this is a personal web page designed to provide you with an in-depth look into my personality, hobbies, as well as my interests. It is a "comprehensive" introduction to my online persona. It also features a collection of educational videos and resources related to my favorite topics; it is therefore considered to be an all-in-one destination for those who share similar passions and want to explore them further.
it also serves as my comfort zone, where i can always return to and add/modify something whenever i feel discomforted or stressed about some certain thing(s).
brought to you by lawi, with the best regards.
have a wonderful day ♥
lawi's note:
⟨ this website is alive, and it grows as time progresses. its growth is a function of two variables, and its value is expressed as the change in time, divided by my laziness squared. ⟩
i call myself lawi. friends use the nicknames law and lawi interchangeably. i am 17 years old. i like math and programming.
i like people who have a good sense of humor.
just kidding. i don't like humor. please don't make any jokes.
here's how you can get in touch with me:
get in touch:
@lawful.11 on instagram
something with a law in it elsewhere.
i also have email.
→ lawscriteria@gmail.com, all lowercase.
if you like to do things formally, here's a web page that'll direct you to my social profiles.
4 favorite series/shows: (unranked)
breaking bad (brba/bromium-barium) the high-stakes american series about a brilliant chemist who is also too egotistical w.n: « i hate mr white. »better call saul (bcs) the breaking bad spin-off w.n: « howard is just vg's punching bag at this point. »
attack on titan (AOT/SNK) the manga, by hajime isayama, and of course, the fan-favorite anime, attack on titan w.n: « beautiful. from the amazingly written eren, to the alliance at the end; all in all, a breathtakingly wonderful show well worth your time. »serial experiments lain, the anime. w.n: « purposefully speechless for this one. it is a true gemstone. »
...
let's all love lain ♥
I also think it is unfair to "rank" these from best to worst, since they are all so different and wonderful in their own unique way. That is why I have sorted them in the order I watched them. i will keep updating this list as i continue to develop this site.
speaking of development, go and read the next article for any updates on the development of this site! [blog]
8.9.24 - (??:??)AM
⟨ my mind is too preoccupied right now to think about chemistry. (who am i kidding? chemistry is the only thing i end up thinking long and hard about. even though i always hate the things that nag me for long. but chemistry feels like a beautiful exception. but seriously, god - why do some things have to be so complicated? Or is that just what life is? -- complicat? And no, that was not a typo. I don't know, but I feel like life should be more complicat than complicated. Don't you? I mean, aren't cats just so cute and adorable and cuschiekutshcyhuuscsyuucutkchshyuuuuuu??? I know we all have our own opinions on things and stuff, but I feel like that fact should be objectively true and realized to be so by everyone. Some things should just be like that, and some things are like that. And the cat fact is undeniably one of those things. I don't know. Just email me or something so we can talk about cats and stuff and things and stuff and be friends and stuff because you're cool if you've read all this) ⟩
launch version: [1.0]: 9.15.24
current version: [1.0]
[1.0]
9/15/24 ⟨ essentially done with the shape of the web page, so here goes publication. ⟩9/29/24 ⟨ currently busy with school, might restart work next month. ⟩
Please note that solutions for problems beyond Calculus II are unlikely to be posted as of now, but don't let that stop you from challenging me!
