I don't know why TH-cam is recommending this old ass video, but you do not need to know this for Calculus 2. To those who are studying for Calculus Final Exams, you can most definitely do it. You got this!! I believe in all of you!! (PS: MIT Integration Bee 2022 is too hard for me. It is monstrosity, very eye unpleasing, and unwelcoming. - A person who likes to go outside.)
Have a look how people in Asia are holding brushes in classic character-writing in elementary school: pretty much the way he does. The aim is to hold the pen perpendicular to the writing surface...
it seems that the difference between taking calculus at more and less prestigious schools comes down to memorization. These are the tables of integrals that most STEM majors fail to memorize.
tips for a problem like #2: ln( f ) = -ln( 1/f ), bc of exponent rules. (check out euler’s proof of “the divergence of the sum of the reciprocals of the primes” for a dope example) also, a more obvious tip is that x = ln(e^x), which helps combine terms here. your guess was right tho, ln(2) is the answer.
@@xxxy912 It's for professionals like engineers, chemists, scientists, etc, calculus and math in general develops problem solving skills and logical thinking part of your brain, which are vital to the professional jobs. After all, we need a logical engineer and a rocket scientist to send rovers to mars.
Sure, but manual integration in itself is useless. If your goal is to get better at logical reasoning, You’ll be better off focusing on proofs, instead of mechanical calculations.
25:02 this is not why you got the wrong answer - you just forgot to redistribute the factor of 1/2 that you took out of ur original quadratic, when you took a 2 out of the expression on the denominator haha - your technique was correct! :D
Q.2 is an improper integral of type integral with limits 0 to infinity . So final answer is infinity , hence its we can say that f(x) is unbounded in the region of integration
@@oliver_pryce no. Replace infinity with a variable, f, and plug in the numbers accordingly. You will get f - ln(f+1) - (0 - ln(2)) --- [ln(e^0 + 1) --> ln(1+1) --> ln(2)] Next set up the limit as f approaches infinity and you should get an answer of ln(2)
@@parthisMC Multiply top and bottom by e^-x. Tgen the function becomes (e^-x)/(1+e^-x). After integrating, the following function will be obtained -ln(1+e^-x). Sub in the limits and the answer will be ln2
It's freshman questions... The books used in IIT are of foreign authors.... Talk about MIT.. They have worlds best minds... So higher level maths is very very hard... Even IIT prof fails to solve them...
@@jackyjack9660 bro honestly for indian students these are cake walk ...we don't need to even solve these many of the results we know directly..every question can be answered in 1 second..not kidding.
25:17 For number 5, im pretty sure that the (1/4) does affect the coefficient of arctan since the formula for arctan integration has (1/a) outside as a coefficient if the whole integral. I solved this on my own since it was the only one i was willing to attempt coz i cant even think about doing the others; either you forgot that you divided the whole bottom part of the integrand by two and have to then move that outside of the integral before integrating, or you forgot that the (1/4) which you added to the (x-(1/2))^2) had to be square rooted to get the integral to the form of (1/(u^2+a^2)) to then use inverse tan (the square root of (1/4) would be (1/2), which would then be set as a, (1/a)=2, and then divide the 2 by the two you pulled out from the integrand earlier. Hats off bro, what you are doing is crazy, as a AP Calc AB student, I can’t even comprehend the other problems.
Chemistry is good and all, the most practical science easily, but it's not as intriguing as math and physics can be. Physics tries to answer questions about the very existence of things, math is a most beautiful structure arising out of baffling simplicity, whose "purity" is tainted every century or so when a field of science requires new tools. But chemistry... it's just chemistry, I guess.
For no.8 when u by parts you get to integrate 2x^2/(1+x^2) this will further become arc(tan). You were actually missing a 2x in that part hence u integrated x/1+x^2 which came as logarithmic function.
