Tuesday, August 15
The instructor wrote the following list of topics:
Computation of limits ... Definition of derivative ... Computation of derivatives ... Log/exp problems ... Continuity & differentiability ... Linear approximation ... Max/min problems ... Optimization ... Related Rates ... Intermediate Value Theorem ... Mean Value Theorem ... Curve sketching ... Riemann sums ... Fundamental Theorem of Calculus ... Area ... Definite integral ... Substitution ... Initial value problemsAll Math 135 final exams seen by the instructor have questions covering all or almost all of these topics. Sometimes, of course, the questions cover more than one of them, etc.
We did a few problems of various types. In particular we found the area under y=2^{x}+x^{4} as x ranges from 0 to 2. That was almost easy. But then I asked people to sketch the graphs of y=x^{2} and y=1x^{2}, and to compute the area of the region enclosed between the curves. For this we needed to think: to translate the statements into pictures, do some algebra, and then compute some definite integrals.
We also discussed a number of problems from the exam handed out last time.
Monday, August 14
We wrote out solutions to all ten of the homework problems and
then discussed some further examples of substitution. You had to be
there to enjoy it!
sin(5x^{4})x^{3}dx
If u=5x^{4}, then du/dx=20x^{3}, du=20x^{3}dx,
and (1/20)du=x^{3}dx. Then
sin(5x^{4})x^{3}dx=
sin(u)(1/20)du=cos(u)(1/20)+C=cos(5x^{4})(1/20)+C.
e^{17x3}x^{2}dx
Here I would try u=17x^{3} so that du/dx=17(3x^{2})
and du=17(3x^{2})dx. But look at the original integral. We
"need" only x^{3}dx, so du=17(3x^{2})dx becomes
(1/[17·3])du=x^{2}dx. The original indefinite integral in
xland, sin(5x^{4})x^{3}dx, becomes the integral
(1/[17·3])e^{u}du. This is easy, and the result
is (1/[17·3])e^{u}+C. Now back to xland:
(1/[17·3])e^{17x3}+C. You can always check
by differentiating.
Comment
As I mentioned in class, students by now (we've worked
through 10 or 15 examples!) should see how the examples are
constructed:
(x^{2}+1)/(2x^{3}+6x)dx
This is, to me, a more complicated integral. I guess my "eye" focuses
on the bottom of the fraction. I wish that were not there. So I will
try u=2x^{3}+6x. This yields du/dx=6x^{2}+6 and then
du=(6x^{2}+6)dx. But the top of the fraction in the original
integral is x^{2}+1 (and there is a dx also, of course). Well,
6x^{2}+6=6(x^{2}+1). So (1/6)du=(x^{2}+1)dx,
and the original integral becomes
(1/u)(1/6)du. The (1/6) is a
multiplicative constant and "comes out" of the indefinite
integral. The remainder is (1/u)du, which is ln(u)+C. So together we get
(1/6)ln(u)+C=(1/6)ln(2x^{3}+6x)+C.
[1/cos(x)]sin(x)dx
Here the psychological (?) effect was to make students "see" the
likelihood of success when the substitution u=cos(x) is used. Then
du=sin(x)dx and du=sin(x)dx. The original integral becomes
[1/u](du)=ln(u)+C. So, back to x's, we have ln(cos(x))+C.
In fact, we have deduced the last entry in the Math 135 antiderivative table, since sine divided by cosine is tangent.
f(x)  f(x) dx=F(x)+C 

tan(x)  ln(cos(x))+C or ln(sec(x))+C 
ln(cos(x)) is the same as ln(sec(x)) since ln(A)=ln(1/A) and 1/cos is sec.
Why the "+C"?
The "+C" helps to remind people that there is a whole family of
antiderivatives. Historically calculus started with problems from
physics and geometry, and the "+C" is used in physics to pick out
preferred antiderivatives. Let me show you an example.
Throwing a ball
Most people who have taken a high school physics course have seen
problems similar to what follows:
I stand on a cliff which is 10 feet high, and then I throw the ball
straight up in the air at 20 feet per second. I will assume that
(earth) gravity pulls the ball down, and the figure I remember for
earth's gravity is that it pulls the ball down at 32 feet per
second^{2}. Suppose y(t) is the height in feet from the ground
level and up is positive. t is the time in seconds after the ball is
thrown.
What is the position of the ball t seconds after
it is thrown?
Solution
How do the statements
in the previous paragraph translate to calcspeak?
English  calcspeak  Physics words 

I stand on a cliff 10 feet high  y(0)=10  This is position. 
I throw the ball straight up in the air at 20 ft/sec.  y´(0)=20  This is velocity. 
Gravity pulls a ball down at 32 ft/sec^{2}.  y´´(t)=32 (the minus sign means down)  This is acceleration. 
Since y´´(t)=32, we get y´(t)=32 dt=32t+C. Now we plug in t=0, 32·0+C=20, so C=20. Therefore y´(t)=32t+20.
The next step backwards is y(t)=32t+10 dt=16t^{2}+20t+C. This C will be identified using y(0)=10: Therefore 16(0)^{2}+20(0)+C=10 and this C is 10.
Therefore y(t)=16t^{2}+20t+10 is a formula giving the height above ground at time t (valid for t≥0).
Some standard questions
Vocabulary
As I mentioned in class, equations such as y(0)=10 and
y´(0)=20 are called initial conditions. In practice,
initial conditions are used to specify one out of a whole bunch of
candidate antiderivatives. The simplest and first models in ecology
and epidemiology and business are frequently differential equations
with initial conditions: these models are called initial value
problems. The initial condition could represent something like the
lemming population at a certain time or the amount of money available
to invest at a certain time. The differential equation is then a way
of describing how the quantity changes with time.
Example #1
Suppose dy/dx=5e^{x/4}+2e^{3x} and y(0)=7. Find a
formula for y(x).
We find the possible formulas by computing the indefinite integral:
5e^{x/4}+2e^{3x}dx.
