x

221

25

2

25

25

25

25

45

25

2

2

55

arcsin

5

5

55arcsin

cossin

2sin5

Quite a bit of work for one integral!Quite a bit of work for one integral!

Examples for you…Examples for you…Here are two for you to work on. Here are two for you to work on.

Notice the subtle difference in the integrand that Notice the subtle difference in the integrand that changes entirely the method used: changes entirely the method used:

dxxx 24 dxxx 22 4

Partial FractionsPartial FractionsLast but not least is integration by Last but not least is integration by partial fractionspartial fractions. This method . This method is based on an algebraic trick. It works to integrate a rational is based on an algebraic trick. It works to integrate a rational function (quotient of polynomials) when the degree of the function (quotient of polynomials) when the degree of the denominator is denominator is greatergreater than the degree of the numerator than the degree of the numerator (otherwise "when in doubt, divide it out") and you can factor (otherwise "when in doubt, divide it out") and you can factor the denominator completely. the denominator completely.

In full generality, partial fractions works when the denominator In full generality, partial fractions works when the denominator has quadratic and repeated factors, but we will consider only has quadratic and repeated factors, but we will consider only the case of distinct linear factors (i.e., the denominator factors the case of distinct linear factors (i.e., the denominator factors into linear factors and they're all different). It also uses the into linear factors and they're all different). It also uses the (easy) fact that (easy) fact that

Cbaxa

dxbax

ln

11

Partial Fractions Partial Fractions (continued)(continued)

The idea of partial fractions is to take a rational function of The idea of partial fractions is to take a rational function of the form the form

(where there can be more or fewer factors in the denominator, (where there can be more or fewer factors in the denominator, but the degree of p(x) must be less than the number of factorsbut the degree of p(x) must be less than the number of factors) ) and rewrite it as a sum: and rewrite it as a sum:

for some constants A, B, C (etc..). for some constants A, B, C (etc..).

)...)()((

)(

cxbxax

xp

...

cx

C

bx

B

ax

A

For instance,For instance,

You can rewrite , You can rewrite ,

first by factoring the denominator:first by factoring the denominator:34

122

xx

x

31

12

xx

x

and then in partial fractions as:and then in partial fractions as:

3125

21

xx

Two questions...Two questions...

?1. 1. WhyWhy do partial fractions? do partial fractions?

2. 2. HowHow to do to do partial partial fractions? fractions?

34

122

xx

x

Two reasons for why:Two reasons for why:First reason:First reason:First reason:First reason:

It helps us to do integrals. It helps us to do integrals.

From the previous example, we see that From the previous example, we see that we don't know how to integrate the we don't know how to integrate the

fraction right away, but ...fraction right away, but ...

dxx

dxx

dxxx

x

1334

12 21

25

2

Cxx )1ln()3ln( 21

25

Two reasons for why:Two reasons for why:Second Second reason:reason:Second Second reason:reason:

Partial fractions helps us to understand the behavior of a rational Partial fractions helps us to understand the behavior of a rational function near its "most interesting” points.function near its "most interesting” points.

For the same example, we graph the function in blue,For the same example, we graph the function in blue,

Let’s take a closer lookLet’s take a closer look

34

122

xx

x

325

x 121

x and the partial fractions and in red:and the partial fractions and in red:

red and blue curvesred and blue curves

Note that one or the other of the red curves mimics Note that one or the other of the red curves mimics the behavior of the blue one at each singularity.the behavior of the blue one at each singularity.

OK, now for the “how”OK, now for the “how”First, we give the official version, then a short-First, we give the official version, then a short-

cut.cut.

Official version:Official version: It is a general fact that the original fraction It is a general fact that the original fraction will break up into a sum with will break up into a sum with one term for each factor in the one term for each factor in the denominatordenominator. So (in the example), write . So (in the example), write

where A and B are to be determined. To determine A and B, where A and B are to be determined. To determine A and B, pick two values of x other than x=-3 or x=-1, substitute them into pick two values of x other than x=-3 or x=-1, substitute them into the equation, and then solve the resulting two equations for the the equation, and then solve the resulting two equations for the two unknowns A and B. For instance, we can put x=0, and get two unknowns A and B. For instance, we can put x=0, and get

and for x=1, get and for x=1, get . .

Solve to get Solve to get AA = 5/2 and = 5/2 and BB = -1/2 as we did before. = -1/2 as we did before.

1334

122

x

B

x

A

xx

x

BA31

31 BA 2

141

83

““Short cut”Short cut”

1. Write the fraction with denominator in factored form, and 1. Write the fraction with denominator in factored form, and leave blanks in the numerators of the partial fractions: leave blanks in the numerators of the partial fractions:

2. To get the numerator that goes over 2. To get the numerator that goes over x x + 3, put your hand + 3, put your hand over the over the x x + 3 factor in the fraction on the left and set + 3 factor in the fraction on the left and set x x = -3 in = -3 in the rest. You should end up with 5/2. the rest. You should end up with 5/2.

3. To get the numerator that goes over 3. To get the numerator that goes over x x + 1, cover the + 1, cover the x x + 1 + 1 and set and set x x = -1 (and you get -1/2). = -1 (and you get -1/2).

It's that simple. It's that simple.

1

__

3

__

)1)(3(

12

xxxx

x

Another example:Another example:

dxxx

x2

2 1

First, the numerator and denominator First, the numerator and denominator have the same degree. So we have to have the same degree. So we have to divide it out before we can do partial divide it out before we can do partial fractions.fractions.