Question 2, for the limit Let L be lim (x-ln(e^x+1)) e^L = e^lim (x-ln(e^x+1)) = lim (e^(x-ln(e^x+1))) = lim (e^x/e^(ln(e^x+1))) = lim (e^x/e^x+1) = 1 And e^L = 1 L = 0
ln(a/b) = -ln(b/a) because you can pull out an exponent of -1 from inside the ln() so ln( e^x / (1+e^x) ) = -ln( 1 + e^-x) evaluating at the boundary you get: lim{t->infinity}[ -ln(1 + 1/t ) + ln(1+1/1) ] = -ln(1) + ln(2) = 0+ln(2) = ln(2) so his guess was right
In many disciplines, log without a specified base is ln(x). You may be confusing it with the more common convention that log(x) is implied to be log base 10
Its half mentally. When sheer practicing these kinds of integrals, I like to memorize the process of certain types of integrals so that when i see a similar type, i know what to do quickly.
@@replicaacliper that's the standard way of evaluating integrals in infinity but I wasn't referring to that but to the fact that power series are only defined for arguments between -1 and 1 exclusive. Thus substituting infinity wouldn't make any sense
@@pinguino55h40 If you take power series based on 1/(1-x) the Best Friend then this can be written as Sum from n=1 to infinity of x^n where x lies between 1 and -1. Outside this interval the power series do not match 1/(1-x).
@@giorgostarnaras5658 You got a point. Is there a way we could use L'hipothals rule (I think that is how you spell it) because ∞ - ∞ is an indeterminate. Would this work? x-ln(x) = x(1-(ln(x)/x)) => lim x * lim (1-(ln(x)/x)) => lim x * lim ((x-ln(x))/x) => lim x * lim ((1-(1/x))/1) (L'hipothal rule thingy) => ∞ * [(1 - 0)/1] = ∞.
@@lohitmurali2137 because it is in indeterminate form and can not be evaluated. The reason is one infinite may be growing faster than the other one so we have no way of knowing the outcome if we try to subtract. In order to evaluate this limit we have to transform it into a indeterminate form that we do know how to handle so we can take advantage of l’hospitals rule (0/0 or infinity/infinity). This can be done a couple of ways and will be left as an exercise for the reader.
I started writing like that since Kindergarten, and then I got too comfortable with it. In 3rd grade, my teacher tried to make me hold it normally, but my handwriting became hella awful like you couldnt even read it. It's just sloppy and a mess. So I just kept my formation to make my handwriting readable.
Look what great job he's doing and sincerely time is important in integration bee you have to write things super fast so that's why go in integration bee video you will see that everyone writing weirdly
It's not that hard most of the question I did in my head without using pen, specially in India any 12th graduate can do this. Keep learning good luck brother
what the hell is that grip bruh , i swear that all white people hold their pens in this weird fashion , i would tell me frinds to wrap their fingers around the pen or pencil like that only to just squeeze it and paralyze their fingers
indetermination, it depends because there are infinities of different magnitudes. You can use different approaches to obtain the final result, but infinity - infinity is not 0. Unless both infinities are the same i guess.
@@talhajat3301 ok i had skipped over that part, sorry, now i see what u mean, ok ok so the way i would solve that is this way: you have X - ln(e^x +1) between 0 and infinity, that means that we just have to replace the X with infinity and then remove the same but replacing with 0 so we get infinite - ln (e^infinite + 1) - [0 - ln(e^0 +1)] so, lets focus on the second part: 0 - ln(e^0 +1) from here, since e^0 = 1, we get that this is equal to -ln(2) now lets focus on the first part: infinite - ln(e^infinite +1) in here, normally we would consider that e^X is an infinity of greater magnitude than just X, because it is, but the difference of ln(e^X) and ln(e^X +1) when X goes to infinity is so small that it leads us to have 2 infinities that are almost the same, like this: Imagine we pick a high value like 9999 for X since we cant go exactly to infinite: X = 9999 ln(e^X) = ln(e^9999) = 9999 (remember that ln is log of base e so to get the solution we ask the question "what number do i have to elevate e to get e^9999, so the answer is just 9999") now, since X and ln(e^X) give us the same result, we no longer have the problem that e^X is a greater infinity than just X, because in out case we have an ln. Now, we have ln(e^X + 1) but that isnt a problem since that +1 is not a big deal. If you use a calculator, for small values you'll see a difference. For example try x=10 in ln(e^x +1) and you'll see how small the difference is already. Since we are going to infinity, they are basically the same number, so X - ln(e^X +1) when going to infinity are just going to cancel out because both infinities are of the same magnitude in this case, so we get this result, that X - ln(e^X +1) = 0 Now, having all of that into account, lets add it up to solve the problem: infinite - ln (e^infinite + 1) - [0 - ln(e^0 +1)] = 0 - - ln(2) = ln(2) So the final answer is ln(2) Personally, i have never ever seen one of these in my life, but it was kinda easy i guess to figure out how it worked, because this exercise seems to have been built around the idea that you spot that both infinities are of the same magnitude. *Side note: yeah they are not exactly the same magnitude of infinity, but as i stated before, you can already see how that +1 in the ln makes a real small difference already when using a value as small as 10. If we go to infinity you will see that the difference will be so small that it wont matter at all.