Each part of this integral can be computed with a substitution (but a
different substitution for each part!):
So for 5e^{x/4}dx please
use u=x/4 so that du/dx=1/4, du=(1/4)dx, and dx=4du. The integral then becomes
20e^{u}du=20e^{u}+C and back in xland this
is 20e^{x/4}+C.
And for 2e^{3x}dx please
use u=3x so that du/dx=3, du=3dx, and dx=(1/3)du, The integral then
becomes (2/3)e^{u}du=(2/3)e^{u}+C=(2/3)e^{3x}+C.
The formula for y(x) is y(x)=20e^{x/4}+(2/3)e^{3x}+C
and the initial condition y(0)=7 becomes
20+(2/3)+C=7, so that C=27(2/3)(79/3) and
y(x)=20e^{x/4}+(2/3)e^{3x}+[79/3].
Example #2
Suppose y´´(x)=3cos(x)+4x^{2}, and we know that
y´(0)=5 and y(0)=2. Find a formula for y(x). Here we need too
antidifferentiations, one going from y´´(x) to y´(x) and the
next, going from y´(x) to y(x). After each transition, we will
use an appropriate initial condition to pick out the specific function
of interest.
Since y´´(x)=3cos(x)+4x^{2},
y´(x)=3cos(x)+4x^{2}dx=3sin(x)+(4/3)x^{3}+C. Since
we know y´(0)=5, if we "plug in" x=0, we get
3sin(0)+(4/3)0^{3}+C=5. This C is 5, and the specific formula
for y´(x) we get is 3sin(x)+(4/3)x^{3}+5.
Since y´(x)=3sin(x)+(4/3)x^{3}+5, y(x)=3sin(x)+(4/3)x^{3}+5 dx=3cos(x)+(1/3)x^{4}+5x+C.
Plug in x=0 and use y(0)=2. The result is
3cos(0)+(1/3)0^{4}+5·0+C=2. Please be careful about
this equation, since the result is 3+C=2, and therefore C=5. The
desired formula is then
y(x)=3cos(x)+(1/3)x^{4}+5x5.
Thursday, August 10
Students volunteered to write solutions of the homework problems. The
word "volunteer" is used ... in the sense of impressment (this means
"enforced service in the army or navy").
Rewriting FTC
FTC stands for Fundamental Theorem of Calculus.
I remarked that most of what is important in a first semester calculus
course is included in FTC and MVT. The latter is the Mean Value
Theorem which I rewrote.
Rewriting MVT
[f(b)f(a)]/[ba]=f´(c) for some c between a and b.
Functions which have zero derivative
If the derivative of a function is always 0, then the MVT tells
us something interesting. We may not know what much about the c which
appears, but "always 0" means f´(c)=0. Then the top of
the fraction on the lefthand side of the equation is 0, so that f(b)
must always equal f(a). This means the function's values are constant
 they can't change! So any function whose derivative is always 0 is
a constant function.
We can restate this using the physical model, where f(t) is the
position of a particle, then f´(t) is the velocity of the
particle. Well, if the particle's velocity is always 0, then "clearly"
(?) the particle never changes position. In the history of math, this
"clearly" took over a century to understand. This is because the
velocity is the instantaneous rate of change of position (the limit!)
and it wasn't clear, really, how to relate the instantaneous rate of
change with the position at two times. The Mean Value Theorem became
an accepted part of calculus quite a while after the subject started.
Functions which have the same derivative
Suppose we know two functions, F(x) and G(x), and that these functions
have the same derivative for all x: F´(x)=h(x) and
G´(x)=h(x) for all x. Well, the difference F(x)G(x) will have
derivative w(x)w(x) and so it is 0 for all x. Therefore F(x)G(x)=C
(some constant) for all x. So F(x)=G(x)+C. The geometric
consequence is shown to the right. I hope that the picture
reflects the following implication:
if two functions have the same
derivative for all x, then the graphs of the functions are up/down
(don't know which) translations of each other. The curves must be
parallel.
All antiderivatives ...
If you know one antiderivative, then you know them all. What I mean is
that if F(x) is an antiderivative of f(x), then all
antiderviatives of F(x) have the form F(x)+C.
Example (stupid)
Well, sqrt(x^{3}+17)9 has derivative equal to
(1/2)(x^{3}+17)^{1/2}. Therefore a complete
description of all antiderivatives of
(1/2)(x^{3}+17)^{1/2} is
sqrt(x^{3}+17)9+C where C is any constant.
People didn't like this because the darn 9 looked silly. Sometimes,
though, you can't detect the "silly" part. We will see further
examples.
Vocabulary
F(x) is called an antiderivative of f(x) if F´(x)=f(x).
F(x) is called an primitive of f(x) if F´(x)=f(x).
F(x) is called an indefinite integral of f(x) if
F´(x)=f(x).
Yes, all of these words and phrases are commonly used for the same
thing: I'm sorry. And there is notation. If F´(x)=f(x) people
frequently write f(x) dx=F(x)+C.
There are no numbers at the top and bottom of the long S, the integral
sign. So the integral does not refer to a definite value or
area or region ... it is indefinite. And most people write "+C" to
remind themselves that "the" antiderivative can refer to any one of a
large family of functions. This turns out to be important in some
other computations. In definite integral computations, most people
omit the +C because evaluating the result at the top minus the result
at the bottom cancels +C's.
