⚠️ For anyone wondering why the integral was only half of the semi-circle - it’s due to the square root in the integral. Because of the square root, y cannot be negative! I should really have removed the bottom half of the semi-circle but I thought it was clear once I explained that we only needed the top half. Hope that clears up any questions!
You have to appreciate that there was no crazy integration technique involved, just pure mathematical reasoning. Understand the problem and use logical reasonings to get to the result. This is what a mathematician does! 10/10
The integretion with the circle got me , wow I thought Ellie was gonna use the trig substitution but wow , that was some gmreally good reasoning , thank you so much !!! I really liked this exercise and your explonation keep it up !!
The engineer in me went: "Oh, it’s one of those edgy problems? Is the solution pi?" Because requiring 10 or 5 digits means the solution has to have that many digits as well and nothing in the equation suggested division by 7 or 9 or the like, that result in those kinds of fractions. Then it’s edgy and the problem is probably about the equation and not the answer being complex.
@@ayuballena8217 To begin with, it's probably just a gag sticker for the maths lab, etc. and not a real password. It's like a Facebook post with "only 3 people in the world could solve this!1!" (it's not that bad but I hope you get my point): it's an equation with a simple solution crafted to look complex and scare off people with low maths skills.
Very briefly explained. I left BSc Math in 2004, but today i gone through such integral solution (by chance) and had tears in my eyes, how we used to solve these equations without any guidance and lack of solution materials notes etc, but internet has made it worldwide source of guidance. Your graphical representation very helpful for solution of such equations. (From Pakistan)
"Truly impressed by the way you used mathematical proofs to solve the integration question! Your logical approach and clear presentation made it easy to follow along. I particularly liked everything you did. Keep up the excellent work! Would love to learn more about your techniques"
Thank you so much Ellie. You are amazing! You break down complex problems into small-sized pieces that anyone can understand. Thanks for sharing your gift and for taking the time to do this. It takes a lot of effort and work to make these types of videos. THANK YOU. You are now my top UA-cam channel. Keep up the amazing work! Thank you again.
I’m surprised of how understandable this was, considering I never did integration. Good job Ellie, In particular I liked the speed of your video. It made it easier to understand your steps.
When I was going to the 9th grade, I was looking forward to learning these things, but I just learned them this year and I love pure mathematics.❤ Thank you Ellie
You simplified the second integral to a primary school formula.😊 As a beginner I evaluate the second integral which led to an arcsin function. After I substituted the boundaries of the integral, it gave me the same result you got but I spent on it a plenty of time😢
You know i saw X^3 multipled by cos and i screamed 0, Then I started to do a trig sub in my head, and you showed me a new simpflication i never though of. Great stuff.
I was a little confused as to why the area is only the top part, but its simply because the square root function means only the positive part of the function for anyone else wondering.
Looking at it quickly. I := [-2,2] is an interval that’s symmetric around 0. So integrals of odd functions [f(x)=-f(-x)] over I will evaluate to zero. Call the result of the integral Q. Q = integral over [-2,2] of q(x)dx, where q:=(x^3*cos(x/2)+1/2)*sqrt(4-x^2). q=(u*v+1/2)*w, where u(x):=x^3 is odd. v(x):=cos(x/2) is even. 1/2 is even. w(x):=sqrt(4-x^2) is even. for functions [even] * [even] = [even]. [odd] * [even] = [odd]. [odd] * [odd] = [even]. u*v*w is odd. - it cancels under the integral. (1/2)*w is even. - it remains. Q = integral over [-2,2] of (1/2)sqrt(4-x^2)dx. Quick change of variables {x=2s, dx=2ds} gives Q = 2*Integral over [-1,1] of sqrt(1-s^2)ds. The integral is the area of the half unit circle. Twice that is the area of the unit circle. Some might know it as π.
For some reason I remember that this was the first integral I was ever exposed to even before taking a calculus class. It doesn’t require very hard integration techniques and just requires logic.
Hey ellie I'm new to you're channel. Nice to meet you. I love mathematics. You're explanation is great. I'll go through you're other math videos and will look forward to more math videos from you :)
Beautifully explained. I'm in my 60's and not done any calculus since my early 20's. Great to have something explained without me losing the plot in less than a minute.
Thank you very much, now I can steal the neighbor's Wi-Fi here with peace of mind. It's brilliant to use the concept of parity to cancel or simplify the integral, I use it a lot in electromagnetism, but I confess I wouldn't know how to solve this integral without you.
