Week 3

Areas between curves

Suppose we want to find the area between two curves, from $x=a$ to $x=b$. This area is represented by the definite integral
$$\int_{a}^{b} (f(x)-g(x)) dx$$
where $f(x)\geq g(x)$ $\forall x \in [a,b]$. $f(x)$ would be the "upper" curve, and $g(x)$ would be the "lower" curve.

Coming up next: volumes.

Week 2 – Part 2

U-Substitution

Remember the chain rule for differentiation?
$$\frac{d}{dx}F(g(x)) = F'(g(x))g'(x)$$
U-substitution is the chain rule done in reverse. Suppose that $F' = f$.
$$\int f(g(x))g'(x) dx = F(g(x)) + C$$
Now let's substitute $u = g(x)$.
$$\int f(g(x))g'(x) dx = F(u) + C$$
But we also know that the following is true.
$$\int f(u) du = F(u) + C$$
Therefore, it should be obvious that
$$\int f(g(x))g'(x) dx = \int f(u) du$$
This means that $du = g'(x) dx$. That's right, we can treat $du$ as a differential. So how does this help us integrate? We can pick an $u$ inside the integrand and find its $du$. Then we substitute the variable $x$ out completely, so our integrand is a function of $u$ and we are integrating with respect to $u$. Once we finish evaluating the integral, we replace $u$ with whatever we picked when we started the integration.

Let's try a simple example. Suppose we have the following integral.
$$\int \cos (2x+3) dx$$
Let's pick $u = 2x+3$. Then $du = 2 dx$. So $dx = \frac{1}{2}du$. Making these substitutions
$$\int \cos (2x+3) dx = \int \cos (u) \frac{1}{2}du$$
But $\frac{1}{2}$ is a constant and may be pulled out of the integral. So
$$\int \cos (u) \frac{1}{2}du = \frac{1}{2} \int \cos (u) du$$
Integrating with regard to $u$, we have
$$\frac{1}{2} \int \cos (u) du = \frac{1}{2}\sin (u) + C$$
Finally, replace $u$ with $2x+3$, which is what we chose earlier.
$$\frac{1}{2}\sin (u) + C =  \frac{1}{2}\sin (2x+3) + C$$
And we are done. Pretty cool, right?

Next week, we start Chapter 6, Applications of Integration. And I am taking my first midterm this week, as well. :'(

Week 2 – Part 1

Last week in Calculus 2, we finished Chapter 5 in Stewart's Calculus. Two of the most important concepts this week were indefinite integrals and u-substitution. Since this will be quite long, I am going to write two posts, each covering one topic.

Indefinite Integrals

An indefinite integral simply represents a "family" of antiderivatives. Suppose $F$ is an antiderivative of $f$. In other words, $F' = f$. Then
$$\int f(x) dx = F(x) + C$$
The $dx$ tells us that we are integrating with respect to $x$, as opposed to another variable. We went over many common indefinite integrals, including this very important one.
$$\int x^n dx = \frac{x^{n+1}}{n+1}$$
for $n\not=-1$. Our professor calls this process "raise and divide." Raise the power $n$ by $1$, and then divide $x^{n+1}$ by the new power $n+1$.

There are many other common indefinite integrals or antiderivatives, which can be found in any calculus textbook. It is very helpful to memorize and know them well. Trust me, that really helps when evaluating integrals.

Stay tuned for my next post!

Calculus 2: Week 1

The first week of classes at my college went by quite quickly. My math class this quarter is Calculus 2. I am taking an evening class, which meets twice a week. This is quite a change from last quarter, when I had Calc 1 each day Monday through Friday.

In class, we started with the concept of an integral, defined as the Riemann sum of areas.

The Definite Integral

Suppose $f$ is defined on $[a,b]$. Let $[a,b]$ be divided into $n$ equal subintervals of width $\Delta x = (b-a)/n$. Now let $x_{0} = a$, $x_{1}$, $x_{2}$, $x_{i} = a + i\Delta x$, ... $x_{n} = b$ be the endpoints of the subintervals, and let $x_{i}*$ be any sample point in the $i$th interval $[x_{i-1}, x_{i}]$. Then

$$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_{i}*)\Delta x$$

The precise meaning of this limit:

$\forall$ $\epsilon >0$ $\exists$ an integer $N$ such that

$$\left|\int_{a}^{b} f(x) dx - \sum_{i=1}^{n} f(x_{i}*)\Delta x \right| < \epsilon$$

 $\forall$ integers $n>N$ and $\forall$ $x_{i}* \in [x_{i-1}, x_{i}]$.


Next, we went over the Fundamental Theorem of Calculus, which shows that differentiation and integration are inverse processes.

Fundamental Theorem of Calculus

Suppose $f$ is continuous on $[a,b]$ and $F' = f$. Then

  $$\frac{d}{dx}\int_{a}^{x} f(t) dt = f(x)$$

$$\int_{a}^{b} f(x) dx = F(b) - F(a)$$


Wow, this took longer than I thought with all the Latex. I will try to make a post like this every week, if I am not too busy. Our class is using Stewart's Calculus (8th Edition), so my formulas, definitions, theorems, etc. for this and other posts are and will be pretty much copied from the textbook.

Coming up soon: indefinite integrals and u-substitution (reverse chain rule).

Some derivatives...

Just testing out Latex on this blog.

$$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$$
$$\frac{d}{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$$
$$\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$
$$\frac{d}{dx}b^x = b^x\ln b$$
$$\frac{d}{dx}\sec x = \sec x\tan x$$
$$\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$$

Looking good!

Microeconomics

Last quarter, I also took an introductory microeconomics course. Besides math, econ is my other favorite thing.

Here are the highlights of the class:

• Supply and demand
• The burden of a tax
• How consumers maximize their utility
• Cost curves of firms
• Market structures: perfect competition, monopolistic competition, oligopoly (my favorite because of the game theory involved), and monopoly
• The role of government
• Inefficiency in markets

This class was so much fun! The professor used really good (and funny!) examples to illustrate the concepts he taught. I am taking macro in the winter.

My grade for Calc 1

Grades were posted for my math class today.  I got an A for Calc 1, which is really deeply satisfying. :)

I am definitely looking forward to Calculus 2, and studying integration.

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