Antiderivative information
Any information about derivatives is information about
antiderivatives: just reverse the connection. So our table of
antiderivatives on the formula sheet can be turned into a table of
antiderivatives. Of course, some folks like the table to look
better. For example, the derivative of x^{n} is
nx^{n1}. "Thus" the antiderivative of nx^{n1} is
x^{n}+C. But people like the table to be more user
friendly. They would prefer to be able to find the antiderivative of
x^{n}. Well, this likely would result from differentiating
x^{n+1}. But that function's derivative is (n+1)x^{n}:
so we need to fix up x^{n+1} by 1/(n+1). Therefore the
antiderivative of x^{n} is [1/(n+1)]x^{n+1}. But when
division cccurs, check the bottom to make sure embarrassing things
aren't happening: hey, this will only work if n is not equal to 1. So
we need to think about the antiderivative of x^{1}=1/x: but
that's ln(x) (how wonderful). Anyway, the child's table of
antiderivatives using the information we have accumulated so far might
look like this:
f(x)  f(x) dx=F(x)+C 

0  C (a constant) 
x^{n} if n not 1  [1/(n+1)]x^{n=1}+C 
x^{1}=1/x  ln(x)+C 
kf(x) k a constant  kF(x)+C if F(x) is one antider. of f(x) 
f(x)+g(x)  F(x)+G(X)+C if F(x), respectively G(x), is one antider. of f(x), respectively G(x). 
sin(x)  cos(x)+C 
cos(x)  sin(x)+C 
e^{x}  e^{x}+C 
a^{x}  [1/ln(a)]a^{x}+C 
Example #1
4x^{6}9x+3+5/x^{8}dx
Here please notice that 3 is 3x^{0} and 5/x^{8} is
5x^{8}. If you have this in mind, then the antiderivative can
just be written out, glancing up at the table for help:
4(1/7)x^{7}9(1/2)x^{2}+3x+5[1/(7)]x^{7}+C.
Example #2
9x^{13}7e^{x}dx=9(1/14)x^{14}7e^{x}+C
Example #3
87sqrt(x)+5sin(x)dx=87(1/[3/2])x^{3/2}5cos(x)+C
Here you need to "decode" sqrt(x): it is x^{1/2} and then this
can be handled with the x^{n} entry in the table. Notice that
the antiderivative of sine gives a minus cosine!
Example #4
(x^{2}+(1/x^{3}))^{2}dx
Here the problem is that ... we may not know what to do! The best
thing is to "square out" the integrand (that's what the function to be
integrated is called). The result:
(x^{2}+(1/x^{3}))^{2}=x^{4}+2(x^{2})(1/x^{3})+1/x^{6}.
Therefore we need to find the antiderivative of
x^{4}+2/x+x^{6}. Each piece of this can be handled by
referring to our table and simple arithmetic. The result is
(1/5)x^{5}+2ln(x)(1/5)x^{5}+C. I think this is a bit
tricky.
Another wrinkle: #1
Let's consider (x^{3}+17)^{8}3x^{2}dx. If you look
long enough at this, maybe you can see the function the integrand
should come from must be something like
(x^{3}+17)^{9}. When this function is differentiated,
the Chain Rule will "spit out" the derivative of the inside
multiplying 9(x^{3}+17)^{8}. The derivative of the
inside is 3x^{2}, so the anbswer is
(1/9)(x^{3}+17)^{9}+C.
Another wrinkle: #2
How about (3x+2)^{100}dx? I could imagine, barely,
computing this integral by "expanding" the 100^{th} power. The
result would be an enormous polynomial, and then I could
antidifferentiate each piece of the polynomial, one by one. Or I could
think a bit: how could we get a function of the form (3x+2)^{100}
as the output of differentiation? Well, maybe one higher power
could be considered. Thus we could guess at (3x+2)^{101}. But
when we differentiate this, the 101 comes out in front, and a 3 comes
out in back (Chain Rule). I can fix up these little problems, and I
bet that
(3x+2)^{100}dx=[1/{101·3}](3x+2)^{101}+C.
You can check this guess by differentiating. The guess
is correct.
Another wrinkle: #3
Now this: 4e^{5x+6}dx. Here again we could guess and
correct. The process will yield (if we make no mistakes!) the
following result: (4/5)e^{5x+6}+C. Again, you can check
this guess by differentiating. The guess is correct.
Substitution
This is sort of the chain rule backwards and is a way of keeping track
of the constants which arise in computations such as those "guesses" I
just wrote. The idea of this method is to prevent mistakes that
guesses (especially those guesses which are too elaborate!) sometimes
cause. Let me go over the three examples that I just did, with a
slightly different language.
Wrinkle #1 rewrinkled
(x^{3}+17)^{8}3x^{2}dx. Take
u=x^{3}+17. Then du/dx=3x^{2} and (if you appreciate
the symbolism!) du=3x^{2}dx. Now let me Translate the
integral I started with from xland to uland. Each piece involving x
needs to be changed to a piece in terms of u. So:
(x^{3}+17)^{8}3x^{2}dx=u^{8}du
Now I will integrate the indefinite integral in u, and the result is
(1/9)u^{9}+C. And now use our dictionary to return to xland
from uland: (1/9)(x^{3}+17)^{9}+C. This is the answer
we guessed at and can confirm by differentiation. I invented this
example to show off the substitution method, and the success therefore
should not be too surprizing.
Wrinkle #2 rewrinkled
(3x+2)^{100}dx: here
u=3x+2 so du/dx=3, du=3dx, and (1/3)du=dx. The integral goes from
xland to uland:
(3x+2)^{100}dx=u^{100}(1/3)du
Now u^{100}(1/3)du=(1/3)u^{100}du=(1/3)(1/{101})u^{101}+C. And back to
xland, using the dictionary for this problem:
(1/3)(1/{101})(3x+2)^{101}+C. I hope that you see the
substitution method is an effort to run the Chain Rule "backwards".
Wrinkle #3 rewrinkled
4e^{5x+6}dx. Here I'll try u=5x+6 so that du/dx=5 and
du=5dx. I notice that I actually have 4dx "left over" so I will modify
the equation du=5dx. It becomes (1/5)du=dx and then (4/5)du=4dx. The
result for translation from uland to xland follows:
4
e^{5x+6}dx=e^{u}(4/5)du
Then the antiderivative is (4/5)e^{u}+C and, back in xland,
this is (4/5)e^{5x+6}+C.
Comments
All of these examples are arranged to work well with this substitution
"trick". Essentially we are working with f(u(x))u´(x)dx and rewriting it as
f(u)du. If we then "happen" to know an antiderivative of f is F,
we can "see" that f(u)du=F(u)+C, and the original indefinite
integral gets solved with f(u(x))u´(x)dx=F(u)+C.