)1(

11

11

122

2

xx

x

xx

x

xx

x

and we can use partial fractions on the and we can use partial fractions on the second termsecond term

Partial fractions gives:Partial fractions gives:

11

)1(

11

x

B

x

A

xx

x

Use either the official or short-cut Use either the official or short-cut method to get A = -1 and B = 2. method to get A = -1 and B = 2. Therefore:Therefore:

Cxxxdxxx

dxxx

x)1ln(2ln

1

211

12

2

An example for you to try:An example for you to try:

FindFind

33

22

dx dx x x ((xx-1)-1)

A) 3/2A) 3/2

B) 4/3B) 4/3

C) ln 2C) ln 2

D) ln 3D) ln 3

E) ln (3/2)E) ln (3/2)

F) ln (4/3)F) ln (4/3)

G) ln (2/3)G) ln (2/3)

H) 3 ln (2)/2H) 3 ln (2)/2

Another example:Another example:44

33

44xx - 6 - 6 x - x - 33xx +2 +222 ==

A) ln (4/3)A) ln (4/3)

B) 2 + arctan (3)B) 2 + arctan (3)

C) ln (9)C) ln (9)

D) ln (12/5)D) ln (12/5)

E) E) /3 - arctan (1/4)/3 - arctan (1/4)

Arc lengthArc length

The length of a curve in the plane is generally difficult The length of a curve in the plane is generally difficult to compute. To do it, you must add up the little “pieces to compute. To do it, you must add up the little “pieces of arc”, ds. A good approximation to ds is given by the of arc”, ds. A good approximation to ds is given by the Pythagorean theorem: Pythagorean theorem:

We can use this to find the length of any graph – We can use this to find the length of any graph – provided we can do the integral that results!provided we can do the integral that results!

dxdx

dydydxds

222 1

Find the arclength of the parabola Find the arclength of the parabola y = xy = x22 for for xx between between

-1 and 1. Since -1 and 1. Since dy / dx = dy / dx = 22xx, the element of arclength is, the element of arclength is

so the total length is:so the total length is:

1

1

241 dxxL

dxx241

dxxL

1

1

241

So far, we have that the length is So far, we have that the length is

To do this integral, we will need a trig To do this integral, we will need a trig substitution. But, appealing to Maple, we substitution. But, appealing to Maple, we get thatget that

)25ln(2

1541

1

1

2

dttL

Can we do the integral?Can we do the integral?

The arc length integral from before was:The arc length integral from before was:

This is a trig substitution integral of the second kind:This is a trig substitution integral of the second kind:

With the identity tanWith the identity tan22 + 1 = sec + 1 = sec22 in mind, let in mind, let

What about What about dt dt ? Since we have that? Since we have that

These substitutions transform the integral intoThese substitutions transform the integral into

This is a tricky integral we This is a tricky integral we need to do by parts!need to do by parts!

dtt

1

1

241

22 tan4 t

tan21t

ddt 221 sec

ddtt 3212 sec41

To integrateTo integrate d 3sec

LetLet ddvu 2sec,sec

ThenThen tan,tansec vdu

dd 23 tansectansecsec

But tanBut tan2 2 = sec= sec22 - 1 , so rewrite the last integral and get - 1 , so rewrite the last integral and get

ddd secsectansecsec 33

Still going….Still going….

ddd secsectansecsec 33

It’s remarkable that we’re almost done. The integral It’s remarkable that we’re almost done. The integral of secant is a known formula, and then you can add of secant is a known formula, and then you can add the integral of secthe integral of sec33 to both sides and get to both sides and get

Cd )tanln(sectansecsec 21

213

So we’ve got this so far forSo we’ve got this so far for dtt 2412 wherewhere tan21t

We need a triangle!We need a triangle!So far,So far,

Cdtt )tanln(sectansec412 21

212

withwith tan21t

t2

1

241 t

From the triangle,From the triangle,

241sec,2tan tt

SoSo Cttttdtt 241ln24141 2

412

412

Definite integral:Definite integral:

So far, we haveSo far, we have

Cttttdtt 241ln4141 2412

212

ThereforeTherefore

2525

541 41

1

1

2 lndxx

To get the answer Maple got before, To get the answer Maple got before, we’d have to rationalize the we’d have to rationalize the numerator inside the logarithm.numerator inside the logarithm.

The area of a surface of revolution is calculated in a mannerThe area of a surface of revolution is calculated in a manner

similar to the volume. The following illustration shows thesimilar to the volume. The following illustration shows the

paraboloid based onparaboloid based on (for x=0..2) that we used before, (for x=0..2) that we used before,

together with one of the circular bands that sweep out its together with one of the circular bands that sweep out its surface area. surface area.

Surface AreaSurface Area

xy

To calculate the surface areaTo calculate the surface areaTo calculate the surface area, we first need to determine To calculate the surface area, we first need to determine

the area of the bands. The one centered at the point (the area of the bands. The one centered at the point (xx,0),0)

has radius has radius and width equal to and width equal to . .

Since we will be integrating with respect to Since we will be integrating with respect to xx (there is a (there is a

band for each band for each xx), we'll factor the ), we'll factor the dxdx out of out of dsds and write and write

. So the area of the band. So the area of the band

centered at (centered at (xx,0) is equal to: ,0) is equal to:

Thus, the total surface area is equal to the integralThus, the total surface area is equal to the integral

x 22 dydxds

dxdx

dyds

2

1

dxx

xd4

112

dxx

x1

4

The surface area The surface area turns out to be:turns out to be:

131333

Good night!….Good night!….

DON’T FORGET:DON’T FORGET:

1. Send in your homework.1. Send in your homework.

2. Meet Nakia and me in the chatroom.2. Meet Nakia and me in the chatroom.

3. Keep in touch by email!3. Keep in touch by email!

4. Have a great week!4. Have a great week!