Calculus July 19-21

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Calculus July 19-21
More on integrals and their applications
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Review of Week 1

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

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Integrals and Areas
A problem that was around long before the
invention of calculus is to find the area of a
general plane region (with curved sides). And
a method of solution that goes all the way back
to Archimedes is to divide the region up into
lots of little regions, so that you can find
the area of almost all of the little regions,
and so that the total area of the ones you
can't measure is very small.
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Ameba
By Newton's time, people realized that it would
be sufficient to handle regions that had three
straight sides and one curved side (or two or one
straight side -- the important thing is that all
but one side is straight). Essentially all
regions can be divided up into such regions.
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These all-but-one-side-straight regions look like
areas under the graphs of functions. And there is
a standard strategy for calculating (at least
approximately) such areas. For instance, to
calculate the area between the graph of y 4x -
x2 and the x axis, we draw it and subdivide it as
follows
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Since the green pieces are all rectangles, their
areas are easy to calculate. The blue parts under
the curve are relatively small, so if we add up
the areas of the rectangles, we won't be far from
the area under the curve. For the record, the
total area of all the green rectangles is
246 25
whereas the actual area under the curve is
Also for the record, 246/25 9.84 while 32/3 is
about 10.6667.
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Area 60 boxes
We can improve the approximation by dividing into
more rectangles Now there are 60 boxes
instead of 20, and their total area is which
is about 10.397. Getting better. We can in fact
take the limit as the number of rectangles goes
to infinity, which will give the same value as
the integral. This was Newton's and Leibniz's
great discovery -- derivatives and integrals are
related and they are related to the area problem.

7018 675
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Limits of Riemann sums

• A kind of limit that comes up occasionally is an
  integral described as the limit of a Riemann sum.
  One way to recognize these is that they are
  generally
• expressed as , where the
• something depends on n as well as on i.

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Green graph

• Again, recall that one way to look at integrals
  is as areas under graphs, and we approximate
  these areas as sums of areas of rectangles.
• This is a picture of
• theright endpoint
• approximation to the
• integral of a
• function.