The method is successful sufficiently often that it is the first
general antidifferentiation "trick" shown to calculus students. There
are several questions which occur almost immediately. How should u be
"chosen"? Well, I usually first try some "mess" whose disappearance
will make the integral "easier". And I look, at the same time, for
other pieces which will combine to be close to the resulting du. And,
no, it doesn't always work. There are integrals which can't be done
using this sort of trick. Then other tricks are tried. And sometimes
they don't work.
Today's last try: sin(7x^{3}+17)x^{2}dx
If u=7x^{3+17}, then du/dx=21x^{2}, and
du=21x^{2}dx. But we have x^{2}dx in our integral, so
we need to play a bit: du=21x^{2}dx turns into
(1/{21})du=x^{2}dx, and
sin(7x^{3}+17)x^{2}dx in xland becomes
sin(u)(1/{21})du in uland. This indefinite integral I can do
by using the table. The result is cos(u)(1/{21})+C and back to
xland: cos(7x^{3}+17)(1/{21})+C.
You can always check the answer. Differentiation is basically easy to do, and so if someone suggests an antiderivative, verifying to suggestion is usually not too hard.
Wednesday, August 9
I urged students to view much of what was discussed today as
entertainment of the most magical kind. The aim is to show how
at least three different civilizations, mostly independently, computed
definite integrals (their integrals were area and volume
numbers). Then I want to show how analyzing a harder problem
amazingly gives what is an easier method (in many examples) for
computing definite integrals.
Back in the good old days (China/Greece/India ...)
This is the approach many very clever people worked on for
centuries. I will start with a problem whose answer we already know,
and hope that we will get the answer we anticipate. I hope then that
the approach will be successful when we try a problem whose answer is
not generally known.
Area of a triangle
Let's find "the area under y=x when x is between 0 and 1". This is the
language used for the area indicated in the picture to the right. It
is a triangle whose vertices are (0,0) and (1,0) and (1,1). The base
and height are both 1 unit long, and the area is half the product of
the base and height, so the area should be 1/2.
I'll split up the interval [0,1] into n equallengthed pieces. So this
means the intervals will be determined by
0 1/n 2/n 3/n ... (ii)/n i/n ... (n1)/n n/n
(yes, that's equal to 1).
The length of each subinterval is 1/n. The i^{th} subinterval
is determined by its endpoints: [(i1)/n,i/n]. I'll approximate the
area of the triangle by "constructing" a bunch of approximating
rectangles and computing the sum of their areas. The i^{th}
approximating rectangle will have the height of the formula y=x at the
righthand endpoint of the subinterval [(i1)/n,i/n], so its height
will be i/n. The area of the i^{th} rectangle will therefore
be height·base, and this is (i/n)·(1/n), which is
i(1/n^{2}). There are n rectangles, and the sum of their areas
is
1·(1/n^{2})+2·(1/n^{2})+3·(1/n^{2})+...+i·(1/n^{2})+...n·(1/n^{2})
Trying to describe this sum in
a culture which does not know algebra is an intriguing and
difficult task: how many
words and sentences and ... would be necessary to describe what we
wrote so compactly above? And, since we do have algebra, I hope that
you can see that 1/n^{2} can be "factored out" and the
expression becomes what follows:
(1+2+3+...+i+...+n)·(1/n^{2}).
For example, 1+2+3+4+5+6, as I computed in class, is equal to 21. But
I am not really interested in what happens for n=6. I want to examine
the "asymptotic behavior" as n gets large. The most efficient way to
do that is to get some neat formula for 1+2+3+...+i+...+n. This is
what's called an arithmetic progression. You may remember a neat
formula, or you might look at the following:
1 + 2 + 3 + 4 + 5 + 6
6 + 5 + 4 + 3 + 2 + 1
If we add up the "columns"  that is, add things vertically, then the
result is
7 + 7 + 7 + 7 + 7 + 7
When the entry on the first line goes up, the entry on the second line
goes down. So the column sums are all the same. There are six of them,
so the sum is 6(7). But this is the sum of two repetitions of
1+2+3+4+5+6, so that sum should be the same as 6(7)/2. Well, this is
correct. And, indeed, 6(7)/2=42/2=21, the same answer as we got by
direct computation. Therefore I can suggest the following formula:
(1+2+3+...+i+...+n=n(n+1)/2: a magic formula!
Now I can write the sum of the areas of the approximating rectangles,
(1+2+3+...+i+...+n)·(1/n^{2}), using this formula. The
result is:
[n(n+1)/2]·(1/n^{2})
We know algebra, and much of the whole course has been an effort to
coax you through the next few equalities.
[n(n+1)/2]·(1/n^{2})=[n^{2}+n]/[2n^{2}]=[1+{1/n}]/2
The last step is accomplished by multiplying the top and bottom of the
expression by 1/n^{2}. But I want to know what happens as
n>infinity. Well, clearly [1+{1/n}]/2>1/2 since
1/n>0. The word clearly is used here not in truth, but
to signal, almost, hey, something significant has happened, and the
whole process may not (probably is not!) "clear".
We computed an expression for the sum of the areas of the approximating rectangles. Then we computed the limit of this expression. The area of the triangle is the limit of the area of the approximating regions, so the area of the triangle should be 1/2. We know this already. I did not use the language of the past few class meetings, but I hope you "see" the appearance of regular partitions, righthand endpoints as sample points, Riemann sums approximating a definite integral, etc.
Area of a parabolic region
I'm going to try the same approach with "the area under
y=x^{2} when x is between 0 and 1". Here the region whose area
I want to compute shares two straight line segment boundaries with the
right triangle (the xaxes and part of the line x=1). The hypotenuse
is replaced by a part of the parabolic arc y=x^{2} connecting
(0,0) and (1,1). My approach to computing this area will be to imitate
the previous work as much as possible. Alll of this could have been
done (with no algebra!!!) by apppropriate representatives from
the three civilizations mention. There will be an interesting
increase in difficulty at one stage, but let me begin:
I'll split up the interval [0,1] into n equallengthed pieces. So this
means the intervals will be determined by
0 1/n 2/n 3/n ... (ii)/n i/n ... (n1)/n n/n
(yes, that's equal to 1).
