|
Home
Course
Information
Assignments
Old Exams
Lecture
Journal
|
|
- Tips on study: You should study the concepts in the textbook and notes (and also this "lecture journal" page) carefully, make sure that you understand the concepts throughoutly (for this purpose, study the examples and the solutions to assignments, and figure out how/why each step
happens -- there is always a reason for each step - otherwise, something must be wrong). After that you can practice,
by trying to do some excercises in the textbook (especially the ones
that are similar to the examples and assignments), you can also try the
old exams. (April 10, 2008) Today
we reviewed our course. Good luck with the exams! (April 8, 2008) Today
we discussed how to solve 2 types of first order differential
equations: separable differential equations and linear
differential equations. - (April 3, 2008) Today
we discussed about
improper integrals. One point we should note is that sometimes we
have to split the integral and calculate each part separately. We
also learn how to test the convergence or divergence of improper
integrals without the need to do the calculation (sometimes this is
even impossible); to choose the one to compare, "looking for the
significant terms in both nominator and denominator" might help.
- (April
1, 2008) We discussed about trigonometric
substitution:
- If the integrand involve a2-
x2 then we may want to subtitute (replace) x = a
sinq where -p/2 £ q
£ p/2
(i.e.
q = arcsin (x/a) ), so that a2-
x2 =a2cos2q and dx = a cosq dq.
- If the integrand involve a2+
x2 then we may want to subtitute (replace) x = a
tanq where -p/2 < q < p/2
(i.e.
q = arctan (x/a) ), so that a2+
x2 =a2sec2q and dx = a sec2q dq .
- If the integrand involve x2-
a2 then we may want to subtitute (replace) x = a
secq where
0 £ q £ p,
q ¹
p/2
(i.e.
q = arcsec (x/a) ), so that x2-
a2 =a2
tan2q and dx = a secq tanqdq.
-
(March
27, 2008) We continued our discussion about integration
of products of powers of sine and cosine
or of powers of tangent and secant (or of powers of cotangent and
cosecant). We aim to apply one of the following:
- M1: If the integral can be write as ò(sin
x)* f(cos x)dx with power of sine is odd,
then it is possible
to substitute
u=cos x. This method is best when power of
sine is also
positive.
-
M2: If the integral can be write as ò(cos
x)* f(sin x)dx with power of cosine is odd,
then it is possible
to substitute
u=sin x. This method is best when power of
cos is also
positive.
-
M3: If the integral can be write as ò(sec
x)* f(tan x)dx with power of sec is even,
then it is possible
to substitute
u=tan x. This method is best when power of
sec is also positive.
- M4: If the integral can be write as ò(sin
x)*(cos
x)*dx with the power of sine and cosine are
even and non-negative, then reducing the power by using sin2x
=(1-cos2x)/2 and cos2x=(1+cos2x)/2.]
We know that (sin x)*(cos
x)* and (tan x)*(sec
x)* are interchangeable. So, how to find out
which of the above methods (M1--M4) is best suitable for a particular
integral:
-
First of all, write the integral as ò(sin
x)*(cos
x)*dx and ò(tan
x)*(sec
x)*dx.
- Then, if the powers of sine and cosine
both are even and non-negative, we use M4. Otherwise:
- Check the corresponding form against the
conditions in M1, M2 or M3, to see if any of them say it is best
suitable, use that method. If there is no method which is best
suitable, then use any method that says possible.
-
(March
25, 2008) We continued our discussion about partial
fractions method. Trick
3:
- If the denominator has only one factor and the
factor is linear, say the denominator is a constant times (x-r)n;
then we can change the variable t = x - r to figure out the
coefficients of partial fractions at once [or use the substitution t =
x - r directly in the integral].
Also, in some very special case, it may happen that the nominator is a
constant times the derivative of the denominator, in which case, we can
use a substitution u = 'the denominator'.
Summarize:
we have a general method of partial fractions to find
integrals of rational functions (... expanding the brackets,
collecting the powers of x together, and equating the coefficients of
the powers of x); in certain case, we have "tricks" to
make life easier.
Note that in Trick 1 and 2, whenever we can extract
out common factor(s) in the "unknown side", it means that we
must be able to extract the same common factor(s) out of
the "known side" <--
this is a good check point.