CALC II, EXAMPLE 1
\[\int \sec^3(x) \, dx = ?\]
Solution: the integrand, \(\sec^3(x)\), can be rewritten as \(\sec(x) \sec^2(x)\):
\[ \int \sec^3(x) \, dx = \int \sec(x) \sec^2(x) \, dx \]
We then utilize the method of integration by parts:
Let \(du = \sec^2(x) \, dx\); \(v = \sec(x)\)
\[ \int v \, du = uv - \int u \, dv \]
\[ \int \sec(x) \sec^2(x) \, dx = \tan(x) \sec(x) - \int \tan^2(x) \sec(x) \, dx \]
\[ = \tan(x) \sec(x) - \int (\sec^3(x) - \sec(x)) \, dx \]
\[ = \tan(x) \sec(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx \]
\[ \int \sec^3(x) \, dx = \tan(x) \sec(x) - \int \sec^3(x) \, dx + \int \sec(x) \, dx \]
We see that there are two \(\int \sec^3(x) \, dx\) terms that are on opposite sides and have opposite signs; thus, we can isolate and combine them to get:
\[ 2 \int \sec^3(x) \, dx = \tan(x) \sec(x) + \int \sec(x) \, dx \]
The integral of \(\sec(x)\) is a common integral, and it is equal to \(\ln |\tan(x) + \sec(x)| + C\). Divide both sides by 2 in order to isolate the unknown:
\[ 2 \int \sec^3(x) \, dx = \tan(x) \sec(x) + \ln |\tan(x) + \sec(x)| \]
\[ \frac{1}{2} 2 \int \sec^3(x) \, dx = \frac{1}{2} \left(\tan(x) \sec(x) + \ln |\tan(x) + \sec(x)|\right) \]
\[ \therefore \int \sec^3(x) \, dx = \frac{1}{2} \tan(x) \sec(x) + \frac{1}{2} \ln |\tan(x) + \sec(x)| + C \]
CALC II, EXAMPLE 2
\[\int \frac{\sin^2(x)}{\cos^3(x)} \, dx = ?\]
Solution: since \(\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}\), we can replace the integrand with \(\frac{\tan^2(x)}{\cos(x)}\) or \(\tan^2(x) \sec(x)\). Substituting in the latter, we get:
\[ \int \frac{\sin^2(x)}{\cos^3(x)} \, dx = \int \tan^2(x) \sec(x) \, dx \]
\[ \int \tan^2(x) \sec(x) \, dx = \int (\sec^2(x) - 1) \sec(x) \, dx = \int (\sec^3(x) - \sec(x)) \, dx \]
\[ = \int \sec^3(x) \, dx - \int \sec(x) \, dx \]
From here on, we can just substitute both integrals with their evaluations as we know them from the previous problem. However, since \(\int \sec^3(x) \, dx\) is not a common integral, we have to assume that we do not know its value. So, there is still more work to do. (Keep in mind that you can skip the following steps and use the expression from Q1/E1, if you have already solved it)
\[ \int \tan^2(x) \sec(x) \, dx = \int \sec^3(x) \, dx - \int \sec(x) \, dx \]
We can use integration by parts again to find another expression for the integral:
Let \(du = \tan(x) \sec(x) \, dx\); \(v = \tan(x)\)
\[ \int \tan^2(x) \sec(x) \, dx = \int \tan(x) \tan(x) \sec(x) \, dx = \tan(x) \sec(x) - \int \sec^3(x) \, dx \]
We then set these two expressions equal to one another:
\[ \int \sec^3(x) \, dx - \int \sec(x) \, dx = \tan(x) \sec(x) - \int \sec^3(x) \, dx \]
And once again, we see that there are two like terms with opposite signs that are on opposite sides of the equation – namely, \(\int \sec^3(x) \, dx\). And so, we isolate and combine them. We then simplify.
\[ 2 \int \sec^3(x) \, dx = \tan(x) \sec(x) + \int \sec(x) \, dx \]
\[ = \frac{1}{2} \left(\tan(x) \sec(x) + \ln |\tan(x) + \sec(x)|\right) \]
Of the two expressions we wrote for the integral, either can be used in the following substitution step. Here, we use the second equation:
\[ \int \tan^2(x) \sec(x) \, dx = \tan(x) \sec(x) - \frac{1}{2} \left(\tan(x) \sec(x) + \frac{1}{2} \ln |\tan(x) + \sec(x)|\right) \]
Simplify.
\[ = \tan(x) \sec(x) - \frac{1}{2} \tan(x) \sec(x) - \frac{1}{2} \ln |\tan(x) + \sec(x)| \]
\[ = \frac{1}{2} \tan(x) \sec(x) - \frac{1}{2} \ln |\tan(x) + \sec(x)| \]
\[ \therefore \int \frac{\sin^2(x)}{\cos^3(x)} \, dx = \frac{1}{2} \tan(x) \sec(x) - \frac{1}{2} \ln |\tan(x) + \sec(x)| + C \]
CALC II, EXAMPLE 3
\[\int e^x \sin(x) \, dx = ?\]
Solution: integrate by parts.