[This paragraph copied
exactly.]
The length of each subinterval is 1/n. The i^{th} subinterval
is determined by its endpoints: [(i1)/n,i/n]. I'll approximate the
area of the parabolic region by
"constructing" a bunch of approximating rectangles and computing the
sum of their areas. The i^{th} approximating rectangle will
have the height of the formula y=x^{2} at the righthand endpoint
of the subinterval [(i1)/n,i/n], so its height will be (i/n)^{2}. The area of the
i^{th} rectangle will therefore be height·base, and
this is (i/n)^{2}·(1/n), which is
i^{2}(1/n^{3}). There
are n rectangles, and the sum of their areas is
1^{2}·(1/n^{3})+2^{2}·(1/n^{3})+3^{2}·(1/n^{3})+...+i^{2}·(1/n^{3})+...+n^{2}·(1/n^{3})
... I hope that
you can see that 1/n^{3} can be "factored out" and the
expression becomes what follows:
(1^{2}+2^{2}+3^{2}+...+i^{2}+...+n^{2})·(1/n^{3}).
[Here some changes needed to be made,
and I tried to indicate them with a change in type font and
color.]
Now let me abandon copying for a while. I'd like to try to find a
compact formula for the sum
1^{2}+2^{2}+3^{2}+...+i^{2}+...+n^{2}.
I computed
1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}
in class. I think the result was 1+4+9+16+25+36=91. I can't just
reverse the sum and add things up, since the gaps between the
successive numbers change. There is a magical formula for this sum. I
found this
link to a web page which discusses how the formula is verified. I
want to use the formula, and am willing to accept it. So:
1^{2}+2^{2}+3^{2}+...+i^{2}+...+n^{2}
is equal to [n(n+1)(2n+1)]/6
I did verify that if we plug in n=6 into [n(n+1)(2n+1)]/6 the result
is [6(7)(13)]/6 which is 7(13) which is 91, and I had added up
1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}
and also got 91 as the answer. No, this is not a "proof" but at least
the formula agrees with the data we have. Anyway, we go back to the
old folks:
Now I can write the sum of the areas of the approximating rectangles,
1^{2}·(1/n^{3})+2^{2}·(1/n^{3})+3^{2}·(1/n^{3})+...+i^{2}·(1/n^{3})+...+n^{2}·(1/n^{3}), using this formula. The
result is:
[n(n+1)(2n+1)]/6·(1/n^{3})
We know algebra, and much of the whole course has been an effort to
coax you through the next few equalities.
[n(n+1)(2n+1)]/6·(1/n^{3})=
n (n+1) (2n+1) (1+{1/n})(2+{1/n})  =  6 n·n·n 6(This was done by dividing each of the factors of the top by one of the n's in the bottom.)
I'll stop copying. What happens as n>infinity? Well, clearly since 1/n>0, our approximating area amount > 2/6=1/3. So 1/3 "is" the area (I guess 1/3 square units, as mentioned in class).
Comments
The analysis of the area problem is very clever. But if other,
similar problems are thought about, then it seems like we will need a
magical formula in every
case. Imagine doing this with trig functions or exponential functions
(the oldtime wizards did handle these cases). But I will show another
approach to this problem.
Another way: do a harder problem
The approach is almost paradoxical. You study a more difficult
problem, and solve that problem. As a result, the problem you started
with is solved easily! The whole process is mechanized, almost. Here
in this case, we want to compute the area of the parabolic region,
which had boundary goiven by the horizontal axis and the parabolic arc
and the vertical line segment. Let me temporarily relabel the
horizontal axis with t (time?). Then the curve becomes
y=t^{2}. I will define a new function, A(x), in the
following way:
A(x) is the area under the parabola from 0 to x. So A(x) is the area bounded by the curve y=t^{2}, y=0, and t=x.We want to compute A(1).
Still photography versus moving pictures?
Maybe you can think about the difference between computing A(1) and
investigating A(x) as a difference between the static/still and
kinetic/moving viewpoints. There are more pictures, and maybe the
insight of how the pictures "connect" will allow us to understand just
one of the pictures better.
What can we say about this function?
I want to learn about A(x). This is what I will learn.
One value of A(x)
I know A(0). This is an almost trivial observation. If I move the
xvalue, the right size of the picture, back towards 0, then the
region just becomes a dot. Its area will be 0. So A(0)=0.
Differentiate it
One thing that is learned in a calculus course is to differentiate
everything in sight. The derivative of A(x) uses A(x+h)A(x) divided
by h. We usually think of h as small. Suppose that h is a small
positive number (the same logic works with some changes of
inequalities if h is negative). Then A(x+h) is the area of a region
which is almost the same as the region whose area is A(x). We just
move the boundary line segment on the right somewhat more to the
right. That is, we move the line segment to x+h. What about
A(x+h)A(x)? This is the area of the region which is below the
parabola, above the horizontal axis, and between x and x+h. What can
we say about it? Since "squaring" is particularly simple function
(well, for example, it is increasing) I know that the height in the
slice between x and x+h will be between x^{2} and
(x+h)^{2}. The width of the slice is h, so that
x^{2}h<A(x+h)A(x)<(x+h)^{2}h
Suppose we
divide this inequality by the positive number h. The result is:
A(x+h)  A(x) x^{2} <  < (x+h)^{2} hWhat happens as h>0? Certainly the x^{2} doesn't change. On the farthest right, consider the term (x+h)^{2}: as h>0, this term must >x^{2}. Huh. Well, the term is the middle is squeezed between two formulas which both >x^{2} as h>0. So it should also >x^{2}. Therefore we have "computed" A´(x): it is x^{2}.
So we can solve the problem!
We now know that:

There is even special notation for "F(b)F(a)" because it occurs so often. Usually people will write F(x)]_{a}^{b} or F(x)_{a}^{b}. The text uses the second version, without the little hooks at the end. The text used when I was taught calculus had the hooks, so surely I will slip sometimes and write them.