You explained the odd and even function better than my math teacher could in a whole semester I appreciate but feel sad for my teacher being horrible at his job 🙃
@@PhalitSehgal with my limited experience of just high school math i notice we never write so neatly ever. "mathematicians are lazy" as one of my teachers said, we just wanna get to the end haha
I am currently starting year 11 maths and this stuff has always seemed super hard to learn but you taught it well so that even I can understand thank you also wondering how hard this question is compared to other
Great video but I really wondered what your approach had been to solve that integral of the semicircle. Obviously with these boundaries your approach was by far the simplest, but could you maybe show with which technique you would approach such an integral? I study medicine but find a lot of joy in doing math, so that would be awesome! I came across that sort of integrals a lot whilst practicing the length of a curve by using integration of line segments…
As a student who's preparing for JEE.... THe first thing i did was ignore the x cube cos x/2 term... odd function, integral with opp limits is 0... then it took 20 seconds.... Depresses me out that JEE turned me from a student who loved mathematics to a robot that calculates this in less than 30 secs
I saw this meme when i was like 11, obviously i had 0 clue what integration was, now I'm 17, preparing for jee, and even tho the question does look scary i considered giving it a try, and the question did trickle down after applying king's lol. Refreshing and nostalgic
I didn't watch the video, just saw the question and here's how I solved it in my head (btw dx is under square root, that's wrong, also first digits, what do you mean picture): 1. Whenever you see integral of f(x) from -a to +a, it can be transformed to integral of [f(x) + f(-x)] from 0 to +a. 2. Applying 1, you would get integral of sqrt(4-x^2) from 0 to 2. 3. You can solve this directly, or by a trig substitution, but also observe that if we set y = sqrt(4-x^2) -> x^2 + y^2 = 2^2, it's the equation of a circle. 4. In the circle, x goes from 0 to 2, and same for y, it also goes from 0 to 2. And we need to get integral of y dx. 5. That's just the area of the first quadrant of that circle. A = pi * 2 * 2 / 4 = pi 6. Alright so, pi is the answer. 7. Here's how you can remember upto 24 digits of pi: How I want a drink alcoholic of course, after the heavy lectures involving quantum mechanics. All of thy geometry, Herr Planck is fairly hard. (just count the no. of letters in each word)
I spotted the odd part straight away and realised the integral was the the area of a quadrant of a circle of radius 2. It would have been better if the password was the LAST 10 digits of the answer though.
any can please clean me out for I1 f(-x) = -f(x) f(x) = f(x) so when we integrate from 2 to -2 f(2) - f(-2) = f(2) - ( -f(2) ) = f(2) + f(2) = 2f(2) so how can I1 = 0 ???????????
Broooo….in integration we calculate area under the function i.e., area under f(x)..one area will be negative valued and other area will be positive valued…or other way is we sum the f(x) values….f(-2) +f(2)+f(1.99)+f(-1.99)….f(1)+f(-1)……(multiplied by infinitesimal x coordinate ) and the f(0) value of odd function is always zero…i think you got the ans.
here before watching this vid, the function could be expressed in form of an odd function and an standard integrand, The -2 to 2 would take care of the odd and we will get pi. Now watching how she does it.
Can some one explain to me why did she consider integrating from -2 to 2 as (1/2 of pi r ^2 )not just ( pi r^2) . What is the reason for including 1/2?
⚠️ For anyone wondering why the integral was only half of the semi-circle - it’s due to the square root in the integral. Because of the square root, y cannot be negative! I should really have removed the bottom half of the semi-circle but I thought it was clear once I explained that we only needed the top half. Hope that clears up any questions!
Yeah, that got me. Thanks for explaining it
Answer was 2π
U Made mistake at calculating the area of semicircle
You solving integrals are satisfying me, and i want to know more about them :O
@@gsrarmy3616the area of the semicircle is 2π, you’re forgetting the factor of ½ that was already on front of the integral
You have to appreciate that there was no crazy integration technique involved, just pure mathematical reasoning. Understand the problem and use logical reasonings to get to the result. This is what a mathematician does! 10/10
The integretion with the circle got me , wow I thought Ellie was gonna use the trig substitution but wow , that was some gmreally good reasoning , thank you so much !!! I really liked this exercise and your explonation keep it up !!
The engineer in me went: "Oh, it’s one of those edgy problems? Is the solution pi?"
Because requiring 10 or 5 digits means the solution has to have that many digits as well and nothing in the equation suggested division by 7 or 9 or the like, that result in those kinds of fractions.
Then it’s edgy and the problem is probably about the equation and not the answer being complex.
@@PH4RXlol same
@@PH4RXwdym “edgy”?
@@ayuballena8217 To begin with, it's probably just a gag sticker for the maths lab, etc. and not a real password.
It's like a Facebook post with "only 3 people in the world could solve this!1!" (it's not that bad but I hope you get my point): it's an equation with a simple solution crafted to look complex and scare off people with low maths skills.