Bruh in india its taught in class 11th or at max they teach u this is class 12th and to be really honest we r taught more intense integration than this in class 12th
MIT lolllllllllllllllll.Questions like this in greece are must before even go to the university.We solve them in highschool.So easyy.Wtf hahaha.U guys are silly asf
I don't know why TH-cam is recommending this old ass video, but you do not need to know this for Calculus 2.
To those who are studying for Calculus Final Exams, you can most definitely do it. You got this!!
I believe in all of you!!
(PS: MIT Integration Bee 2022 is too hard for me. It is monstrosity, very eye unpleasing, and unwelcoming. - A person who likes to go outside.)
thx makes me feel better lol
Silver the hedgehoggggg
Hello
Woah hello there. Do you intend to make videos again?
@@kanishkachakraborty Yea......if I can ;_;
Neither can I wrap my head around the way he's holding his pen nor around the integration.
*I can neither wrap my head around his pen-holding, nor the integration
@@savitar725 Nobody cares
Have a look how people in Asia are holding brushes in classic character-writing in elementary school: pretty much the way he does. The aim is to hold the pen perpendicular to the writing surface...
I think he was holding the pen between his ring finger and pinky, rather than his index and middle fingers.
@Savitar consider therapy
The way he holds his pen 💀💀💀
?
He may be an osu! player
The way he managed to solve num 6 gave me epilepsy
@@alphonse6259 he may use vaxei grip OwO
Lmao I write the exact same way never seen anyone else do it like this 😂
it seems that the difference between taking calculus at more and less prestigious schools comes down to memorization. These are the tables of integrals that most STEM majors fail to memorize.
You are so briliant. I'm really impressed by how you can solve those questions. Keep it up, man.
Thank you!!!! It's experience and passion.
Not that brilliant, pretty basic integrals
@@Robert-ko2pq bet you can do better, Robert
@@Robert-ko2pq 🤡
@@shmevanriceballz2857 They'are basic...
tips for a problem like #2:
ln( f ) = -ln( 1/f ), bc of exponent rules. (check out euler’s proof of “the divergence of the sum of the reciprocals of the primes” for a dope example)
also, a more obvious tip is that x = ln(e^x), which helps combine terms here.
your guess was right tho, ln(2) is the answer.
substitute 1/x or multiply and divide by e^-x and use geometric series and you get the sum for log2
I used to be really fast at integration and derivatives. Just takes practice and time. Becomes second nature eventually.
Useless skill tbh
@@xxxy912 explain
For research purposes, it's not that useful. Rigor is more important unless you are into applied Mathematics.
@@xxxy912
It's for professionals like engineers, chemists, scientists, etc, calculus and math in general develops problem solving skills and logical thinking part of your brain, which are vital to the professional jobs.
After all, we need a logical engineer and a rocket scientist to send rovers to mars.
Sure, but manual integration in itself is useless. If your goal is to get better at logical reasoning, You’ll be better off focusing on proofs, instead of mechanical calculations.
