@@diondredunigan5282 no just differential. how it works here (or at my high school at least) is we have the 1 highest level math course in the last year of high school, and that single course is split between differential calculus and an intro to linear algebra
Woahhhhh! These sums are so cool ! The standard of integration bee has gone to another level in the recent years !! Wow ! Thank you for uploading this video sir, there are very less videos of integration bee on TH-cam.
Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023" You can simply find it!
That skill is called integration. Essentially, you can imagine a line on a graph, and they use calculus skills to find the area between the line and the y=0 and between the x values given on that giant swirly S. The line on the graph is defined by the equation after that big S
Explanation for Tiebreaker: denote by I_k the integral of (lnx)^k with these limits (we will insert k=2020 at the end) plug in u=log(x) the integral transforms to -infinity to 0 of u^k*e^u integrate by parts differentiating the u^k part, you will get a similar integral but a factor of k comes out and the power decreases to k-1, this gives us a recursion formula: I_k=-k*I_(k-1), I_1=-1 can be checked and thus, I_k=(-1)^k*k! so I_2020=2020!
This is kinda hard for me to understand but what do I know I’m only in calc 1, kudos to you man, u should put this on katex so it’s easier to read, really appreciate your work in making this more easy to understand
@@Teracosa it's a u-substitution that changes the expression and bounds u=log(x) e^u=x e^u du = dx log(0)=-∞ log(1)=0 So then log(x)^2020 dx from 0 to 1 becomes log(e^u)^2020 e^u du = u^2020 e^u du from -∞ to 0
@@xinpingdonohoe3978 I don't understand why log(0) = -infinity and instead is not undefined. Is it undefined as it heads toward infinity? Would log(-1) = -infinity as well?
takes too long because water film sticks on the boards which makes them "unusable" for as long as the water dries, it will chocke out the chalk making whatever you write extremely thin and hard too see
With that kind of blackboard, and that angle. I don't think It'll be easy to read what the other guy wrote. Edit: by blackboard, I mean, dusty blackboard, full of chalk dust
But I think students at top American universities are the best in the world. A disproportionate share of the top innovators of the world come from like 10 American universities.
You just need to apply integration by parts, take u=1 and v=log^2020(x) One can generalise this as If I(n) = integration from 0 to 1 log^n(x) then I(n)= -n I(n-1) hence I(n) = (-1)^n n! .
@@jr13763 Neither contestant had seen the problem before. They guessed that given the complicated form of the functions resembling inverses of each other that integral had to be the area of this elbow shape: math.stackexchange.com/questions/3732846/uc-berkeley-integral-problem-show-that-int-02-pi-frac-min-sin-x-cos-x
It is a trick ,the given integral contains the sum of a function and its inverse.Such a sum when evaluated between the point of intersection of both the functions equals twice the area of the trapezium formed by (1,1) ,(2,2) ,(1,0) and (2,0) .
Bruh... Did u even see what that indian(on left) was doing?? He had no idea of wtf is to be done... In Q1 he is making a hopeless triangle that we learn in 10th grade and in Q4:wtf is he even trying to do with those random graphs that he made. U can easily hear ppl making fun of him for trying to act like a smarty pants... This is what makes me feel bad about Indians(mind u NOT ALL) ,empty vessels that make alot of noise. I'm sry to day this.. I myself am an Indian but not proud of how ppl behave as Indians.
@@farhaanshaikh2873 dude, at that point if all of them are in MIT, its sufficient to say that most would've been able to solve, given enough time. But what matters is the ability to remain calm in such pressure and time crunch, AND the sudden striking power when the solution just comes to your mind
Itay Cohen to generalise this problem let a be a constant, e(x) be an even function and o(x) be an odd function. The integral of [e(x)]/{1+a^[o(x)]} dx from -b to b equals the integral of e(x) dx from 0 to b. To prove this substitute u=-x into the original integral, simplify by using the properties of even and odd functions, multiply top and bottom by a^[o(u)], replace u with x (it’s a dummy variable), then add this to the original integral.
