My MTH240 Semester II Reflection Assignment
Giovanni Bianchi
January-March 2025
1
Introduction
In this assignment, I will be writing weekly about how my learning of Calculus
II is going. My intention is to finish writing on time to submit the assignment
for my grade. However, I will also be using this assignment as a way for me to
retain and review course content weekly rather than forget about it.
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Weekly Reflection
In this Section, I will write a weekly reflection on my learning.
2.1
Week 1
This week, I was taught Integration By Parts as the main topic.
I went to class Monday afternoon ready to start learning immediately. My professor did a very brief review of some key skills, focusing mainly on u-substitution
before moving forward into the new lesson. This is when I began learning integration by parts. We learned the formulas and then proceeded with practice
problems to aid our learning.
The integration by parts formula for an indefinite integral is
Z
Z
udv = uv − vdu
The above formula is the anti-derivative of the integrand.
The integration by parts formula for a definite integral is
Z b
Z
udv = uv −
a
b
vdu
a
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When I went to class on Tuesday, we were taught more about trigonometric
integrals and how we can use trigonometric identities to solve them. I learned
that the main identities that get used are the half-angle identities, so I must
ensure that I memorize those, as well as the product to sum identities. I am
optimistic about my ability to perform well with what I have learned so far, I
just need to go over the topics and practice more.
2.2
Week 2
This week, I was taught a couple more methods to solve integrals. The first
method was to use Trigonometric Substitution, which can be used to solve
more than just integrals.
This substitution is used when two squared terms are added together under a
root function, one of which contains a squared variable. The variable can then
be replaced with a trigonometric function, and the equation can then be solved
using Pythagorean identities. Along with this, we also went over how to ’complete the square’ so that we can use this new substitution to complete problems.
On Tuesday, I went to my first tutorial for the class, which helped me better
understand how to complete some of the problems from last week and use the
new methods that I have been taught.
Later I went to my second lecture of the week, where I was taught integration
using partial fractions. This method helps to solve integrals that are presented
as rational functions by converting them into products and sums. We also reviewed long division, which I did forget admittedly; however, now that I have
gone over it once more, there should be no further issues there. There are various
opportunities to make errors in the integration process using partial fractions,
so I must ensure that I familiarize myself with this technique much more before
my lab quiz next week.
2.3
Week 3
During this week’s class, we spent a lot of time discussing Improper Integrals.
An improper integral occurs when the function being integrated has a limit for
which we must account. These limits can be integer values or infinities.
Most of the time, these limits will not be given to us, and we are to examine
the function to decide if there is a limit or not. For example, if the function is
a rational function, the denominator can not be zero, and the limit would be
approaching whatever value makes the denominator zero. To show this with an
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integral, we write the limit using a variable not found in the function, and we
use that variable to represent the value which will be substituted into the upper
or lower bound of the integral. If the integral is not bounded by the limit, then
it must be split into two integrals so that the limit can be found in the bounds.
We were also introduced to new terminology related to these types of integral.
An integral is Convergent if the limit exists, and is Divergent if the limit does
not exist or is evaluated to be an infinity. If the integral converges, we must
state which integer it converges to (i.e., what the limit is equal to).
2.4
Week 4
This week, we both started and finished unit 4, which pertained to differential equations and initial value problems. With initial value-problems, usually
a differential equation is given, and you must solve for the original equation.
However, since integrating leaves you with a C term, initial (x,y) values are
given in order to be able to solve for the C term.
The second part of unit 4 which was evaluating differential equations required
a little more thinking. There are two types of differential equations. The first
type are Separable Differential Equations, which is when the equation can be
expressed as the multiplication or division of x and y. The second type is a
Linear Differential Equation, which is when the equation can only be expressed
as the addition of the x and y functions. The linear equations are much harder
to solve, and you need to find the integrating factor to be able to solve.
The other lesson we started was lesson 5.1, which was n introduction to sequences. Sequences are essentially equations used to create numerical patterns,
similar to something done in elementary school, only on a much larger and more
complex scale. So far, we have been given sets of number, and have been asked
to solve for the sequence it stems from.
2.5
Week 5
Week five was focused around Infinite Series and 4 out of ten methods to solve
them. First off, a series is the sum of the terms in a sequence, denoted with the
sigma symbol. In this unit we are learning about specifically series that have
an infinite amount of terms.
The first method is used when a series is a Geometric Series. This is when
the series can be written as a constant times another constant to the power of
n, arn . A geometric series is said to convergent if the absolute value of r is less
than 1, otherwise it is divergent. To find the sum of a geometric series, it must
be a convergent series. The formula for the sum is: term1/1 − r.