The integrals were amazing💯
Thanks for sharing it🙏🏼
Great job sir good to see more maths TH-camrs!!
This video makes Calc AB feel so easy
Bro solved the first integral in like 1 minute but then struggled for 30s on ln(infinity)
it's a limit with [infinity - infinity] so there's the problem
25:02 this is not why you got the wrong answer - you just forgot to redistribute the factor of 1/2 that you took out of ur original quadratic, when you took a 2 out of the expression on the denominator haha - your technique was correct! :D
Q.2 is an improper integral of type integral with limits 0 to infinity .
So final answer is infinity , hence its we can say that f(x) is unbounded in the region of integration
I plugged it in to an integral calculator and got ln(2) - the same as him. Ngl I thought it woulda been 0 but ig I’m just dumb
x - ln(e^x + 1) = ln(e^x / (e^x + 1) ) which limits to ln1 = 0 at infinity
@@oliver_pryce no. Replace infinity with a variable, f, and plug in the numbers accordingly. You will get f - ln(f+1) - (0 - ln(2)) --- [ln(e^0 + 1) --> ln(1+1) --> ln(2)]
Next set up the limit as f approaches infinity and you should get an answer of ln(2)
@@parthisMC Multiply top and bottom by e^-x. Tgen the function becomes (e^-x)/(1+e^-x). After integrating, the following function will be obtained -ln(1+e^-x). Sub in the limits and the answer will be ln2
@@parthisMC Yes that's what i said i just didnt minus the -ln2 because there wasnt a problem with that part
I have never met someone who holds their pencil the same way as me 😭
this is the most chaotic video i've ever seen. from the pencil grip to the fire alarm battery.
Feels good when MIT questions seems doable when you are preparing for IIT..
They are actually very easy ..not even mains level ....
It's meant to be how fast you can solve it not whether u can solve it or not
yeah but these are very easy by MIT standards, almost every freshman can do this quiz perfectly.
It's freshman questions...
The books used in IIT are of foreign authors.... Talk about MIT.. They have worlds best minds... So higher level maths is very very hard... Even IIT prof fails to solve them...
@@jackyjack9660 bro honestly for indian students these are cake walk ...we don't need to even solve these many of the results we know directly..every question can be answered in 1 second..not kidding.
Anyone just got this recommended to them now
Whoa this is trippy, this was recommended to me about 2 hours ago too!!
Just before my Calc BC exam too!
Can do crazy math, but can't change the fire alarm battery.
Как я сюда попал😳
Никогда не видел, чтобы люди так держали ручку)
I'm more concerned about his pencil grip than the problem
How does this man only have 572 subscribers?
damn my short term memory so bad, I checked five times before writing-***594
Watching others to do calculus is just a way to cure my Insomnia.
same xD
i'm still in calc 2 rn and i've basically forgotten all the rules my final's next week 😭
surprised the lead didn't break, drilling that pencil hard into the paper
That's one of the most bizarre ways to hold a pencil I've ever seen.
you're smart, ill give you that.
but watching how you hold that pencil will give me nightmares tonight.
25:17 For number 5, im pretty sure that the (1/4) does affect the coefficient of arctan since the formula for arctan integration has (1/a) outside as a coefficient if the whole integral. I solved this on my own since it was the only one i was willing to attempt coz i cant even think about doing the others; either you forgot that you divided the whole bottom part of the integrand by two and have to then move that outside of the integral before integrating, or you forgot that the (1/4) which you added to the (x-(1/2))^2) had to be square rooted to get the integral to the form of (1/(u^2+a^2)) to then use inverse tan (the square root of (1/4) would be (1/2), which would then be set as a, (1/a)=2, and then divide the 2 by the two you pulled out from the integrand earlier. Hats off bro, what you are doing is crazy, as a AP Calc AB student, I can’t even comprehend the other problems.
yeah man, chemistry rocks! the best profession one can choose!
Chemistry is good and all, the most practical science easily, but it's not as intriguing as math and physics can be. Physics tries to answer questions about the very existence of things, math is a most beautiful structure arising out of baffling simplicity, whose "purity" is tainted every century or so when a field of science requires new tools. But chemistry... it's just chemistry, I guess.
