Class 1 - Overview / Fundamental Theorem of Calculus (Tuesday, 9/5/2017)
Welcome to Math 116! My name is GilYoung Cheong, a third year PhD student in mathematics at the University of Michigan. In U of M, it is usual for graduate students/post doctoral researchers to teach intro-level courses, while tenured faculty members teach more advanced courses.
The following are class notes for Tuesday, Sept 5, 2017. They roughly cover Section 5.1, 5.2, and 5.3 of your book.
We now start studying some mathematics.
Exercise*. Solve Number 1 of Exam 1, Winter 2017.
You will find the following hints useful:
Hint for Part (a): Using chain rule, differentiate $f(\ln(x))$ with respect to $x$, and use what you get in computing the integral, using the fundamental theorem.
Hint for Part (b): Using product rule, note that $(xf'(x))' = f'(x) + xf''(x)$. Taking the integral operator appearing in the problem this implies that
\[\int_{0}^{4}(xf'(x))'dx = \int_{0}^{4}f'(x)dx + \int_{0}^{4}xf''(x)dx.\]
Now, notice that you can compute the two of the three integrals appearing in the above identity, using the fundamental theorem.
Hint for Part (c): Using chain rule, differentiate $f(x)^{3}/3$ in $x$ and use what you get in computing the integral, using the fundamental theorem.
The following are class notes for Tuesday, Sept 5, 2017. They roughly cover Section 5.1, 5.2, and 5.3 of your book.
Logistics
The place you can see everything. The course website will consists of the most important information including various policies, exam dates, team homeworks, or past exam data. You must skim this in the first week because you may get penalty by not knowing relevant information.
Class hours. We meet Tuesdays, Thursdays, and Fridays from 11:40 AM to 1:00 PM. In Wolverine Access, it is written that we start at 11:30 AM, but we start 10 minutes later than that by what is called Michigan time rule. My guess is that this a marketing strategy for U of M, but I am not sure about the origination of Michigan time. It creates the confusion because on exam dates, Michigan time does not apply.
The location we meet is 2437 Mason Hall.
The location we meet is 2437 Mason Hall.
Office hours. My office hours are
- Tuesdays and Thursdays 1:00 PM - Noon (just talk to me right after class; my office is 5832 East Hall);
- Wednesdays 2:00 PM - 3:00 PM at Math Lab (located at the basement of East Hall).
You should let me know ahead if you come to my office hours. The best is to start talking to me right after class. However, if for some reason you need to visit me at 1:30 PM, then please let me know after class that you want to visit, or you can inform me by e-mail. This drastically reduces chance that I miss your visit.
Grading components. Your grade will essentially be determined by uniform (grading) component. 95% of the uniform component will be determined by uniform exams, which I will explain soon. 5% of the uniform component will be determined by web homework.
There is a complementary grading component to the uniform component, called section (grading) component. This will be mainly determined by quizzes and team homeworks. Unless you lack a significant amount of effort in quizzes and team homeworks, the section component will not affect your grade. With a fair process (which I will not reveal but whose record will be kept on Canvas) of recording points on the section component, a very few students will have 1/3 of grade up (e.g., B- to B). If you do not participate well in your team homeworks or miss quizzes without telling me, 1/3 of your grade will be lowered (e.g., A to B+).
Uniform exams. These exams are the most important parts of this course, in terms of your grade. Please check the date and time on the course website. You must let me know if you will have conflict on those dates right now, because they are are almost impossible to be changed.
As mentioned before, the uniform exams consists of 95% of the uniform component in determining your grade. 25% will be from Exam 1, 30% will be from Exam 2, and 40% will be from Exam 3 (a.k.a Final exam).
Gateway exams. There will be two Gateway exams, which I will announce several weeks later. Each one is a computer-based and Pass/Fail exam. You will have to pass this in a specific period of time (TBA). Not passing each Gateway exam will decrease 1/3 of your letter grade (e.g., B- to C).
Quizzes. The quizzes will be announced in advance so that you have some practice before taking them. They will be challenging, but because they only count as section components, you can just try as hard as you can without being too worried about affecting your grade. They exist for you to be vigilant about your mathematical skills 24/7.
Team homework. Please read the tutorial for the team homework.
Student data sheet. Fill out your student data sheet. This often helps me understand what background you have, and why you take this course. I believe that having conversations with students are important in their successful completion of the course, and this is the first step.
We now start studying some mathematics.
Special functions we study
A function $f$ is a mapping an object $x$ to a unique object $y$, which we often write as $y = f(x)$ or $f : x \mapsto y$.