The second method for solving infinite series is the Test for Divergence. This
can be used for any series to attempt to find if a series is divergent. If the limit
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of the series does not equal zero, then the series is divergent. However, if it
equals zero, then another method must be tried to continue solving the series.
It does not mean the series is convergent!.
The third method is used when a series is a Telescoping Series. This is
when most of the terms in a series end up canceling each other out by way of
addition/subtraction. In order for this to be true of a series, there must be
a subtraction sign. Reminder than lnx/a = lnx − lna. The partial sum of a
telescoping series can be found by writing out the first few terms and last few
terms (using n as the last term) to figure out the pattern of eliminated elements.
From there, eliminate all canceling terms, and you’ll be left with a partial sum
of the series. Then, to find the Sum of the whole series, take the limit of the
partial sum. If it exists, then the series is convergent, if it is infinite then it’s
divergent.
The fourth method learned was The Integral Test. In order to use this test,
the function that is part of the series must be positive, continuous and NOT
increasing. If it fits these conditions, then go ahead and take the infinite integral
of the function, with the lower bound being your first n-value. If the integral is
convergent, then so is the series. If the integral is divergent, then the series is
as well. Keep in mind that the sum of the series is not equal to the value of the
integral.
2.6
Week 6
The fifth method is the Direct Comparison Test, which is the same as the
comparison test for integrals. The trick is to pick a similar series, either a
smaller series that diverges or a larger series that converges.
The sixth method is the Limit Comparison Test. For this test, you take the
series you are given and divide it by any series that is known to be convergent
or divergent. Take the limit of that fraction. If the answer is a finite positive
number, then both series have the same behavior. If not, then try a series with
the opposite behavior as the first series you tried. (NOTE: the two series in the
fractions do not have to have anything in common.)
2.7
Week 7
The seventh method is the Convergence Test for Alternating Series. This
should only be used for series that have an alternating sign for every term (ie.
(−1)n or cos(nPI)). The rest of the series can be labeled bn. If the following two
conditions are met, then the series is convergent. If not, the test is inconclusive.
1. the limit as bn approaches infinity = 0
2. the derivative of bn is less than 0
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We also then learned about Absolute Convergence, which is when the absolute value of a series is convergent. If this is true, then the series is absolutely
convergent, but if this is not true, and the regular series converges, then it is
said to be conditionally convergent. It is also known that if a series is absolutely
convergent, then the regular series must be convergent as well. We can use this
to find whether a series is convergent.
For example: if the series is cos(n)/n2 then cos(n) can be both positive and
negative values, making it hard to solve. If we use the absolute convergence
test, we can eliminate the negative outcomes, and solve to find that the series
is absolutely convergent, meaning the regular series converges as well. (very
useful for sine and cosine) If the series is not absolutely convergent, then
the test is inconclusive.
The ninth test is the Ratio Test, which is useful when the series is a fraction
and almost always used when you see n!. This is how the test works: for a series
Σan find (as n approaches ∞)lim abs(a(n + 1)/an).
If the limit is less than 1, the series is absolutely convergent.
If the limit is greater than 1 or ∞, then the series is divergent.
If the limit = 1, the ratio test is inconclusive.
The tenth test is the Root Test. It is used when an entire series is raised to
an exponent of n. To use this test, simply place a root of n over the limit of
the series (as it approaches infinity) to cancel the exponent. Then evaluate the
limit (same as ratio test)
If the limit is less than 1, the series is absolutely convergent.
If the limit is greater than 1 or ∞, then the series is divergent.
If the limit = 1, the ratio test is inconclusive.
2.8
Week 8
This week, we did a review and continuation of learning ratio, root and absolute
convergence test. After that, we began the next unit with Power Series. A
power series is in the form Σcn(x − a)n = c0 + c1(x − a) + c2(x − a)2 + .... This
is a power series centered at ’a’. When evaluating the power series, use either
the Root Test or Ratio Test (most common) and set it to less than 1. This
will give you an interval, which is called the interval of convergence. The radius
of convergence is therefore the distance between the two intervals.
It is also important to note that using the equation for geometric series, we can
manipulate functions to write them as sums (or series). Along with this, the
derivatives and integrals of the functions and series’ can be used to help find
the other. The derivative and integral is always taken in terms of x (not n).
NOTE: the radius of convergence does not change with the integral or derivative, but the endpoints must be tested to see if they are included.
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2.9
Week 9
In this week, we learned about the Properties of Power Series, which includes addition, subtraction and multiplication of Power Series. For the addition
or subtraction of power series, if both series are approaching the same value and
their limits exist, then the limit of the addition/subtraction of the series’ can be
found by the same addition/subtraction of the convergent values for each series.
The interval of convergence of the added series, is found from the intervals of the
added two series. Using them, whatever values in the interval can be reached
by both series becomes the new interval.