Call the integral A. Do a substitution x -> -x in the given form of A. This leads to an almost identical form for A, except the exponent in the denominator picks up a minus sign. Now add these two nearly identical expressions for A to get 2A = integral from -infty to infty of sum of two nearly identical fractions. Now, if you factor out 2020^(-|x|) from the fractions, it turns out that the remaining two nearly identical fractions sum to unity (exactly one). You can check this by simply forming a common denominator and directly adding them. In general, the following identity holds: 1/(1 + p) + 1/(1 + p^{-1}) = 1
What he's done in question 1, first he replaces x to -x then add the two integral leaving behind only upper part of the integral and by symmetry he then turns down the limits to 0 to infinite and cancel the 2 both side, and then he just left with one integral which is of the type a^-x with limits 0 to infinite where "a" is 2020 and by logarithm method he got the answer
Nope, it just looked simple, it was pretty long, if u would do it in a legit way atleast 2 pages will be filled. That guy just guessed it, coz there was a pattern
@@kelvinella The tiebreaker takes 2 seconds. Simply recognize the given integral as the definition of (-1)^n Gamma(n+1) for n = 2020. (partly joking lol)
Lmao "there's still two more questions headass" - 5:20
LMAO
“Best out of 5 and he has 2, that means he won!!!” The real math genius in this video 5:20
he thought the one on the right won
The one on the right won it all. Hurt so much
the cameraman and his friends are the embodiment of my reaction to this video
I’ve only taken high school differential calculus and I’m actually so excited to (hopefully) one day be able to do stuff like this 👍
Did your course not teach integral calculus also?
@@diondredunigan5282 no just differential. how it works here (or at my high school at least) is we have the 1 highest level math course in the last year of high school, and that single course is split between differential calculus and an intro to linear algebra
@INERT same here in Australia we have to do differential calculus, integration, differential equations and vectors
@@mangomanlassi7779well your integrals are way easier than this tho
Woahhhhh! These sums are so cool ! The standard of integration bee has gone to another level in the recent years !! Wow !
Thank you for uploading this video sir, there are very less videos of integration bee on TH-cam.
Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023"
You can simply find it!
i legit dont know wtf they doing,but m happy for them tho.
solving integrals
Was a party!🧐
That skill is called integration. Essentially, you can imagine a line on a graph, and they use calculus skills to find the area between the line and the y=0 and between the x values given on that giant swirly S. The line on the graph is defined by the equation after that big S
@@jacksoncasper8428 I really don't know all that. Idk if i'm dumb or my school is bad. I'm a highschool student btw.
@@jaranis9273 it's part of calculus, which you don't have to take in high school i think
The integral of x squared dx from zero to one >>> having solved this in one hour made me feel like I'm a genius.
Crikey
Those dusters are hopeless
Explanation for Tiebreaker:
denote by I_k the integral of (lnx)^k with these limits (we will insert k=2020 at the end)
plug in u=log(x) the integral transforms to -infinity to 0 of u^k*e^u
integrate by parts differentiating the u^k part, you will get a similar integral but a factor of k comes out and the power decreases to k-1, this gives us a recursion formula: I_k=-k*I_(k-1), I_1=-1 can be checked
and thus, I_k=(-1)^k*k! so I_2020=2020!
This is kinda hard for me to understand but what do I know I’m only in calc 1, kudos to you man, u should put this on katex so it’s easier to read, really appreciate your work in making this more easy to understand
Gamma function
Why does substituting u=log(x) entail that transformation?
@@Teracosa it's a u-substitution that changes the expression and bounds
u=log(x)
e^u=x
e^u du = dx
log(0)=-∞
log(1)=0
So then log(x)^2020 dx from 0 to 1 becomes log(e^u)^2020 e^u du = u^2020 e^u du from -∞ to 0
@@xinpingdonohoe3978 I don't understand why log(0) = -infinity and instead is not undefined. Is it undefined as it heads toward infinity? Would log(-1) = -infinity as well?
Wow thanks for 100k views everyone...honestly didn't expect this to be so popular!
head butt
Really hope one day I’ll participate this seems like so much fun!
Seems like the definition of "fun" varies a lot between people...
Funny :)
@@iliveinyourwalls5193 definitely
It is so painful to watch people tryna write on those hella dirty blackboards. Why dont they just clean them with wet towels and dry them afterwards?
takes too long because water film sticks on the boards which makes them "unusable" for as long as the water dries, it will chocke out the chalk making whatever you write extremely thin and hard too see
@@FF-pv7htwhy not just do it at the end of the day then?
They should have digital boards at Berkeley, but they like to live in 1970's
@@yinvara what tells you they don’t?
@@martaecala Nothing. Yet I can't assume that they do, especially given their condition in the video
I feel slightly more confident knowing that I also got 1/2, and it was correct. So much integrating by parts, but worth it.
they spend the 3 first minutes just on writing the integral expression :')
En sí el ejercicio es largo y algo molesto de escribir, pero tiene un truco para desarrollarlo rápidamente
Ok this was funny so I am adding it
Jee Aspirants: Ye me kar leta hun aap jake dream 11 par team bana lo
lol you are right
Those two guys were really good and we had an exciting finish.
holy dorkfest ... getting your math fix ... and all this to earn a cap ... that's so rewarding
Some people don't care about money as much and put their education on top because they find it fun.
all i see are are near sighted people competing to see who's glasses are better.