Functions with no singularity. There are many functions in the universe, but we restrict ourselves studying functions that send real numbers to real numbers. Among them, the following are the ones of our most interest:
The above functions are differentiable everywhere. This means that we can take the derivative $f'(x) = \frac{d}{dx}f(x)$ of a function $f(x)$ at every $x$. Explicitly, we can consider the following:
Derivative of polynomials. For $n \geq 1$, we have
Since $d/dx$ respects addition and scalar multiplication (i.e., $d/dx$ is a linear operation), we have
For example, if $f(x) = 3x + 1$, then $f'(x) = 3$, the slope of the line $y = 3x + 1$.
Derivative of cosine and sine functions. We have
and
Exercise. Recall the quotient rule (i.e., $(f/g)' = (f'g - fg')/g^{2}$). We define the tangent $x$ to be
whenever $\cos(x) \neq 0$. Consider the function $f$ given by $f(x) = \tan(x)$. Compute $f'(x)$.
Exercise. For which $x$, is $\cos(x) = 0$? Find them all. (Suggestion: you may want to review math material prior to this course to draw the graph $y = \cos(x)$).
Exercise. Compute $\frac{d}{dx}(1/x)$.
Derivative of exponential functions. For any $a > 0$, we have
Exercise. Using the above fact, compute $\frac{d}{dx}e^{x}$, when
(Hint: $\ln(1) = 0$, and the definition of $e$ is not so important in doing this exercise.)
Functions with singularity. Although the functions we have seen so far are nice, we will often see a combination of such functions $f(x)$ that may not be differentiable or continuous at some $x$. Such an $x$ is usually called "singularity", but you don't need to know this terminology for this course.
For example, the function $f(x) = 1/x$ is differentiable at every $x$ except $x = 0$. In fact, we cannot even talk about what "$1/0$", in our world of real numbers. Here is why: if there is such a thing as $c = 1/0$, then multiply $0$ to the both sides to get
so we must have $0 = 1$. This is a nonsense, so the assumption that "the real number $c = 1/0$ exists" must be false. In general, if $f(x)$ and $g(x)$ are polynomials the function defined by $x \mapsto f(x)/g(x)$ is differentiable everywhere except at the roots of $g$. Such a function is called a rational function.
Exercise. At which $x$ is the function $f(x) = 1/(1 + x^{2})$ differentiable?
As you could figure out in previous exercises, the function $f(x) = \tan(x)$ is differentiable at every $x$ except when $x$ is $(2n - 1)\pi/2$ for some integer $n$. For such $x$, you can check that
where $\sec(x) = 1/\cos(x)$. (When you check this, it is likely that you will need $\cos^{2}(x) + \sin^{2}(x) = 1$, which is a version of Pythagorean theorem.)
\[0.32835 = \dfrac{99 \cdot 100 \cdot 199}{6 \cdot 100^{3}} \leq \int_{0}^{1}x^{2}dx \leq \dfrac{100 \cdot 101 \cdot 201}{6 \cdot 100^{3}} = 0.33835,\]
which supports my claim that the answer for the middle should be $1/3 = 0.3333333 \dots.$
We can make this even more precise. If we replace $100$ with general $n \geq 1$ in the above argument, we get
\[\dfrac{(n-1)n(2n - 1)}{6 n^{3}} \leq \int_{0}^{1}x^{2}dx \leq \dfrac{n(n+1)(2n+1)}{6 n^{3}}.\]
Thus, we have
\[\dfrac{1}{3} = \lim_{n \rightarrow \infty}\dfrac{(n-1)n(2n - 1)}{6n^{3}} \leq \int_{0}^{1}x^{2}dx \leq \lim_{n \rightarrow \infty}\dfrac{n(n+1)(2n+1)}{6n^{3}} = \dfrac{1}{3}.\]
Therefore, we must have
\[\int_{0}^{1}x^{2}dx = \dfrac{1}{3},\]
as desired.
Functions with no singularity. There are many functions in the universe, but we restrict ourselves studying functions that send real numbers to real numbers. Among them, the following are the ones of our most interest:
- Polynomial functions. These are the functions of the form $f(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0}$ where $a_{n}, \dots, a_{1}, a_{0}$ are real numbers.
- Cosine functions. These are the functions given by $\theta \mapsto \cos(\theta)$, meaning the length divided by the hypotenuse of a right triangle of angle $\theta$ (presented with radian convention).
- Sine functions. These are the functions given by $\theta \mapsto \sin(\theta)$, meaning the height divided by the hypotenuse of a right triangle of angle $\theta$ (presented with radian convention).
- Exponential functions. Such functions are given by $x \mapsto a^{x}$ for some fixed positive real number $a$.