Any antiderivative?
We decided in the previously discussed area problem that the area,
which equals _{0}^{1}x^{2} dx, must be 1/3. We
actually found A(x)=(1/3)x^{3} and got 1/3 because it was
A(1). But what if I wanted (?) to use F(x)=(1/3)x^{3}+73 as my
antiderviative of x^{2}. Then the FTC says that the area
should be F(1)F(0). But this is
((1/3)1^{3}+73)((1/3)0^{3}+73) and the 73's just cancel each other, and we again
get 1/3 as the answer. So in computing the definite integral, the
constants involved in specifying the antiderivative cancel. The
constants are important in connection with other computations, as you
will see.
Example #1
x^{5} on [2,5]
Example #2: how big is the hump?
Look at the curve y=sin(x).
What is the area under the first bump
of this sine curve?
The accompanying picture is supposed to be
very sloppy. I wanted to emphasize the need to have some size of the
answer. The answer should not be very large (456 is too darn
large). The answer should not be very small (1/456 is too
darn small).
Although I sincerely hope that in most situations in the future you will have computational help (machines!) you should have some idea of the size of your answer, so that when the machine reports something really far from the correct answer, you'll know enough to worry a bit. So I try to imagine the answer. Look at the (better!) picture to the right. The sine bump is inside a box whose height is 1 and whose width is Pi, so the area, when we compute it, should be less than Pi. Also the area under the bump contains a triangle (as shown) whose height is 1 which has a base of width Pi, so the area of the bump should be more than Pi/2. Now maybe I will compute the area.
So I need _{0}^{Pi}sin(x) dx. I know a function whose derivative is sin(x). Such a function is cos(x) (keep track of minus signs!). So the area is cos(x)_{0}^{Pi}. This is cos(Pi)[cos(0)]. This is (1)(1)=2 (keep track of minus signs!). You can check that 2 is between Pi and Pi/2 as we thought it should be.
Example #3
I computed this definite integral:
_{1}^{1}x^{3}x dx. According to
FTC, I can evaluate this by first finding an antiderivative, and I
guess the
antiderivative:(1/4)x^{4}(1/2)x^{2}. So I must
compute (1/4)x^{4}(1/2)x^{2}_{1}^{1}=((1/4)1^{4}(1/2)1^{2})((1/4)(1)^{4}(1/2)(1)^{2})=(1/4)(1/4)=0. I don't think errors were made. Is
there no area "under" this curve?
The answer is that we computed a definite integral. Remember that the definite integral counts area below the horizontal axis negatively. The function f(x)=x^{3}x is antisymmetric or odd. Its graph is shown to the right. The geometric area inside the two bumps is the same, but the signed area in the lefthand bump is positive, and the signed area in the righthand bump is negative. The result when they are combined (in the definite integral from 1 to 1) is 0.
The instructor is not too smart ...
When I discussed the problem involving f(x)=x^{3}x in class,
I tried to sketch a graph rapidly. So I wrote x^{3}x=0 and
then x(x^{2}1)=0 and then x(x1)(x+1)=0. There are roots at
0 and 1 and 1. There aren't any other roots (a polynomial of degree 3
can only have 3 roots) and the polynomial changes sign as the x
"crosses" each of the roots. So I drew bumps in the wrong order (minus
the truth). Oh well, I do apologize but the idea is still "correct":
the geometric areas in the picture cancel and the signed or oriented
areas reported by the definite integral are, together, equal to 0.
Example #4
Let look at y=x(x1)(x3), slightly changed from the example I
actually did in class. I wrote this already factored, and the roots
are 0 and 1 and 3. A picture is shown to the right. Can we find the
geometric area (not the signed area!) enclosed by this curve
and the xaxis? I look at the picture and see that the definite
integral from 0 to 1 will be positive and will equal the geometric
area. The definite integral from 1 to 4 will be negative, and if the
sign is "corrected" (take away the minus!) the result will be the
geometric area. If we computed the integral over the interval [0,3]
some of the positive and negative values will cancel and the result
will not be what we wnat. So I'll compute two integrals. In both cases
we'll need an antiderivative of x(x1)(x3). But when the polynomial
is written in factored form, I can't readily guess an
antiderivative. Let me multiply out:
x(x1)(x3)=x(x^{2}4x+3)=x^{3}4x^{2}+3x
Now I "guess" an antiderivative:
(1/4)x^{4}(4/3)x^{3}+(3/2)x^{2}.
Note There isn't that much guessing going on. I hope you
already suspect this. An organized method for obtaining
some antiderivatives will be presented in the next class.
We compute:
_{0}^{1}x(x1)(x3)dx=_{0}^{1}x^{3}4x^{2}+3x dx=
(1/4)x^{4}(4/3)x^{3}+(3/2)x^{2}_{0}^{1}=
((1/4)1^{4}(4/3)1^{3}+(3/2)1^{2})((1/4)0^{4}(4/3)0^{3}+(3/2)0^{2})=(1/4)(4/3)+(3/2)={316+18}/12=5/12.
_{1}^{3}x(x1)(x3)dx=_{1}^{3}x^{3}4x^{2}+3x dx=
(1/4)x^{4}(4/3)x^{3}+(3/2)x^{2}_{1}^{3}=
((1/4)3^{4}(4/3)3^{3}+(3/2)3^{2})((1/4)1^{4}(4/3)1^{3}+(3/2)1^{2})=
((81/4)(108/3)+(27/2))((1/4)1^{4}(4/3)1^{3}+(3/2)1^{2})=
({243432+162}/12)({316+18}/12)=(27/12)(5/12)=32/12.
Well, maybe not doing this example in class was a good idea: the numbers are not pleasant. But please not that the first definite integral has value 5/12, positive as it should be. The second definite integral is 32/12, and this is a negative "area", below the horizontal axis. The geometric area bounded by the line and curve must then be (5/12)+(32/12)=37/12. I made several errors with the arithmetic (signs and fractions!) in finishing this example. Oh well.