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approximating
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Example...
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solution
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Area between two curves
A standard kind of problem is to find the area
above one curve and below another (or to the left
of one curve and to the right of another). This
is easy using integrals. Note that the "area
between a curve and the axis" is a special case
of this problem where one of the curves simply
has the equation y 0 (or perhaps x0 )
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Solving area problems
1. Graph the equations if possible 2. Find
points of intersection of the curves to
determine limits of integration, if none are
given 3. Integrate the top curve's function
minus the bottom curve's (or right curve minus
left curve).
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Example
Find the area between the graphs of ysin(x) and
yx(p-x)
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Its easy to see that the curves intersect on
the x-axis, and the values of x are 0 and p.
The parabola is on top, so we integrate
And this is the area between the two curves.
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An Area Question
Find the area of the region bounded by the curves
y4x2 and yx23. A. 1/2 B. 1 C. 3/2 D. 2
E. 5/2 F. 3 G.7/2 H. 4
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Position, velocity, and acceleration
Since velocity is the derivative of position and
acceleration is the derivative of velocity,
Velocity is the integral of acceleration, and
position is the integral of velocity. (Of
course, you must know starting values of position
and/or velocity to determine the constant of
integration.)
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Example...
An object moves in a force field so that its
acceleration at time t is a(t) t -t12
(meters per second squared). Assuming the object
is moving at a speed of 5 meters per second at
time t0, determine how far it travels in the
first 10 seconds.
2
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Solution...
First we determine the velocity, by integrating
the acceleration. Because v(0) 5, we can write
the velocity v(t) as 5 a definite integral, as
follows 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
goes from 0 to 10. But dx v(t) dt. So we can
write (distance traveled between t0 and t10)
3950/3 1316.666...
meters .
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Surfaces of revolutionVolume
A "surface of revolution" is formed when a curve
is revolved around a line (usually the x or y
axis). The curve 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.
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Volumes
You might already be familiar with finding
volumes of revolution. Once a surface is
formed by rotating around the x-axis, you can
sweep out the volume it encloses with disks
perpendicular to the x axis.
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Here is the surface formed...
Here is the surface formed by revolving
around the x axis for x between 0 and 2, showing
one of the disks that sweep out the volume
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To calculate the volume
enclosed inside the surface, we need to
add up the volumes of all the disks. The disks
are (approximately) cylinders turned sideways,
and the disk centered at (x,0) has radius and
width (or height) dx. The volume of the disk is
thus To find the total volume of the solid we
have to integrate this quantity for x from 0 to
2. We get
, or
cubic units
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A formula for volume
In general, if the piece of the graph of the
function of y f (x) between x a and x b is
revolved around the x axis, the volume inside the
resulting solid of revolution is calculated as
The same sort of formula applies if we
rotate the region between the y-axis and a curve
around the y-axis (just change all the x's to
y's).
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The same region rotated around the y axis
A different kind of problem is to rotate the
region between a curve and the x axis around the
y axis (or vice versa). For instance, let's look
at the same region (between y0 and y for x
between 0 and 2), but rotated around the y axis
instead
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The generating curve
Here is the surface being swept out by the
generating curve
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Washers
We could sweep out this volume with washers
with inner radius y2 and outer radius 2 as y
goes from 0 to
Each washer is (approximately) a cylinder with a
hole in the middle. The volume of such a washer
is then the volume of the big cylinder minus the
volume of the hole.
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The volume of the washers...
For the washer centered at the point (0, y), the
radius of the outside cylinder is always equal
to 2, and the radius of the hole is equal to the
corresponding x (which, since , is equal to
y2 ). And the height of the washer is equal
to dy. So the volume of the washer is
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Therefore the volume of the entire solid is
cubic units
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Cylindrical shells
Another way to sweep out this volume is with
"cylindrical shells".
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The volume of a cylindrical shell
Each cylindrical shell, if you cut it along a
vertical line, can be laid out as a rectangular
box, with length , with width and with
thickness dx. The volume of the
cylindrical shell that goes through the point
(x,0) is thus So, we can calculate the volume of
the entire solid to be
cubic units, which agrees with the answer we got
the other way.
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Other volumes with known cross sections
Another family of volume problems involves
volumes of three-dimensional objects whose
cross-sections in some direction all have the
same shape. For example Calculate the volume of
the solid S if the base of S is the triangular
region with vertices (0,0), (2,0) and (0,1) and
cross sections perpendicular to the x-axis are
semicircles. First, we have to visualize the
solid. Here is the base triangle, with a few
vertical lines drawn
on it (perpendicular to the x-axis).
These will be diameters of the semicircles in the
solid.
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3-D
Now, we'll make the three-dimensional plot that
has this triangle as the base and the
semi-circular cross sections.
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3-D BOX
From that point of view you can see some of the
base as well as the cross section. We'll sweep
out the volume with slices perpendicular to the
x-axis, each will look like half a disk
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The volume of that object
Since the line connecting the two points (0,1)
and (2,0) has equation y 1 - x/2, the centers
of the half-disks are at the points (x, 1/2 -
x/4), and their radii are likewise 1/2 - x/4.
Therefore the little bit of volume at x is half
the volume of a cylinder of radius 1/2 - x/4 and
height dx, namely Therefore, the
volume of the solid S is
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Finally...
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 radius 1/2.
Using the formula for a cone, we
arrive at the same answer, cubic units.
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Methods of integration
Before we get too involved with applications of
the integral, we have to make sure we're good at
calculating antiderivatives. There are four basic
tricks that you have to learn (and hundreds of ad
hoc ones that only work in special situations)
1. Integration by substitution (chain rule in
reverse) 2. Trigonometric substitutions (using
trig identities to your advantage) 3. Partial
fractions (an algebraic trick that is good for
more than doing integrals) 4. Integration by
parts (the product rule in reverse) We'll did 1
last week, and well do the others this week.
LOTS of practice is needed to master these!
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Integration by parts
This is another way to try and integrate
products. It is in fact the opposite of the
product rule for derivatives Product rule for
derivatives d(uv) u dv v du . Integration
by parts Use this when you have a product
under the integral sign, and it appears that
integrating one factor and differentiating the
other will make the resulting integral easier.
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Example
-- There's no other rule for this, so we
try parts.
If we differentiate x2 , we'll get 2x which seems
simpler. And integrating ex doesn't change
anything. Let u x and dv ex dx. Then du 2x
dx and v ex. The integration by parts formula
then gives us
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Continuing We can use parts again on this
integral (with u 2x and dv ex dx) to get
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You try one...
contrast this with
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XOXXOX
There is a method called the "tic-tac-toe" method
for integration by parts. Different strategies
for choosing u and dv are called for to integrate
x ln x or x tan-1 x or
something like ex sin x
x
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A problem for you...
p
x cos(x) dx
Find the value of
0
A) p B) 2p C) 2 D) 0
E) -2 F) 1 G) 1/2 H) p/2
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And another
1
x ln x dx
Evaluate
0
A) -1/4 B) -1/2 C) 0 D) 1/4 E) 1/2
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Trigonomeric substitutions
These are just substitutions like the ordinary
substitution method, but some of them are a
little surprising. There are two kinds To
integrate products of powers of sine and cosine
(like sin (x) cos (x) ) . You need the
identities sin (x) cos (x) 1
(everybody knows that one!), and the double
angle formulas cos (x) and sin (x)