Amazing! This looks so fun, I’m going on to uni later this year, hopefully they have stuff like this where I’m going
Did they have stuff like that?
How' s uni going now?
Legends say that guy in a gray shirt still walks back and forth
The first question isn't that difficult once you break it up into smaller integrals
I passed my calculus subject but I dunno if i can solve what are they solving XD
They're allowed to LOOK at their competitor's board???!?!?!??! Why??????
With that kind of blackboard, and that angle.
I don't think It'll be easy to read what the other guy wrote.
Edit: by blackboard, I mean, dusty blackboard, full of chalk dust
20:15 i love how this guy just holds up his brown bag XD
Excelentísimo, por el vídeo cómo en mis tiempos. SALUDOS,!!!!!!
Estoy eguro de no ser el unico que está aquí por cosa de broncano
now I really wanna see what problem 2 was
@ThePhysicistCuber Hello! I wrote the problems for this bee. The second problem in the finals round was x^2/(x^4-x^2+1) over R
And the answer is ( 1/4(3)½ )ln|(x²-root3 x+1)/(x²+root(3)+1)| +(1/2).arctan(x-1/x) ?
And most important +C
Me: still struggling to solve first linear degree
I can't see the question clearly 😭
We studied mathematics in Iraqi universities at a top level..
Didn’t ask
But I think students at top American universities are the best in the world. A disproportionate share of the top innovators of the world come from like 10 American universities.
@@historyandculture562 Brain drain and picking Americans for innovation in priority thus they have more funds and a higher advantage.
Before watch this video: oh, I'm interested in integral. I want to see this video.
When I see first problem: ?
What was the answer to problem 3??
these make the ap calc exam look like preschool
I can smell the chalk through my screen lol
can anyone explain how did he solve the last math?
You just need to apply integration by parts, take u=1 and v=log^2020(x)
One can generalise this as
If I(n) = integration from 0 to 1 log^n(x) then I(n)= -n I(n-1)
hence I(n) = (-1)^n n! .
@@adarshrajshrivastava1113 How do I even get started on learning all of this?
@@David-f9z8e start small, learn ap calc, and then start doing more harder problems
How can i Take part of this class
dang... that integral sure is a monster
That black board is literally white
If you don't wear glasses, you haven't got a chance.
Indians are everywhere. Omni present👏
Oh wow my favorite professor in the background :o
How can you do that problem 3 in like one second in your head??
They probably solved that problem before and memorized the answer to it.
@@jr13763 Neither contestant had seen the problem before. They guessed that given the complicated form of the functions resembling inverses of each other that integral had to be the area of this elbow shape: math.stackexchange.com/questions/3732846/uc-berkeley-integral-problem-show-that-int-02-pi-frac-min-sin-x-cos-x
It is a trick ,the given integral contains the sum of a function and its inverse.Such a sum when evaluated between the point of intersection of both the functions equals twice the area of the trapezium formed by (1,1) ,(2,2) ,(1,0) and (2,0) .
@@jr13763 its not JEE question paper or CBSE boards paper that you memorize and get full marks
dude
most of the participants are indian.
something as an indian i am extremely proud of...
hope one day ill participate...
Bruh... Did u even see what that indian(on left) was doing?? He had no idea of wtf is to be done... In Q1 he is making a hopeless triangle that we learn in 10th grade and in Q4:wtf is he even trying to do with those random graphs that he made. U can easily hear ppl making fun of him for trying to act like a smarty pants... This is what makes me feel bad about Indians(mind u NOT ALL) ,empty vessels that make alot of noise. I'm sry to day this.. I myself am an Indian but not proud of how ppl behave as Indians.
@@farhaanshaikh2873 dude, at that point if all of them are in MIT, its sufficient to say that most would've been able to solve, given enough time. But what matters is the ability to remain calm in such pressure and time crunch, AND the sudden striking power when the solution just comes to your mind
@1 2 same
@1 2 as you should lol!
@1 2 lmao
who won the point for Qn 3?
Can someone explain how did he solve the first one ?
Itay Cohen to generalise this problem let a be a constant, e(x) be an even function and o(x) be an odd function. The integral of [e(x)]/{1+a^[o(x)]} dx from -b to b equals the integral of e(x) dx from 0 to b. To prove this substitute u=-x into the original integral, simplify by using the properties of even and odd functions, multiply top and bottom by a^[o(u)], replace u with x (it’s a dummy variable), then add this to the original integral.
