What follows below reflects uncategorized recent activity in the courses I am teaching, with the most recent activity at the top. This information can be categorized by course and type by selecting the "courses" link above.

Integral Approximations

Section: 
7.5
Date: 
Tuesday, September 7, 2010 - 14:00

Math 2300 Section 005 – Calculus II – Fall 2010

Approximation Notes and Examples – September 3 & 7, 2010

Main Idea: If we are trying to approximate the integral

\int _{a}^{b}f(x)\, dx

and we are subdividing the interval [a,b] into n\in\mathbb{Z}^{+} (n is a positive integer) subintervals, then we have five methods at our disposal. The first three are geometrically defined (LEFT, RIGHT and TRAP), while the fourth and fifth are a modification of the other methods (MID and SIMP). The bottom diagram shows the difference in LEFT, RIGHT, TRAP and MID when f(x)=\sin(x)+1, [a,b]=[1,5] and n=4.

tikz graph

Partial Fractions Quiz

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, September 3, 2010

  1. Explain how you know that an integration problem might involve integration by parts. That is, what clues you in to the idea that integration by parts might be a valid approach to solving an integral?

    Solution: The main thing I am looking for here is something along the lines of the following: The integrand appears to be a product of two functions u and v^{{\prime}}, where we can find u^{{\prime}} and v and the integral of u^{{\prime}}v is in some way easier than the original integral. But, there are other instances when integration by parts comes up that don't look like this, such as when integrating \ln x.

  2. Solve the integral, showing all calculations that lead to your result. (You may use a calculator, but your calculations need to be capable of being followed as if the integral is computed by hand.)

    \int\frac{3}{(x-2)(x-5)}\, dx

    Solution: This is an obvious partial fractions problem. First, write

    \int\frac{3}{(x-2)(x-5)}\, dx=3\int\frac{1}{(x-2)(x-5)}\, dx.

    and then compute the partial fractions.

    \frac{1}{(x-2)(x-5)}=\frac{A}{x-2}+\frac{B}{x-5}

    Multiplying on both sides by (x-2)(x-5) and then putting the polynomial on the right in standard form gives

    \displaystyle 1 \displaystyle=A(x-5)+B(x-2)
    \displaystyle 1 \displaystyle=(A+B)x-5A-2B

    This results in the system of equations:

    \displaystyle A+B \displaystyle=0
    \displaystyle-5A-2B \displaystyle=1

    which has the solution A=-\frac{1}{3} and B=\frac{1}{3}. So, we have

    \displaystyle 3\int\frac{1}{(x-2)(x-5)}\, dx \displaystyle=3\int\frac{-1/3}{x-2}+\frac{1/3}{x-5}\, dx
    \displaystyle=-\int\frac{1}{x-2}\, dx+\int\frac{1}{x-5}\, dx
    \displaystyle=-\ln|x-2|+\ln|x-5|+C.

Trig Substitution

Section: 
7.4
Date: 
Friday, September 3, 2010 - 14:00

Math 2300 Section 005 – Calculus II – Fall 2010

Trigonometric Substitution Examples – September 3, 2010

The main idea of using trigonometric substitution is that a substitution plus a trig identity can be used together to simplify an integral. There are three primary categories into which trig substitutions fall, listed here. (I'm listing them all here since your book really only does two of them.) As we discussed in class, the domain needs to be restricted as the evaluation of the integral will depend on an inverse trig function.

When You See Try to Substitute Domain Use the Identity
\sqrt{a^{2}-x^{2}} x=a\sin\theta -\frac{\pi}{2}\le\theta\le\frac{\pi}{2} 1-\sin^{2}\theta=\cos^{2}\theta
a^{2}+x^{2} or \sqrt{a^{2}+x^{2}} x=a\tan\theta -\frac{\pi}{2}<\theta<\frac{\pi}{2} 1+\tan^{2}\theta=\sec^{2}\theta

Review: Completing the Square

Section: 
Review for 7.4
Date: 
Thursday, September 2, 2010 - 14:00

Math 2300 Section 005 – Calculus II – Fall 2010

Completing the Square Examples – September 2, 2010

  1. Put the quadratic f(x)=x^{2}-2x+5 in vertex form. Explain the transformations required to translate g(x)=x^{2} into f(x), allowing the graph of f(x) to be given.

    Solution:

    \displaystyle f(x) \displaystyle=x^{2}-2x+5
    \displaystyle f(x) \displaystyle=x^{2}-2x+1-1+5
    \displaystyle f(x) \displaystyle=(x^{2}-2x+1)+4
    \displaystyle f(x) \displaystyle=(x-1)^{2}+4.

    Starting with g(x)=x^{2}, we first shift one unit right and obtain (x-1)^{2}, then shift four units up and obtain (x-1)^{2}+4.

  2. Find the center and radius of the circle

    x^{2}-6x+y^{2}+8y-20=2.

    Solution: By completing the square on each variable, we have (notice that this time we are adding the required constants on both sides of the equation, instead of adding and subtracting on the same side)

    \displaystyle x^{2}-6x+y^{2}+8y-20 \displaystyle=2
    \displaystyle x^{2}-6x+9\quad+\quad y^{2}+8y+16 \displaystyle=2+20+16
    \displaystyle(x-3)^{2}+(y+4)^{2} \displaystyle=38.

    The center is thus (3,-4) and the radius is r=\sqrt{38}.

Worksheet 2

Solutions have been posted to worksheet #2.