Yeah mr white, yeah science!
6:18 it is actually 2ln2
On number 20 Im pretty sure you could just set u = x^4 and utilize the gamma function to obtain 1/4 * gamma(3/2) = sqrt(pi)/8
I love integration... Would've loved to do something like this at Georgia Tech.
fix yo fire alarm batteries
this video discourages me from taking engineering at as a higher education major
solution of problem 4,8 are wrong.. 8th is x*log(x^2+1)-2x+2arctan(x).. d(log(x^2+1))/dx=2x/(x^2+1)..he missed 2x in num..
For no.8 when u by parts you get to integrate 2x^2/(1+x^2) this will further become arc(tan).
You were actually missing a 2x in that part hence u integrated x/1+x^2 which came as logarithmic function.
Lmao this mf out here checking the answers
yes its true
Change the battery in you smoke detector! ...you must have the same one as me 😀
I’m on maximum render distance but I still fail to see how most of the people that work apply this everyday.
They don’t, it’s mostly for “”fun””.
why is bro holding his pencil like that
Huge respect!!
Just look at how he holds the pen, guy’s a nut
Question 2, for the limit
Let L be lim (x-ln(e^x+1))
e^L = e^lim (x-ln(e^x+1)) = lim (e^(x-ln(e^x+1))) = lim (e^x/e^(ln(e^x+1))) = lim (e^x/e^x+1) = 1
And e^L = 1 L = 0
Is this English
Don't forget the lower bound. It's 0 - [0 - ln(e⁰ + 1)] = ln(2)
@@gamerdio2503 yeah but I was just calculating the limit for infinity. Sorry I didn't mention this in my comment
The pen hold is giving me anxiety dude 😬😬😬😬
What a strange way to hold a pen. Just imagine how much quicker he'd be if he held it properly.
someone else tested it and apparently it's actually faster
They way you hold the pen Has nothing to do with speed and depends on the person.
Wtf has that pencil ever done to you?!?!
Integrate this: CSC^3 x dx.
Integrals get a bit hazy after the sophomore year doesn't it XD
him: ezy
me: the fuck
yeah i should probably retake calculus.
For #2 the antiderivative can be rewritten as log[e^x / (1+e^x)] which approaches log [1] as x goes to infinity.
remember that you have to plug in 0 in the integral, and then evaluate the limit approaching infinite oWo
ln(a/b) = -ln(b/a) because you can pull out an exponent of -1 from inside the ln()
so ln( e^x / (1+e^x) ) = -ln( 1 + e^-x)
evaluating at the boundary you get:
lim{t->infinity}[ -ln(1 + 1/t ) + ln(1+1/1) ] = -ln(1) + ln(2) = 0+ln(2) = ln(2)
so his guess was right
ok how he hold his pen like that tho
Omfg if you don’t fix that smoke alarm-
I cant ;_;
The smoke detector has got my attention 😵💫
And I thought S.D = sqrt(Σx^2-((Σx)^2/n)/n-1) was something
I didn't know that ln(x) and log(x) are the same, lol. I don't think this guy qualified.
In many disciplines, log without a specified base is ln(x). You may be confusing it with the more common convention that log(x) is implied to be log base 10
Lol this is not what I needed to see before my Calc BC exam.
10:42 Aren’t they all though? xd.
30:02 nasty indeed
me staring at the test and trying to use u = x 🙃🙃
тебя держат в заложниках?
Now i thing jee advance guyes have to make vedio about how fast they do.
I have a fear of JEE students ;-;
Что? Это экзамен в MIT? Советские школьники такое с пелёнок решали
bro I have no idea whats going on like im ngl
I don't even know wtf is even happening. Too much memorizing.
the title of this video gives me anxiety
do you do this mentally? how? sheer practice?
Its half mentally. When sheer practicing these kinds of integrals, I like to memorize the process of certain types of integrals so that when i see a similar type, i know what to do quickly.