Let \(du = e^x \, dx\); \(v = \sin(x)\)
\[ \int e^x \sin(x) \, dx = e^x \sin(x) - \int e^x \cos(x) \, dx \]
\[ = e^x \sin(x) - e^x \cos(x) + \int -e^x \sin(x) \, dx \]
\[ \int e^x \sin(x) \, dx = e^x \sin(x) - e^x \cos(x) - \int e^x \sin(x) \, dx \]
\[ 2 \int e^x \sin(x) \, dx = e^x \sin(x) - e^x \cos(x) \]
\[ \int e^x \sin(x) \, dx = \frac{1}{2} \left(e^x \sin(x) - e^x \cos(x)\right) \]
\[ \therefore \int e^x \sin(x) \, dx = \frac{1}{2} e^x (\sin(x) - \cos(x)) + C \]
CALC II, EXAMPLE 4
\[\int \frac{1}{x^2 - x - 1} \, dx = ?\]
Solution: complete the square on the denominator, and factorize:
\[ x^2 - x - 1 = x^2 - 1 - \left(\frac{5}{4} - \frac{1}{4}\right) = x^2 - x + \frac{1}{4} - \frac{5}{4} = \left(x - \frac{1}{2}\right)^2 - \frac{5}{4} \]
\[ \int \frac{1}{x^2 - x - 1} \, dx = \int \frac{1}{\left(x - \frac{1}{2}\right)^2 - \frac{5}{4}} \, dx \]
Let \(u = x - \frac{1}{2}\), hence \(du = dx\)
\[ \int \frac{1}{\left(x - \frac{1}{2}\right)^2 - \frac{5}{4}} \, dx = \int \frac{1}{u^2 - \frac{5}{4}} \, du \]
The denominator is now a difference of squares, and so we factorize to get two linear factors. We then integrate using partial fractions and the fact \(\int \frac{1}{l} \, dl = \ln |l|\) to get the final answer.
\[ \int \frac{1}{u^2 - \frac{5}{4}} \, du = \int \frac{1}{\left(u - \frac{\sqrt{5}}{2}\right)\left(u + \frac{\sqrt{5}}{2}\right)} \, du \]
\[ \int \frac{1}{\left(u - \frac{\sqrt{5}}{2}\right)\left(u + \frac{\sqrt{5}}{2}\right)} \, du = \int \frac{A}{u - \frac{\sqrt{5}}{2}} \, du + \int \frac{B}{u + \frac{\sqrt{5}}{2}} \, du \]
Thus:
\[ A \left(u + \frac{\sqrt{5}}{2}\right) + B \left(u - \frac{\sqrt{5}}{2}\right) = 1 \]
Setting \(u = \frac{\sqrt{5}}{2}\) gives \(\sqrt{5} A = 1\) so \(A = \frac{1}{\sqrt{5}}\), and setting \(u = -\frac{\sqrt{5}}{2}\) gives \(-\sqrt{5} B = 1\) so \(B = -\frac{1}{\sqrt{5}}\).
\[ \int \frac{A}{\left(u - \frac{\sqrt{5}}{2}\right)} \, du + \int \frac{B}{\left(u + \frac{\sqrt{5}}{2}\right)} \, du = \frac{1}{\sqrt{5}} \int \frac{1}{u - \frac{\sqrt{5}}{2}} \, du - \frac{1}{\sqrt{5}} \int \frac{1}{u + \frac{\sqrt{5}}{2}} \, du \]
\[ = \frac{1}{\sqrt{5}} \ln \left| u - \frac{\sqrt{5}}{2} \right| - \frac{1}{\sqrt{5}} \ln \left| u + \frac{\sqrt{5}}{2} \right| \]
\[ = \frac{1}{\sqrt{5}} \left( \ln \left| x - \frac{1 + \sqrt{5}}{2} \right| - \ln \left| x - \frac{1 - \sqrt{5}}{2} \right| \right) \]
\[ = \frac{1}{\sqrt{5}} \ln \left| \frac{2x - 1 + \sqrt{5}}{2x - 1 - \sqrt{5}} \right| \]
\[ = \frac{1}{\sqrt{5}} \ln \left| 1 + \frac{2 \sqrt{5}}{2x - 1 - \sqrt{5}} \right| \]
\[ \therefore \int \frac{1}{x^2 - x - 1} \, dx = \frac{1}{\sqrt{5}} \ln \left| 1 + \frac{2 \sqrt{5}}{2x - 1 - \sqrt{5}} \right| + C \]