Tuesday, August 8
We went over some homework problems. Specifically, we looked at
problems 1 and 6 and 7 and 8. In each of these problems, we were given
the function f(x)=x^{2}+x and the interval [0,3]. We were then
given various partitions and sample points, and asked to write
(actually, to compute!) the Riemann sum associated with these
choices.
I'll discuss problem 6, because I asked some additional questions which were relevant to an important assertion I made later in the day's presentation.
Problem 6 of chapter 8
In problem 6, we were given the function f(x)=x^{2}+x and the
interval [0,3]. The problem asked about the Riemann sum which resulted
from using a regular partition with six subintervals, and the
choice of same points was determined by using the righthand endpoints
of each subinterval.
A regular partition has equal subintervals. since the length of [0,3]
is 3, the length of the subintervals of a regular partition with six
subintervals is 3/6: that's onehalf. The partition points will be
{0,1/2, 1, 3/2, 2, 5/2, 6}: there are seven of these, one more than
the number of subintervals requested. What about the sample points?
They will be 1/2 and 1 and 3/2 and 2 and 5/2 and 3 respectively: these
are the righthand endpoints of each subinterval. So the Riemann sum
will be this:
((1/2)^{2}+(1/2))·({1/2}0)+((1)^{2}+(1))·(1{1/2})+((3/2)^{2}+(3/2))·({3/2}1)+((2)^{2}+(2))·(2{3/2})+((5/2)^{2}+(5/2))·({5/2}2)+((3)^{2}+(3))·(3{5/2})
We were also requested to sketch y=x^{2}+x on the interval
[0,3], and draw areas corresponding to the Riemann sum. This is what
I've attempted to do in the picture shown. I mentioned in class that
the curve y=x^{2}+x goes through (0,0) and (3,12). I also
mentioned (hey, y´=2x+1, positive in [0,3]) that the curve is
increasing between these points. I also mentioned (hey,
y´´=2, also positive in [0,3]) that the curve is concave up
between these points. So the graph drawn is qualitatively correct. I
should mention that the darn curve does go through the correct points,
so it is more or less quantitatively correct, also (note that the
vertical and horizontal axes have different scales, however).
A question which is not in the book
Suppose we wanted to compute the area bounded by y=x^{2}+x,
x=0, x=3, and the xaxis. This area is usually called the area "under
y=x^{2}+x between 0 and 3". How close is this specific
Riemann sum to the actual area? Let me ask two questions:
And then problem 7 ...
The same function, and the same interval, and a regular partition with
100 subdivisions, and the sample points at the righthand endpoints of
the subintervals. Well, what happens? The subinterval width is now
.03, because we are dividing the interval into 100 equal parts.
The Riemann sum itself is something like:
SUM as i goes from 1 to 100 of
(({3/100}i)^{2}+({3/100}i))·(3/100)
In the i^{th} subinterval, the righthand endpoint is (3/100)
(the width) multiplied by the number of the subinterval, and the
result is {3/100}i. Since we're "starting" from 0, this is the correct
"formula".
What about the error?
Now if I try to draw a picture similar to what's above, well, the
details are too darn difficult. But the idea is the same. There
are 100 error pieces. They all have width 3/100 and they all can be
moved to the right to stack up inside a rectangle of width 3/100 and
height, still, 12. Now we can see again that the Riemann sum is an
overestimate that the overestimate is no bigger than
12·(3/100), which is 36/100, about onethird. This is certainly
less than the error of the approximation in problem 6, which we
estimated as 6.
Huh: a mystery function!
The instructor now asked students: what if
f(x)=sqrt(x^{5}+x^{2}+x+1)? The correct response of
the students might have been, well, what if? But we tried to graph
this function on the interval [0,1]. The strangeness of the function
was partially explained by the "nice" facts: f(0)=1 and f(1)=2. I
defined the function so that it had nice values at the endpoints. Then
I remarked that this function was increasing (calculus tells me this,
since f´(x)=(1/2)(STUFF)^{1/2}(5x^{4}+2x+1),
and this is certainly positive in [0,1]). The graph is also concave
up, but I don't care since I really don't want to find another
derivative. In fact, a machinegenerated graph is shown to the right.
Such a simple graph. What if, for some really silly reason, we want to find the area "under" y=f(x) on the interval [0,1]. Again, this phrase refers to the area bounded by the graph y=f(x) and the lines x=0, x=1, and the xaxis (y=0). I do not believe that any application I know of would require computation of this area. But there are much more complicated "areas" (not really areas, but see below, please!) which people very much want to compute. Well, I turn to the program which handles much of the symbolic computation that I want to do, since my brain is small. I ask the program (I'll show you later, really) to compute this area. The program, Maple, is one of the standard programs (similar to Mathematica) used all over the world to do lots of computations. The program is like a very powerful version of the TI89. Its design was begun in 1980, and it has been developed by literally tens of thousands of academic and industrial people since then. Well, this wonderful, wonderful program, in its latest version (less than a year old), can't compute this area: really. Let me ask something more modest.
Can we get a Riemann sum for this f(x) on the interval [0,1] which is
within .01 of the true value? More precisely, can we describe such a
sum, which I'll then have a machine compute?
Abstraction (but just a bit!)
Well, let's try to
approximate this area by a Riemann sum. And maybe we can find a
Riemann sum which is within .01 if we are a little bit careful. We
will begin with a regular partition with n subintervals. We will
choose the n later. Since the whole interval, [0,1], has length 10=1,
the width of each subinterval is 1/n. Let's also, just to be definite
and simple, choose the righthand endpoint in each subinterval as our
sample points. So, since we are "starting" at 0, the first righthand
endpoint of a subinterval will be 1/n, and the second will be
(1/n)·2, the third will be (1/n)·3, etc. The
i^{th} righthand endpoint will be (1/n)·i. The i will
go from 1 to n. Each piece of the Riemann sum will look like f((1/n)·i)·(1/n). We
are supposed to add up these pieces as i goes from 1 to n.
And the error here?