Partial Fractions

Section: 
7.4
Date: 
Wednesday, September 1, 2010 - 14:00

Math 2300 Section 005 – Calculus II – Fall 2010

Method of Partial Fractions Examples – September 1, 2010

General Rules for the Method of Partial Fractions:
Assuming you're given something of the form

\int\frac{P(x)}{Q(x)}\, dx

where P(x) and Q(x) are polynomials in the variable x, the following represents an approach to solving this integral.

  • If the degree of P(x) is greater than the degree of Q(x), perform long division first. This will result in a polynomial in x plus some simpler fraction. You may need to use the method of partial fractions on that simpler fraction, but you can now pull the polynomial off and integrate that individually. We'll continue assuming that P(x) and Q(x) are the numerator and demoninator of that simpler fraction.

  • If Q(x) only has linear factors with single multiplicities (each factor only occurs once), say something like

    Q(x)=(x-1)(x+1)x

    then start by including each factor once in the partial fraction expansion. In this case, we'd have

    \frac{P(x)}{Q(x)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x}.

    Our goal will be to calculate A, B and C.

  • If Q(x) has repeated linear factors, then include each factor once for each multiplicity up to the total multiplicity of that factor. For instance, if Q(x)=x(x-1)^{4} then the factor (x-1) has multiplicity 4, and we would use

    \frac{P(x)}{Q(x)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}+\frac{D}{(x-1)^{3}}+\frac{E}{(x-1)^{4}}.
  • If Q(x) has an irreducible quadratic, something like Q(x)=(x-1)(x^{2}+1), then that quadratic should be under a linear factor for the method of partial fractions. We would use

    \frac{P(x)}{Q(x)}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+1}.
  • Now here is the key idea: Clear denominators (by multiplying by each of the factors that occurs in the denominators of the fractions on the right) and you will then have two polynomials. If you multiply out the polynomial on the right, and collect terms by degree of the variable used (x in this case), then you'll know what all of the coefficients are… after solving some system of linear equations.

    For instance, in the example above where

    \frac{P(x)}{Q(x)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x}

    clearing denominators on the right cancels out Q(x) and gives us

    \displaystyle P(x) \displaystyle=A(x+1)x+B(x-1)x+C(x-1)(x+1)
    \displaystyle=(A+B+C)x^{2}+(A-B)x-C

    Then, the x^{2} coefficient of P(x) is equal to A+B+C, the x coefficient is A-B and the constant term is -C. From this, we can figure out what A, B and C are.

2nd Week Survey

Due Date: 
Tuesday, August 31, 2010 - 17:30 - Monday, September 6, 2010 - 23:55

This survey will replace a quiz score during the semester. I won't see your individual responses, but I will receive a list of my students that have taken the survey. Below is the e-mail I received from Louisa and Ryan. I'll discuss this in class on Wednesday, September 1.

--------------------------------------------------

Dear Math 2300 students,

Here is the link to the course survey for you to complete by next Tuesday, September 7th. Your participation is really helpful because understanding more about students' beliefs about mathematics and how they evolve helps us in our teaching. On this survey, we are asking you to express “your own beliefs” about mathematics. It will only take about 15-20 minutes of your time.

http://www.zoomerang.com/Survey/WEB22B5EXTNC25

Your instructor will not see your individual responses, only whether you participated in the survey and a summary of the data for the class as a whole.

Thank you for your time.

Best regards,

Louisa Harris & Ryan Grover

Book Assignment 2

Due Date: 
Tuesday, September 7, 2010 - 16:00

Section 7.3: Problems 52 and 54
Section 7.4: Problems 56 and 64
Section 7.5: Problems 20 and 22

Table of Integrals

Section: 
7.3
Date: 
Monday, August 30, 2010 - 14:00

Ryan Grover taught this lecture as I was out sick.

Friday Quiz 1

MATH2300-005 – Fall 2010 – University of Colorado

Quiz - Friday, August 27, 2009

  1. Explain how you know that an integration problem might involve substitution. That is, what clues you in to the idea that substitution might be a valid approach to solving an integral?

    Solution: I will accept answers that make sense here. What I would write is as follows. The biggest hint that an integral might involve substitution is when the integral contains an inside function and that inside function's derivative (up to some constant) is somewhere in the outside of the integrand.

  2. Solve the integral, showing all calculations that lead to your result. (You may use a calculator, but your calculations need to be capable of being followed as if the integral is computed by hand.)

    \int _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx

    Solution: There are several ways to solve this. We know what the curve \sin x looks like, and integrating it from -n to n for any real number n will give us zero. If we scale that curve by \pi vertically and compress horizontally by a factor of \pi^{2}, we expect the same to be true. So, right away, we know that we'll end with zero. Using the substitution w=\pi^{2}x we find the following.

    \displaystyle\int _{{-\pi}}^{\pi}\pi\sin(\pi^{2}x)\, dx \displaystyle=\frac{1}{2}\int _{{-\pi^{3}}}^{{\pi^{3}}}\sin(w)\, dw
    \displaystyle=\frac{1}{2}\left[-\cos w\right]_{{-\pi^{3}}}^{{\pi^{3}}}
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(-\pi^{3})\right]
    \displaystyle=\frac{1}{2}\left[-\cos(\pi^{3})+\cos(\pi^{3})\right]
    \displaystyle=0.

    The second to last step is true because \cos(\theta)=\cos(-\theta) for all \theta.

© 2011 Jason B. Hill. All Rights Reserved.