There are cases where the integrand is not a rational function,
however, after a substitution, the integral become integral of a
rational function.
Trigonometric
integration: The aim of this part is to calculate
integrals of products of powers of sine and cosine
or integrals of products of powers of tangent and secant
or integrals of products of powers of cotangent and cosecant.
These three forms are interchangeable:
- To rewrite either of the forms (tan x)*(sec
x)* or (cot x)*(csc
x)* as (sin x)*(cos
x)* is easy.
- To rewrite (sin x)*(cos
x)* as (tan x)*(sec
x)* then all (sin
x)* must go into (tan
x)*
---- it is because (sec x)
is (cos
x)-1 so there can be no (sin x) in (sec
x)*.
- Similarly, to rewrite the form (sin x)*(cos
x)* as the form (cot x)*(csc
x)* then all (cos
x)* must go into (cot
x)* ----- because (csc x)
is (sin
x)-1.
- [This one is rarely useful.] To interchange
between (tan x)*(sec
x)* and (cot x)*(csc
x)* we can go via the form (sin x)*(cos
x)* first.
We will see that,
depending on cases, we may prefer one form to the others.
-
(March
20, 2008) We discussed about integration of rational
functions by partial fractions. We also discussed today two "tricks" to
find out some (or even all) of the coefficients (before expanding the
brackets and equating the corresponding powers of x). Trick 1:
- For each linear factor x-r (of the given
denominator), we can let x = r in
both sides of the equation (the effect is that almost all terms should
become 0); this way we
will find one coefficient right away. Do this for all
distinct linear factors.
- Substitute the coefficients we've just found, and
rearrange the equation so that on one side we collect
all "known
terms", and the unknow ones remain on the other side.
- Simplify the "known side" (but don't expand the
"unknown side" yet).
- On the "unknown side", we shall see that we can
extract out common factors (the above linear factors). Try to extract
these factors out of the "known side" (by some division
algorithm, and it must be
possible without any
reminder, i.e. the reminder must be 0).
- After cancell out the common factors, we have a
new equation where it may be possible to repeat the above steps for the
new equation (as long as we still have our linear factor(s) left in the
'unknow side').
(Repeat use of) Trick 1 can help us to find all coefficients
relating to the linear factors. Trick 2
(after finding all coefficients relating to the
linear
factors using Trick 1 as above - including the cancelling out of the
common factors step):
- In the case where there is only one quadratic factor left,
we can use division algorithm to find out about the coefficients
(divides the "known side" by the quadratic factor to find the
remainder, rearrange the new "known one" to "known side" and divides
both sides by the common factor, which is the quadratic factor itself,
and keep doing this).
Here is an example illustrating
this trick.
-
(March
18, 2008) Today,
we continued our discussion about integration by parts: tabular process
to ease our calculation when we can write the function under integral
into product of a polynomial and another function whose antiderivatives
can be found easily. Sometimes, we need to use substitution first,
before applying integration by parts. An application of integration by
parts is to derive recursive formulae to calculating integrals; an
example is ò(x2+1)ndx
(in the class, I made a mistake in calculation, missing some n in the
last several equations).
-
(March
13, 2008) We
first reviewed basic way to deal with an integral:
- See if the integral matches any standard
formula.
- See if the integral can be simplified by a
substitution.
- See if the integral can be written as sum of
a number of integrals where some (or
all) of them can be processed as above. Deal with each one of them
separately!
The following can also be helpful:
- Reduce the fraction if the function under the
integral is the
quotient
of two polynomials such that the degree of the nominator
is greater than or equal the degree of the
denominator: we
divide the nominator by the denominator to find the quotient and the
remainder.
- If there is a quadratic form under the
denominator
or under the root, then completing the square. Then make a
substitution so that the quadratic form becomes one of the forms
±u2 ± a2
where a is a constant.
- Separate the fraction; the aim is again to
express
the integral as sum of a number of integrals (as above) and
deal with each one of them separately.
We then discussed about the technique of integration by parts.
It is
the reverse process of product rule for differentiation. The formula is
òudv = uv - òvdu. Remarks:
- Write the function under the integral as product
of 2 parts
where you can find easily antiderivative of one of them.
- If possible, try to make a choice so that after
applying integration by parts, the new integral is simpler.
