Quaecumque VeraDepartment of Mathematical and Statistical Sciences

MATH 115 (V1) - Winter 2008


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  • 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 =a2cos2and dx = a cosq dq.
    •  If the integrand involve a2+ x2 then we may want to subtitute (replace) x = a tanq  where    -p/2 < < p/2  (i.e. q = arctan (x/a) ), so that a2+ x2 =a2sec2and dx = a sec2dq .
    •  If the integrand involve x2- a2 then we may want to subtitute (replace) x = a secq  where          0  q  p,   p/2  (i.e. q = arcsec (x/a) ), so that x2- a2 =a2 tan2and 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 (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 (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 (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 (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 (sin x)*(cos x)*dx and (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 (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 udv = uv - vdu. 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).
  • (January 31, 2008)  Today, we compared the difference between disk/washer method and cylindrical shell method in calculating the solid generated by rotating a region about a line (axis of rotation):

    • In disk/washer method, we cut the region at a line segment perpendicular to the the axis of rotation. After rotation the trace of this segment is a disk or washer. We need to find the formula for the area of this disk (or washer), so we need to find the radius of the disk (or inner and outer radii of the washer). The formula of volume of the solid would be 
      • either ab(disk or washer area)dx  where [a,b] is the range for x; 
      • or cd(disk or washer area)dy  where  [c,d] is the range for y.
    • In shell method, we cut the region at a line segment parallel to the the axis of rotation. After rotation the trace of this segment is a cylindrical shell. We need to find the formula for the area of this shell, to do so we need to find the radius of the shell and the height of the shell. The formula for the area of the shell is then 2p(shell radius)(shell height). Thus, the formula of volume of  the solid would be 
      • either ab2p(shell radius)(shell height)dx  where [a,b] is the range for x; 
      • or cd2p(shell radius)(shell height)dy  where  [c,d] is the range for y.
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.