1
1
The first,
3
6
2
2
1cos(2x) 2
2
1-cos(2x) 2
2
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The trick is as follows...
(a) If both the power of sine and the power of
cosine are even, then use the double angle
formulas to divide both in half. Keep doing this
until at least one of the powers is odd. (b)
Once at least one of the powers is odd, say it's
the power of cosine, then let u sin( -- ) --
then use the Pythagorean identity to convert all
but one power of cosine to sines, and the last
power of cosine will be the du in a substitution.
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Lets do one...
Both the powers are even, so use the double angle
formula trick
In the last two terms, use the identities again
to get whats on the next slide.
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Trig identity city!
and (finally!) this is something we can
integrate!
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and at last
WOW!!
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Heres one with an odd power for you to do
A) 2/15 B) 4/15 C) 2/5 D) 8/15
E) 2/3 F) 4/5 G) 14/15
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Try this one
A) 2 B) p C) p - 1/2 D) 2 p E)
3p/8
p
cos x dx
4
0
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The second,
2
2
-- and more important, kind of trig substitution
happens when there is a sum or difference of
squares in the integrand. Usually one of the
squares is a constant and the other involves the
variable. If some other substitution doesn't
suggest itself first, then try the Pythagorean
trig identity that has the same pattern of signs
as the one in the problem.
Either
,
, or
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Example
If you think about it, the substitution u
5 - x2 wont work, because of the extra factor of
x in the numerator.
But 5 - x2 is vaguely reminiscent of 1 - sin2,
so let
Then
and
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We make all the substitutions and get
This is a trig integral of the other sort . And
we can use the double-angle formula to get
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Now we have to get back to xs
So far,
where
therefore
and so