The multiplication of Power Series is much more complicated and would result
in having to multiply and add together all values of one series with all values
of the other. This of course is not possible because the series would continue
indefinitely. To represent this multiplication, we can use the first 3-5 values of
each series and multiply them all together and add them up. This would leave
us with a set of values. If it is possible to get a pattern from this set of values
and write it as a series, then do so, otherwise, leave it as it is. This can be
done if both series converge to values on the a common interval. The product
series would then also converge on the same interval. This is called the CauchyProduct.
The last thing we learned was Taylor and Maclaurin Series. If a power series
has the representation of Σcn(x − a)n with it’s center of approximation at a, the
coefficient cn would be f (n )(a)/n! (the nth derivative). This is called the Taylor series of f(x) at a. If a=0 for a Taylor Series, this is called a Maclaurin series.
There is also a binomial series which is represented as Σ(kchoosen)xn . As a
function this looks like (1 + x)k . It is convergent when absolute x is less than 1.
2.10
Week 10
During week 10, we began learning about Multi-variable Functions, and how
to represent them. The domain of a multi-variable function can be understood
as an area (for 2 variables), volume (for 3 variables) and so on. In this course,
we are only going to be focusing on 2 variable functions, because the concepts
apply across the board. These graphs would be 3D functions. We also found
the limits of multi-variable functions, by using the same methods we use for
single variable functions. We can also try to prove that the limit does not exist
by trying to find two functions (ex. y=x) that pass through the point that the
limit approaches. If the limit of those two functions are not equal, then the
limit does not exist.
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2.11
Week 11
In the second last week of my learning, we did more examples with limits of
multi-variable functions, which helped me to better practice and understand
what needs to be done in order to properly solve questions. We then also went
over Partial Derivatives. This is simply referencing the different possible
derivatives of multi-variable functions. Since there are multiple different derivatives for multi-variable functions, we can not use the f’(x) notation, so for a
function such as z=f(x,y) we represent the derivatives as ∂z/∂x (in terms of x)
and ∂z/∂y (in terms of y). When taking the partial derivatives, the process is
basically the exact same as a single variable function. We only take the derivative in terms of 1 variable, which means we take any other variables as constants
in order to perform derivatives. This presents a small challenge because it is
difficult to read letters and treat them just as numbers when trying to perform
derivatives, but with more practice, it became easier to comprehend and complete.
There is also the concept of higher order derivatives, such as second or third
derivatives. In the case of a two variable function z=f(x,y), there are 4 second
derivatives: ∂ 2 z/(∂x ∗ ∂x), ∂ 2 z/(∂x ∗ ∂y), ∂ 2 z/(∂y ∗ ∂x) and ∂ 2 z/(∂y ∗ ∂y). In
this notation, the first derivative is taken for the furthest right variable, and is
continued for second, third, fourth, etc. derivative from to the left. It is also
valuable to note that ∂ 2 z/(∂x ∗ ∂y) and ∂ 2 z/(∂y ∗ ∂x) will more often than not
yield the exact same function, therefore we only need to show work for one of
them.
2.12
Week 12
In this last week, we went over the chain rule for multi-variable functions. The
theory is fairly simple, however it gets somewhat complicated when it is applied
to functions. The chain rule is used to find the derivative, or partial derivatives
of multi-variable functions when the variables in the function are representative
of another function, of different variables. For example if:
z=f(x,y)
x=g(t) y=h(t)
then dz/dt = ∂z/∂x ∗ dx/dt + ∂z/∂y * dy/dt
The same would apply if x and y represented more multi-variable functions.
When applying this theorem, it is important to make a tree diagram to visualize all the different ways a variable can be obtained.
The other thing we learned was the Implicit Function Theorem. This states
that if we have a function:
F(x,y,z) = 0 and we want to find a partial derivative
∂z/∂x = -∂F/∂x / ∂F/∂z
This helps to make implicit differentiation with multi-variable functions easier.
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3
Procedural and Conceptual Understanding of
Calculus
Although it may prove difficult, it is very important to be able to understand
calculus concepts, especially in the field of engineering. There are many situations that may lead to more abstract mathematical thinking, in which we would
need to not only understand, but also be able to apply calculations in given
circumstances. I personally do not believe that either concepts or calculations
are more important than the other when it comes to understanding calculus.
The way I see it, if you do not understand the concepts, you can not apply
any calculations, because you will not know what to use, or when to use it.
However, it is also very important to have good knowledge of performing calculations, because if you understand concepts, but are unable to apply them,
then your knowledge is rendered nearly useless, unless you are only dealing with
theoretical situations.
In other words, if you are going to have knowledge of calculus at all, I believe
that it is of the utmost importance to not only be able to understand concepts,
but also be able to perform calculations to back up your knowledge.