The above functions are differentiable everywhere. This means that we can take the derivative $f'(x) = \frac{d}{dx}f(x)$ of a function $f(x)$ at every $x$. Explicitly, we can consider the following:
Derivative of polynomials. For $n \geq 1$, we have
$\dfrac{d}{dx}x^{n} = nx^{n-1}.$
Since $d/dx$ respects addition and scalar multiplication (i.e., $d/dx$ is a linear operation), we have
$\dfrac{d}{dx}(a_{n}x^{n} + a_{n-1}x^{n-1} \cdots + a_{2}x^{2} + a_{1}x + a_{0}) = na_{n}x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_{2}x + a_{1}.$
For example, if $f(x) = 3x + 1$, then $f'(x) = 3$, the slope of the line $y = 3x + 1$.
Derivative of cosine and sine functions. We have
$\dfrac{d}{dx}(\cos(x)) = -\sin(x)$
and
$\dfrac{d}{dx}(\sin(x)) = \cos(x).$
Exercise. Recall the quotient rule (i.e., $(f/g)' = (f'g - fg')/g^{2}$). We define the tangent $x$ to be
$\tan(x) := \dfrac{\sin(x)}{\cos(x)},$
whenever $\cos(x) \neq 0$. Consider the function $f$ given by $f(x) = \tan(x)$. Compute $f'(x)$.
Exercise. For which $x$, is $\cos(x) = 0$? Find them all. (Suggestion: you may want to review math material prior to this course to draw the graph $y = \cos(x)$).
Exercise. Compute $\frac{d}{dx}(1/x)$.
Derivative of exponential functions. For any $a > 0$, we have
$\dfrac{d}{dx}a^{x} = a^{x}\ln(a)$.
Exercise. Using the above fact, compute $\frac{d}{dx}e^{x}$, when
$e := \lim_{n \rightarrow 0}(1 + 1/n)^{n}.$
(Hint: $\ln(1) = 0$, and the definition of $e$ is not so important in doing this exercise.)
Functions with singularity. Although the functions we have seen so far are nice, we will often see a combination of such functions $f(x)$ that may not be differentiable or continuous at some $x$. Such an $x$ is usually called "singularity", but you don't need to know this terminology for this course.
For example, the function $f(x) = 1/x$ is differentiable at every $x$ except $x = 0$. In fact, we cannot even talk about what "$1/0$", in our world of real numbers. Here is why: if there is such a thing as $c = 1/0$, then multiply $0$ to the both sides to get
$0 = 0 \cdot c = 0 \cdot (1/0) = 1,$
so we must have $0 = 1$. This is a nonsense, so the assumption that "the real number $c = 1/0$ exists" must be false. In general, if $f(x)$ and $g(x)$ are polynomials the function defined by $x \mapsto f(x)/g(x)$ is differentiable everywhere except at the roots of $g$. Such a function is called a rational function.
Exercise. At which $x$ is the function $f(x) = 1/(1 + x^{2})$ differentiable?
As you could figure out in previous exercises, the function $f(x) = \tan(x)$ is differentiable at every $x$ except when $x$ is $(2n - 1)\pi/2$ for some integer $n$. For such $x$, you can check that
$f'(x) = \sec^{2}(x),$
where $\sec(x) = 1/\cos(x)$. (When you check this, it is likely that you will need $\cos^{2}(x) + \sin^{2}(x) = 1$, which is a version of Pythagorean theorem.)
Some hints of integration
Draw the line $y = x$. You can then compute the area between this line and the $x$-axis from $x = 0$ to $x = 1$, because you learned how to compute the area of a triangle before. The answer is $1/2$. What if I ask you the same question with $y = x^{2}$ instead of $y = x$? Can you compute the area? One of the major goals that people have developed the theory of integration is to compute areas under non-linear curves given by the special (differentiable) functions.
One way to approach this problem is to cut the area vertically into many pieces and approximate each piece by a rectangle. As a notation, we denote the area we want to compute as
Cutting the interval $[0, 1] = \{x : 0 \leq x \leq 1\}$ will yield the following $100$ subintervals of $[0, 1]$:
For each interval $[k/100, (k+1)/100]$ where $0 \leq k \leq 99$, we can consider two kinds of rectangles that approach the area $\int_{k/100}^{(k+1)/100}x^{2}dx$:
Because the function $x \mapsto x^{2}$ is increasing when $0 \leq x \leq 1$, we can compare the two kinds of rectangles and the area we want to compute as follows:
\[\left(\dfrac{k}{100}\right)^{2}\dfrac{1}{100} \leq \int_{k/100}^{(k+1)/100}x^{2} dx \leq \left(\dfrac{k + 1}{100}\right)^{2}\dfrac{1}{100}.\]
Adding these up from $k = 0$ to $k = 99$, we have
\[\sum_{k=0}^{100}\left(\dfrac{k}{100}\right)^{2}\dfrac{1}{100} \leq \int_{0}^{1}x^{2}dx \leq \sum_{k=0}^{100}\left(\dfrac{k+1}{100}\right)^{2}\dfrac{1}{100}.\]
Using the identity $\sum_{k=0}^{m}k^{2} = m(m+1)(2m+1)/6$, the above becomesOne way to approach this problem is to cut the area vertically into many pieces and approximate each piece by a rectangle. As a notation, we denote the area we want to compute as
$\displaystyle\int_{0}^{1}x^{2} dx.$
Cutting the interval $[0, 1] = \{x : 0 \leq x \leq 1\}$ will yield the following $100$ subintervals of $[0, 1]$:
$[0, 1/100], [1/100, 2/100], [2/100, 3/100], \dots, [98/100, 99/100], [99/100, 1].$
For each interval $[k/100, (k+1)/100]$ where $0 \leq k \leq 99$, we can consider two kinds of rectangles that approach the area $\int_{k/100}^{(k+1)/100}x^{2}dx$:
- Left-hand height: we take the interval $[k/100, (k+1)/100]$ as the base and $(k/100)^{2}$ as the height of the rectangle. Thus, the area of the rectangle is $(k/100)^{2}(1/100)$.