Also,
And from the triangle,
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The conclusion is
Quite a bit of work for one integral!
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Examples for you
Here are two for you to work on. Notice the
subtle difference in the integrand that changes
entirely the method used
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Partial Fractions
Last but not least is integration by partial
fractions. This method is based on an algebraic
trick. It works to integrate a rational function
(quotient of polynomials) when the degree of the
denominator is greater than the degree of the
numerator (otherwise "when in doubt, divide it
out") and you can factor the denominator
completely. In full generality, partial
fractions works when the denominator has
quadratic and repeated factors, but we will
consider only the case of distinct linear factors
(i.e., the denominator factors into linear
factors and they're all different). It also uses
the (easy) fact that
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Partial Fractions (continued)
The idea of partial fractions is to take a
rational function of the form (where there can
be more or fewer factors in the denominator, but
the degree of p(x) must be less than the number
of factors) and rewrite it as a sum for some
constants A, B, C (etc..).
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For instance,
You can rewrite ,
first by factoring the denominator
and then in partial fractions as
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Two questions...
1. Why do partial fractions? 2. How to do
partial fractions?
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Two reasons for why
First reason
It helps us to do integrals. From the previous
example, we see that we don't know how to
integrate the fraction
right away, but ...
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Two reasons for why
Second reason
Partial fractions helps us to understand the
behavior of a rational function near its "most
interesting points. For the same example, we
graph the function in blue,
and the partial fractions and
in red
Lets take a closer look
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red and blue curves
Note that one or the other of the red curves
mimics the behavior of the blue one at each
singularity.
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OK, now for the how
First, we give the official version, then a
short-cut. Official version It is a general
fact that the original fraction will break up
into a sum with one term for each factor in the
denominator. So (in the example), write 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 the equation, and then solve
the resulting two equations for the two unknowns
A and B. For instance, we can put x0, and get

and for x1, get
.
Solve to get A 5/2 and B -1/2 as we did
before.
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Short cut
1. Write the fraction with denominator in
factored form, and leave blanks in the numerators
of the partial fractions 2. To get the
numerator that goes over x 3, put your hand
over the x 3 factor in the fraction on the left
and set x -3 in the rest. You should end up
with 5/2. 3. To get the numerator that goes
over x 1, cover the x 1 and set x -1 (and
you get -1/2). It's that simple.
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Another example
First, the numerator and denominator have the
same degree. So we have to divide it out before
we can do partial fractions.
and we can use partial fractions on the second
term
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Partial fractions gives
Use either the official or short-cut method to
get A -1 and B 2. Therefore
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An example for you to try
3
dx x (x-1)
Find
2
A) 3/2 B) 4/3 C) ln 2 D) ln 3
E) ln (3/2) F) ln (4/3) G) ln (2/3) H) 3
ln (2)/2
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Another example
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4x - 6 x - 3x 2

2
3
A) ln (4/3) B) 2 arctan (3) C) ln (9) D)
ln (12/5) E) p/3 - arctan (1/4)
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Arc length
The length of a curve in the plane is generally
difficult to compute. To do it, you must add up
the little pieces of arc, ds. A good
approximation to ds is given by the Pythagorean
theorem We can use this to find the length
of any graph provided we can do the integral
that results!
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Example
Find the arclength of the parabola y x2 for x
between -1 and 1. Since dy / dx 2x, the
element of arclength is
so the total length is
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Conclusion
So far, we have that the length is
To do this integral, we will need a trig
substitution. But, appealing to Maple, we get that
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Can we do the integral?
The arc length integral from before was
This is a trig substitution integral of the
second kind
With the identity tan2 1 sec2 in mind, let
What about dt ? Since we
have that
These substitutions transform the integral into
This is a tricky integral we need to do by parts!
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To integrate
Let
Then
But tan2 sec2 - 1 , so rewrite the last
integral and get
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Still going.
Its remarkable that were almost done. The
integral of secant is a known formula, and then
you can add the integral of sec3 to both sides
and get
So weve got this so far for
where
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We need a triangle!
So far,
with
From the triangle,
So
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Definite integral
So far, we have
Therefore
To get the answer Maple got before, wed have to
rationalize the numerator inside the logarithm.
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Surface Area
The area of a surface of revolution is calculated
in a manner similar to the volume. The following
illustration shows the paraboloid based on (for
x0..2) that we used before, together with one of
the circular bands that sweep out its surface
area.
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To calculate the surface area
To calculate the surface area, we first need to
determine the area of the bands. The one
centered at the point (x,0) has radius
and width equal to .
Since we will be integrating with respect to x
(there is a band for each x), we'll factor the dx
out of ds and write .
So the area of the band centered at (x,0) is
equal to Thus, the total surface area is
equal to the integral
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The surface area turns out to be
13 3
p
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Good night!.
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email! 4. Have a great week!