@@Silver-cu5up where do you practice there intergral
@@zzzzzzzjsjyue2175 I just search for every Integration Bee past problems from any university
Were you also using a DnD character sheet to cover the test paper?
Yea lol
Didn't he get #8 wrong?
Number 20 can be solved with power series right?
nope cuz it goes from 0 to infinity, the power series is not defined for the upper bound
@@pinguino55h40 if you consider the integral as the limit of a definite integral then it should be, no?
@@replicaacliper that's the standard way of evaluating integrals in infinity but I wasn't referring to that but to the fact that power series are only defined for arguments between -1 and 1 exclusive. Thus substituting infinity wouldn't make any sense
@@pinguino55h40 the power series for e^x converges for all x
@@pinguino55h40 If you take power series based on 1/(1-x) the Best Friend then this can be written as Sum from n=1 to infinity of x^n where x lies between 1 and -1. Outside this interval the power series do not match 1/(1-x).
I may be too late, but I think the paper in right at the beginning is a D&D player sheet, am I right?
Yes xD
can i have the work sheet
i think it will be perfect study material for calc 1
no way people take that interpretation, robots out in na KEKW
TH-cam recommended me this lol.
In question 2, I don't think answer will be ln(2)
If it was e^-x, then I would have considered it.
It is ln(2)
26:21 It doesn't you're right
Jee aspirant : damm this is ncert level.
... 😂
y u gotta do this?
Yea no one cares
JEE ASPIRANTS....ASSEMBLE!😏
Lost me at the first one
For (2), isn't it 0 because ∞ - ln ∞ = ∞ - ∞ = 0. I may be wrong though.
you can't subtract infinities, it may be 0 in the end but you probably need to do some factoring and solve the limits
@@giorgostarnaras5658 You got a point. Is there a way we could use L'hipothals rule (I think that is how you spell it) because ∞ - ∞ is an indeterminate. Would this work? x-ln(x) = x(1-(ln(x)/x)) => lim x * lim (1-(ln(x)/x)) =>
lim x * lim ((x-ln(x))/x) => lim x * lim ((1-(1/x))/1) (L'hipothal rule thingy) => ∞ * [(1 - 0)/1] = ∞.
You are wrong
@@alexisrodriguez7127 Why though. I am sort of confused?
@@lohitmurali2137 because it is in indeterminate form and can not be evaluated. The reason is one infinite may be growing faster than the other one so we have no way of knowing the outcome if we try to subtract. In order to evaluate this limit we have to transform it into a indeterminate form that we do know how to handle so we can take advantage of l’hospitals rule (0/0 or infinity/infinity). This can be done a couple of ways and will be left as an exercise for the reader.
Почему в задании не указано омнование логарифмов типа (log) понятно что такое lnX или lna а какое ооснование подразумевается под log&
Log это по основание 10
log только по основание 10, lg все другие (от 0 до 1 и от 1 до бесконечности) и ln по основание е
В Америке log без указания основания часто означает, что речь идет о натуральном логарифме (у нас пишут Ln)
1:54 For number 2, when evaluating the upper limit, you can just view it as ln(e^x) becuase the +1 drops off as x -> infinity, so its x - x = 0
Integration chemistry mein kab aane lag gaya
Gawd i was so good at integration in high school
in the 2nd one inf+ln(2) is still infinity
It's not ∞ + ln(2) tho, it's ∞ - ∞
impressive, but how can you write like that? No offense intended. I hold my hand the other way around
I started writing like that since Kindergarten, and then I got too comfortable with it. In 3rd grade, my teacher tried to make me hold it normally, but my handwriting became hella awful like you couldnt even read it. It's just sloppy and a mess. So I just kept my formation to make my handwriting readable.
@@Silver-cu5up Ok! If it works for you, that's all that matters
How did you practice for something like this?
Solving integrals from other past problems in other universities really help me experience all kinds of integration tricks.
Quite sure that the only type of bra you know is algebra...
Nvm kidding but he has great integration skills!!!!!
Who the hell taught you how to hold a pen?