- Right-hand height: we take the interval $[k/100, (k+1)/100]$ as the base and $((k+1)/100)^{2}$ as the height of the rectangle. Thus, the area of the rectangle is $((k+1)/100)^{2}(1/100)$.
Because the function $x \mapsto x^{2}$ is increasing when $0 \leq x \leq 1$, we can compare the two kinds of rectangles and the area we want to compute as follows:
\[\left(\dfrac{k}{100}\right)^{2}\dfrac{1}{100} \leq \int_{k/100}^{(k+1)/100}x^{2} dx \leq \left(\dfrac{k + 1}{100}\right)^{2}\dfrac{1}{100}.\]
Adding these up from $k = 0$ to $k = 99$, we have
\[\sum_{k=0}^{100}\left(\dfrac{k}{100}\right)^{2}\dfrac{1}{100} \leq \int_{0}^{1}x^{2}dx \leq \sum_{k=0}^{100}\left(\dfrac{k+1}{100}\right)^{2}\dfrac{1}{100}.\]
\[0.32835 = \dfrac{99 \cdot 100 \cdot 199}{6 \cdot 100^{3}} \leq \int_{0}^{1}x^{2}dx \leq \dfrac{100 \cdot 101 \cdot 201}{6 \cdot 100^{3}} = 0.33835,\]
which supports my claim that the answer for the middle should be $1/3 = 0.3333333 \dots.$
We can make this even more precise. If we replace $100$ with general $n \geq 1$ in the above argument, we get
\[\dfrac{(n-1)n(2n - 1)}{6 n^{3}} \leq \int_{0}^{1}x^{2}dx \leq \dfrac{n(n+1)(2n+1)}{6 n^{3}}.\]
Thus, we have
\[\dfrac{1}{3} = \lim_{n \rightarrow \infty}\dfrac{(n-1)n(2n - 1)}{6n^{3}} \leq \int_{0}^{1}x^{2}dx \leq \lim_{n \rightarrow \infty}\dfrac{n(n+1)(2n+1)}{6n^{3}} = \dfrac{1}{3}.\]
Therefore, we must have
\[\int_{0}^{1}x^{2}dx = \dfrac{1}{3},\]
as desired.
Fundamental Theorem of Calculus
If you really work hard to understand the area computation we gave in the previous section, you can realize that the same type of argument would give
\[\int_{0}^{1}x^{n}dx = \dfrac{1}{n+1},\]
for any $n \geq 1$, or better yet, you will realize
\[\int_{a}^{b}x^{n}dx = \dfrac{a^{n+1}}{n+1} - \dfrac{b^{n+1}}{n+1}.\]
With more theoretical arguments, this actually generalizes to the following:
Fundamental Theorem of Calculus. If $f$ is a differentiable function whose derivative is continuous, then for any real numbers $a$ and $b$, we have
\[\int_{a}^{b}f'(x)dx = f(b) - f(a).\]
Exercise*. Solve Number 1 of Exam 1, Winter 2017.
You will find the following hints useful:
Hint for Part (a): Using chain rule, differentiate $f(\ln(x))$ with respect to $x$, and use what you get in computing the integral, using the fundamental theorem.
Hint for Part (b): Using product rule, note that $(xf'(x))' = f'(x) + xf''(x)$. Taking the integral operator appearing in the problem this implies that
\[\int_{0}^{4}(xf'(x))'dx = \int_{0}^{4}f'(x)dx + \int_{0}^{4}xf''(x)dx.\]
Now, notice that you can compute the two of the three integrals appearing in the above identity, using the fundamental theorem.
Hint for Part (c): Using chain rule, differentiate $f(x)^{3}/3$ in $x$ and use what you get in computing the integral, using the fundamental theorem.
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