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Use of Calculus in Other Subjects
Over the past school year, I have been able to see the use of calculus in more
courses than just MTH140 and MTH240. The main courses I have been able
to see the use of calculus are in PCS211 and PCS125, the first year engineering physics courses. Although we have not used calculus very much for solving
problems, during the lessons the professors would explain equations and show
how they were derived by using concepts learned in my calculus courses. The
main physics concepts where calculus was applied were when we were learning
about fields, both gravitational and magnetic. Integrals were applied and used
in these cases to help solve for the magnitude of such fields. The concepts of
fields were taught to me in high school, however I did not have the knowledge
of calculus that I have now, so it is now easier to understand these physics
concepts since I am able to understand more complex derivations for important
equations. Due to this additional knowledge, I can now more easily solve and
comprehend problems involving fields and therefore calculus as well.
5
Overcoming Challenges in Learning MTH240
Calculus II
In my studies of calculus and complex numbers, I have come to rely on a few
different learning strategies to help me overcome challenges. I have also been
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able to use the help of technologies, as well as the help of friends and classmates
to aid me in my understanding of MTH240.
5.1
Learning Strategies
My first strategy is something that seems obvious, but I know that a lot of people struggle with it. This is the commitment to not skipping a calculus lecture.
I know personally that there are many days when I get very tired from classes,
but I need to keep in mind the importance of showing up to every class, because
I will not receive a better explanation of the topics elsewhere. Being present in
every class helps me to ensure that I get familiar with all the course content,
even on days that I am tired. It is important to at least show up and listen
to what my prof has to say, so that I can save myself from having to learn the
content last minute before the exam.
Another strategy I have found to be important and very helpful is making my
own set of notes outside of my lectures. This allows me to go over all the course
content with great detail, and then taking the most important key points to
write down in my own separate notes. These notes then become very helpful
before an assessment as I am able to look over all the content that I have seen to
be important without wasting time trying to go through the textbook or all the
course notes the night before an exam. It allows me to be able to have a quick
look at the important things an hour before an exam, just to keep everything
fresh in my mind.
5.2
Tools and Technology
Although I did not use technologies too much in my math learning journey,
there were times when it was very useful to me. I personally found Desmos to
be the most useful tool because of its ability to graph complex functions and
graphs that I would not be able to visualize otherwise. This became useful to
me when doing homework, or reading through lesson slides, as I was able to
better understand the functions I was working with, and therefore had an easier
time solving the problems.
5.3
Collaboration and Learning
I believe that the most important strategy which helped in my calculus journey
was the aid of peers. Although it is very important to be able to solve problems
on our own, part of the process is learning from other people as well. There
were many times when my friends and I would book out study rooms so that
we could work on calculus problems together, and go over basic concepts. One
of the best ways this helped me personally was that I was able to help teach
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other people how to solve problems. Being able to teach others helps both individuals to gain a better knowledge of the content. People can build off of
each others strengths and help overcome weaknesses as well. If someone does
not understand something, then they can have their peers help them through
it, rather than sitting alone not getting anywhere. I feel as though this was
the most useful asset to me during my education journey because when I would
explain concepts to people, I then explained them to myself as well, making sure
I am constantly studying and reminding myself of important notes and concepts.
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Personal Growth and Confidence
In my process of learning calculus, I have come to grow a lot from where I was
at the beginning of the semester. I was very unsure of how much time and effort
to put into my courses, and how to properly balance them with personal and
social time as well. As the semester continued, I began to realize that there is
no set amount of time we can assign to learning, because we have to continue
until we feel confident enough in our performance and application of our learning. My learning strategy of writing my own study notes started in my math
courses, and has come to expand into every one of my other courses, which
has significantly helped me to perform better on assessments. Working through
MTH240 has caused me to better organize myself in both my academics and
personal life as well. I have picked up the habit of taking note of my tasks,
and crossing them out once they have been accomplished. This has helped me
to never miss a quiz or assignment, which is difficult to do with all the assignments and assessments that come with engineering courses. This has bled into
my personal life as well to help me ensure I do not forget to accomplish any tasks.
As I have worked harder during the semester, and gained a better knowledge
of calculus, my confidence in solving problems of all kinds has increased. I
feel more confident to tackle mathematics problems for which there may be no
guaranteed solution because of all that I have learned. I have come to find
problem solving more intriguing than I originally thought it to be, and I have
found a passion for using what I have learned to further my knowledge in all
areas. After writing this reflection, I feel more confident and prepared for my
upcoming final exam because I have been reflecting weekly on what we have
learned, which has helped me to solidify it in my mind. After further reflection
with this last section, I have also come to realize that I have the skills I need to
succeed in my courses, and I am eager to display my skills and end the semester
strong.
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