@@calcul8er205 of course. How did I not see that 🥴
@@MrCrackheadst im surprised u didnt see that, its trivial :v
Solutions?
Is this math department?
How old are you guys?
wat is that first integral!!!!
I love how at the last the winner was showing the method
Pie are not square, pie are round. Cornbread are square.
you could have a square pie if you dont like them round ones
Everything is possible when it comes to pies
where does the denominator go for q1?
Call the integral A. Do a substitution x -> -x in the given form of A. This leads to an almost identical form for A, except the exponent in the denominator picks up a minus sign. Now add these two nearly identical expressions for A to get 2A = integral from -infty to infty of sum of two nearly identical fractions. Now, if you factor out 2020^(-|x|) from the fractions, it turns out that the remaining two nearly identical fractions sum to unity (exactly one). You can check this by simply forming a common denominator and directly adding them.
In general, the following identity holds: 1/(1 + p) + 1/(1 + p^{-1}) = 1
@@neelmodi5791 thank u so much!
What he's done in question 1, first he replaces x to -x then add the two integral leaving behind only upper part of the integral and by symmetry he then turns down the limits to 0 to infinite and cancel the 2 both side, and then he just left with one integral which is of the type a^-x with limits 0 to infinite where "a" is 2020 and by logarithm method he got the answer
I also got an answer for 5 question, in this ques we have to perform integration by parts 2 times and then Laplace transform
Sixth question is on the gamma function, so it's easy
4:58 WTF 😳
sheeeeeesh
shout out mr brookner he taught me abstract algebra
love the commentary
i dont even know a single thing they did ;-;
That's ok most of the people in the audience didn't know either.
Few no jee comparison comments
jee aint hasd as this one
Mit bee wale me bhar bhar ke the…
All 5 were easy
Like si vienes por EL PACHANGAS
Tiebreaker was a pretty easy one!!
Nope, it just looked simple, it was pretty long, if u would do it in a legit way atleast 2 pages will be filled. That guy just guessed it, coz there was a pattern
@@rishipoonia7374 but why is there a need for that.?cuz after writing first few ones it just becomes obvious ,and it's meant to be solved fast as well
the problem is too high
Phenomenal work
You couldn’t print the question for them? That’s idiotic. The other kid could see properly and wasted a lot of time.
the tie breaker is the easiest LOL, done in like 15 seconds
i dont think so
@@frosty8655 let -t=log(x), then it becomes gamma(2021). simple
@@kelvinella The tiebreaker takes 2 seconds. Simply recognize the given integral as the definition of (-1)^n Gamma(n+1) for n = 2020. (partly joking lol)
FGNKFGK WHY IS IT SO FUN
He is Indian
Solo quiero ser uno de ellos
Mannnnnn i love this video!
Yo hice mi primera integral a mis 2 años
Indians are rocking everywhere 🤘🤘🤘
برافو عليك يااخي👍👍
Como eu vim parar aqui kkkk
indians and asians need another category is not fair
which semester are these students in ?
They were both second semester freshmen at berkeley at this point in time
@@VedantKumud23 for 2nd semester that's hardcore....
Bro all these questions are mid compared to jee advanced
After covid teachers are looking to bring there students back to the classrooms. 🤔🤷🏽♂️✌🏽👍🏻🎓🎉
Competition Competition Everywhere 😭😭
Whats the trick behind the 3rd one?
smth to do with reflecting/creating a rectangle whose area is easy to solve for iirc
check here: math.stackexchange.com/questions/3630192/how-can-we-evaluate-this-integral-from-the-2020-uc-berkeley-integration-bee
@5:57 biomajor -> DNA🤣🤣
I solved in 2 mins and ans is 0
How the hell did the white guy beat all those asians?
Hadyn's a smart guy
Exactly!!!
He beat them cause he’s smart. What a stupid comment, that’s honestly kinda racist
Wow! I have no idea how he solved the first integral.
yup..
I only see Indians and Chinese
go bears
4:55
Let's go rohit gang
So this is what needed to survive on earth 👍
If u say headass in college
All I see is glasses 😂😂😂
5:55 making fun of bio majors lol
That smart guy seems to be Indian
the best of the best
Why do you think he is Indian? Are Indians very good mathematicians?
So it's all about science 😅
I could solve only 3 of these
nice question
Any 3 minute can solve it in indian jees.
IIT aspirants be like:- We solved this when we were in our school right?
Ghanta
nope
Frickin mathematicians...
무슨 대한민국 이과 고삼이 푸는문제를 신나게 풀고있노
4:58 it's easy , any good JEE aspirants can do it easily:))
@@Salmanul_ exactly
@@6Scarfy99 answer is 2?
Help,explain please
Ima memorize it