- If a choice is not working, try another one.
- Remember,
come back to where you started is not always bad! Sometimes, we can
find an unknown integral by rearranging that integral to one side of
the
equation.
-
(March
11, 2008) We
discussed about the inverses of hyperbolic functions,
their derivatives and application to integration. Here
is a summary
about inverse hyperbolic functions.
-
(March
6, 2008) We
discussed about the inverses of cotangent, secant, cosecant functions,
their derivatives and application to integration (of arcsecant). Here
is a summary
about inverse trigonometric functions.
We then discussed about hyperbolic functions and their derivatives.
Hyperbolic functions are very similar to trigonometric functions (but
there are also some differences). Here is a summary
about hyperbolic functions.
-
(March
4, 2008) We discussed about the inverses of sine,
cosine and tangent functions, their derivatives and application to
integration.
-
(February
28, 2008) We continued our review. Then, we discussed
about exponential growth and decay (§7.5). Finally, we discussed a
notion to
compare the growth of functions f(x) and g(x) as x ®¥ (§7.6).
-
(February
26, 2008) Today, we discussed about derivatives of
logarithm base a and how to find limit of power f(x)g(x).
Then, we had a review.
-
(February
14, 2008) Today, we discussed about natural exponentia
function as well as exponentia and logarithm functions of base a. This
also enable us to deal with functions like xa
for any real number a or f(x)g(x).
-
(February
12, 2008) Today,
we discussed about the derivatives of inverse functions. Then, we gave
a definition of natural logarithm. We deduced from the
definition
some simple properties as well as finding the derivatives of
logarithmic functions. The usefulness of logarithm is because it
transfers multiplication into addition, division into subtraction ...
For example, we can use logarithm to find the derivatives of
functions which are expressed as a complicate product and/or
quotient.
-
(February
07, 2008) We continue our discussion about area of surface
of revolution. Here is the basic principle. We then discussed
about one-to-one functions and their inverses (with a small exploration
on how to plot a graph).
-
(February
05, 2008) We discussed briefly about differential. We
then, using differential to formulate formula for the length
of curves; here is the summary.
Finally,
we discussed the area of surface created by rotating a curve
(C) about an axis. Here
is the slides in today's lecture. The general
formula for the area of this surface of revolution is
where  is again
the differential of the length of the curve (C). Then, we discussed
about parametric curves and how to finds their lengths.
-
(January
29, 2008) We discussed how to calculate the volume of a
solid of revolution by cylindrical shell method. Demos gallery for shell
method and here is a nice visual explanation.
-
(January
24, 2008) We
briefly discussed the way to calculate area of region whose boundaries
are better to be expressed as functions of y. We then moved to a
different type of application: calculating the volume of a
solid
if we know the crossectional-area A(x) [or A(y)]. Particularly, if our
solid is a solid of revolution, the crossection is either a disk or a
washer, the formula for crossectional-area is easy to find.
Demos for disk
and washer
methods; another
one.
-
(January
22, 2008) We discussed how to use integration to calculate
the area of a region bounded between curves in various situations.
-
(January
17, 2008) Today,
we discussed indefinite integral, subsitution rule for both definite
and indefinite integrals. Remember, substitution is often useful when
you want to simplified a problem. We also have a glance at how to
integrate power of sin and cosin functions.
-
(January
15, 2008) We
continued our discussion of definite integral: as total change of a
quantity when we know the rate of change of that quantity, area when
function is negative (or both negative and positive). We also discussed
the Fundametal Theorem of Calculus, which formally demonstrates that
integration and derivation are reverse of each other.
-
(January 10, 2008)
We discussed L'Hopital's rule for indeterminate form ∞/∞, and how to
change from other indeterminate forms to either 0/0
form or ∞/∞ form. Here
is an
example we discussed in the class. We
also recalled the definition of definite integral of a function f(x);
as
well as definite integeral's meaning as the area of a region on
xy-plane when f(x)≥0.
-
(January 8, 2008)
Today, we reviewed some basic facts about differentiation: chain rule,
product rule, quotient rule, and antiderivatives of a function. We then
discussed Cauchy's Mean Value Theorem. Finally, we discussed
L'Hopital's rule for indeterminate form 0/0.
|