Look what great job he's doing and sincerely time is important in integration bee you have to write things super fast so that's why go in integration bee video you will see that everyone writing weirdly
@@aweebthatlovesmath4220 really? Isn’t holding a pen normally faster anyway? Btw nice username.
Easy qs for a Class 12th Indian I’d say
Dude. Learn how to hold the pencil properly. Watched less than a minute and turned off.
nice grip
It's not that hard most of the question I did in my head without using pen, specially in India any 12th graduate can do this. Keep learning good luck brother
Who asked
@@ecuas_7 it's YT not your home 🏠
We gotta find who tf asked
what the hell is that grip bruh , i swear that all white people hold their pens in this weird fashion , i would tell me frinds to wrap their fingers around the pen or pencil like that only to just squeeze it and paralyze their fingers
What’s infinity - infinity
indetermination, it depends because there are infinities of different magnitudes. You can use different approaches to obtain the final result, but infinity - infinity is not 0. Unless both infinities are the same i guess.
@@AlFredo-sx2yy yeah but what would I do for infinity - infinity in this integration problem
@@talhajat3301 give me the time stamp i dont know which one you are reffering to
@@AlFredo-sx2yy 1:37 is around when the problem becomes infinity-infinity, it’s q2
@@talhajat3301 ok i had skipped over that part, sorry, now i see what u mean, ok ok so the way i would solve that is this way:
you have X - ln(e^x +1) between 0 and infinity, that means that we just have to replace the X with infinity and then remove the same but replacing with 0 so we get
infinite - ln (e^infinite + 1) - [0 - ln(e^0 +1)]
so, lets focus on the second part:
0 - ln(e^0 +1) from here, since e^0 = 1, we get that this is equal to -ln(2)
now lets focus on the first part:
infinite - ln(e^infinite +1) in here, normally we would consider that e^X is an infinity of greater magnitude than just X, because it is, but the difference of ln(e^X) and ln(e^X +1) when X goes to infinity is so small that it leads us to have 2 infinities that are almost the same, like this:
Imagine we pick a high value like 9999 for X since we cant go exactly to infinite:
X = 9999
ln(e^X) = ln(e^9999) = 9999 (remember that ln is log of base e so to get the solution we ask the question "what number do i have to elevate e to get e^9999, so the answer is just 9999")
now, since X and ln(e^X) give us the same result, we no longer have the problem that e^X is a greater infinity than just X, because in out case we have an ln.
Now, we have ln(e^X + 1) but that isnt a problem since that +1 is not a big deal. If you use a calculator, for small values you'll see a difference.
For example try x=10 in ln(e^x +1) and you'll see how small the difference is already. Since we are going to infinity, they are basically the same number, so X - ln(e^X +1) when going to infinity are just going to cancel out because both infinities are of the same magnitude in this case, so we get this result, that X - ln(e^X +1) = 0
Now, having all of that into account, lets add it up to solve the problem:
infinite - ln (e^infinite + 1) - [0 - ln(e^0 +1)] = 0 - - ln(2) = ln(2)
So the final answer is ln(2)
Personally, i have never ever seen one of these in my life, but it was kinda easy i guess to figure out how it worked, because this exercise seems to have been built around the idea that you spot that both infinities are of the same magnitude.
*Side note: yeah they are not exactly the same magnitude of infinity, but as i stated before, you can already see how that +1 in the ln makes a real small difference already when using a value as small as 10. If we go to infinity you will see that the difference will be so small that it wont matter at all.
Bruh in india its taught in class 11th or at max they teach u this is class 12th and to be really honest we r taught more intense integration than this in class 12th
Average indian JEE flexer
Harder =/= better
Take this from ur fav Indian american
Why is he holding his pencil like a sped?
So hard
4:32
damn..
الطلاب السادس شلونكم 😂😂😂😭
Noice
MIT lolllllllllllllllll.Questions like this in greece are must before even go to the university.We solve them in highschool.So easyy.Wtf hahaha.U guys are silly asf
wow bro you’re so cool! You really showing them yankees how much intellectually superior you are! Want a cookie?
Are you in MIT?