I did not try to draw a picture with lots and lots (n large!)
of thin rectangles. I did try to draw a graph which has the area shown
and shows what I think is one "typical" rectangle whose area
contributes to the Riemann sum. The error is again pushed over to the
right. As Mr. Acosta observed, these
little errors will line up nicely inside a rectangle which is itself
fairly thin (the width is 1/n) and which is 21 high. So the total
error will be less than (1/n)(21) and this is 1/n.
If I want to have this error less than .01: well, .01 is 1/100 and so the error will be less than 1/100 if I choose n to be, say, 100. (I am sorry almost that there are so many coincidences although the numbers work out very simply.) Then the Riemann sum for the function f(x) on the interval [0,1] associated to the regular partition with 100 subdivisions and with the choice of the righthand endpoints as sample points must be within .01 of the "true area".
Here is the actual command and the response of the Maple program which computes this Riemann sum:
add(sqrt((.01*i)^5+(.01*i)^2+(.01*i)+1)*.01,i=1..100); 1.391610100I asked the program to time this command. Errr ... the response I got to that request was the disconcerting time = 0.00 which doesn't mean the command took no time (!) but that the time needed was less than onehundredth of a second. Sigh. If I wanted more accuracy, using this idea of approximating area by Riemann sums, of less than .001, I could alter the command above by changing the ".01" to ".001" and the "100" to "1000". The result is 1.387097725 and took the program .07 seconds. With 10,000 terms, the result was 1.386647601 (.67 seconds!). The true answer, using other techniques, is 1.386597600 (to ten digit accuracy)
What you should know ...
You should know that fairly simple ideas can be used to compute these
Riemann sums and that, using widely available programs, the results
don't take long. I am more interested in the conceptual background of
these sums (not only area, but also blood flow, distance, future value
of an income stream, etc.).
The definite integral
I hope you are willing to believe, after the rather involved
discussion about error, that if the partition of the interval is
chosen to be "fine enough" (so the lengths of the subintervals are all
small) then the Riemann sums will all be close to something. This is
true, and the "something" has a name.
As the number of points in the partition increases, as the length of the subintervals approaches 0, if the function involved is fairly nice (let's say continuous, although the function can have some jumps and breaks) then the Riemann sums, for any possible selection of sample points, approach a unique limit, which is called the definite integral of f(x) from a to b. The "any possible" is italicized because in practice when data is gathered it may be difficult to get exactly the sample points desired, so the "any possible" says that, more or less, the specific choice of sample points doesn't matter too much!
Here is the rather wild traditional notation. Riemann sums themselves
are frequently abbreviated with capital Greek sigmas as I mentioned
last time. The notation for the definite integral is the
following:
_{a}^{b}
f(x) dx
The long "S" is an abbreviation for Sum. So you can look at this notation and see
that it is an abbreviation for summing up, as x goes from a to b, the
"rectangles" of height f(x) and width dx. Here "dx" stands for some
really really really small width. Yes, we will relate it to the other
uses of dx in this course.
You can think that the definite integral stands for "area" but that isn't quite correct. Let me look at some examples.
Some definite integrals
I drew the graph shown to the right, and stated that this was a graph
of y=f(x). I then asked for the following definite integrals:
_{1}^{3}f(x) dx
_{2}^{4}f(x) dx
_{3}^{5}f(x) dx
_{4}^{6}f(x) dx
Since I "identified" pieces of the curve as simple shapes,
probably the integrals can be computed.
I deliberately did not
write a formula or formulas (since this function would need a
piecewise definition!) for f(x), since I wanted people to think about
the concepts involved.
_{1}^{3}f(x) dx
Here the region involved is just a rectangle.
The definite integral's approximating Riemann sums divide this region
into many thin rectangles with height 1 and tiny width. They "add up"
to the whole rectangle.
The base is 2 units long, and the
height is 1 unit, so that the area is 2. This is the value of the
definite integral.
_{2}^{4}f(x) dx
The region here is a square and a triangle joined on a side.
The thin Riemann sum rectangles exactly fill up the square, and in the
triangular part I hope you can see the "error" will be some fuzzy
stuff around the diagonal line segment, and this error>0. So we
will just get the area of the square and the triangle as the limits.
The
square is 1 by 1 unit and has area 1. The triangle, with base and
height both 1, has area 1/2. Therefore the definite integral,
in this case measuring the total area, has value 1.5.
_{3}^{5}f(x) dx
The geometry is clear, but what the definite integral computes
is sometimes not exactly what the geometry seems to say. The Riemann sum
definition of the definite integral is a sum of values of f(x)
multiplied by litte changes in x, dx. This makes area above the horizontal
axis count positively and area below that axis, negatively. The
picture below shows the signed area
which a definite integral would compute.
The sign of the definite integral in applications can have meaning
which is central to the computation: are those dollar amounts profit
or loss? Is there a net amount of fluid being pumped around the body,
or is there a backflow?
In this case with this f(x) and this interval, the definite
integral counts the triangle with positive area and the quarter
circle with negative area. So the value of the definite
integral is 1/2 Pi(1)^{2}/4. This is a negative number.
_{4}^{6}f(x) dx
Here all of the curve is below the axis, and we will "count" all of
the area negatively. This is half a circle of radius 1. The value of
the definite integral is Pi(1)^{2}/2.
So when the definite integral is interpreted as an area, it actually computes what's called an oriented area or a signed area. Regions above the xaxis count positively and regions below the xaxis count negatively. Profit and loss are different quantities.
What's next?
Tomorrow I hope to show you one method that Chinese and Greek and
Indian people used to compute the actual values of definite
integrals. To them the definite integrals represented the values of
areas or, sometimes, volumes. The methods they used were extremely
clever but perhaps very idosyncratic. By this I mean that the methods
were special to almost every computation. (Look, please, in chapter 9
of the text.) It turns out that there's a more systematic method. This
is the Fundamental Theorem of Calculus. (Look, please, in chapter 10
of the text.)
Maintained by greenfie@math.rutgers.edu and last modified 7/25/2006.