I love that you used the semi circle, it brought back a long lost memory from Calc AB before I learned trig substitution. Great video!
I appreciate just how knowledgeable and cohesive you are to be able to hand guide me through this. takes a lot to simplify haha
Very briefly explained. I left BSc Math in 2004, but today i gone through such integral solution (by chance) and had tears in my eyes, how we used to solve these equations without any guidance and lack of solution materials notes etc, but internet has made it worldwide source of guidance. Your graphical representation very helpful for solution of such equations.
(From Pakistan)
"Truly impressed by the way you used mathematical proofs to solve the integration question! Your logical approach and clear presentation made it easy to follow along. I particularly liked everything you did. Keep up the excellent work! Would love to learn more about your techniques"
Thank you so much Ellie. You are amazing! You break down complex problems into small-sized pieces that anyone can understand. Thanks for sharing your gift and for taking the time to do this. It takes a lot of effort and work to make these types of videos. THANK YOU. You are now my top UA-cam channel. Keep up the amazing work! Thank you again.
I’m surprised of how understandable this was, considering I never did integration. Good job Ellie, In particular I liked the speed of your video. It made it easier to understand your steps.
Thanks sister now I can learn maths from you with free Wi-Fi 😊
When I was going to the 9th grade, I was looking forward to learning these things, but I just learned them this year and I love pure mathematics.❤
Thank you Ellie
Earned another sub and liked. A great researcher but also a great teacher. Explained it so well and with clarity.
This was a beautiful solution to the Free WiFi problem. I enjoyed watching this video a lot.
loved your way of explaining, its simple jet effective.
With the dx inside one can interpret it as fractional derivative. And the answer is (-4/(3Pi))*(8^(1/2)-1) in that case.
You simplified the second integral to a primary school formula.😊 As a beginner I evaluate the second integral which led to an arcsin function. After I substituted the boundaries of the integral, it gave me the same result you got but I spent on it a plenty of time😢
You know i saw X^3 multipled by cos and i screamed 0, Then I started to do a trig sub in my head, and you showed me a new simpflication i never though of. Great stuff.
I was a little confused as to why the area is only the top part, but its simply because the square root function means only the positive part of the function for anyone else wondering.
Ah yes, sorry i missed the explanation of that out! I’ll add a pinned comment for anyone else wondering ☺️
Both can be considered
y^2+x^2=r^2, x^2-r^2=y^2, y=±(x^2-r^2), so the positive bit is above the x axis
Thanks a lot. Now I can log in to the wifi.
Top tier solving. Process explained beautifully. Earned a sub.
Smart approach.
Love it, good explanation!
fun fact: if you have to solve an integral for wifi password, the answer is either pi or e. trust me
Even though not a maths student, I still watch your videos and I Love your accent, it's really very sweet🧡
I appreciate it for the effort you put on this, explained it so perfectly!
Taking area of circle is a great idea
Looking at it quickly.
I := [-2,2] is an interval that’s symmetric around 0.
So integrals of odd functions [f(x)=-f(-x)] over I will evaluate to zero.
Call the result of the integral Q.
Q = integral over [-2,2] of q(x)dx,
where q:=(x^3*cos(x/2)+1/2)*sqrt(4-x^2).
q=(u*v+1/2)*w, where
u(x):=x^3 is odd.
v(x):=cos(x/2) is even.
1/2 is even.
w(x):=sqrt(4-x^2) is even.
for functions
[even] * [even] = [even].
[odd] * [even] = [odd].
[odd] * [odd] = [even].
u*v*w is odd. - it cancels under the integral.
(1/2)*w is even. - it remains.
Q = integral over [-2,2] of (1/2)sqrt(4-x^2)dx.
Quick change of variables {x=2s, dx=2ds} gives
Q = 2*Integral over [-1,1] of sqrt(1-s^2)ds.
The integral is the area of the half unit circle. Twice that is the area of the unit circle.
Some might know it as π.
For some reason I remember that this was the first integral I was ever exposed to even before taking a calculus class. It doesn’t require very hard integration techniques and just requires logic.
Very nice perfformance Ellie
I love it so much , please continue
Hey ellie I'm new to you're channel. Nice to meet you. I love mathematics. You're explanation is great. I'll go through you're other math videos and will look forward to more math videos from you :)
Beautifully explained. I'm in my 60's and not done any calculus since my early 20's. Great to have something explained without me losing the plot in less than a minute.
Thank you very much, now I can steal the neighbor's Wi-Fi here with peace of mind. It's brilliant to use the concept of parity to cancel or simplify the integral, I use it a lot in electromagnetism, but I confess I wouldn't know how to solve this integral without you.
Your clarity and detail of explanation is really good! Very well explained. Something many mathematicians are not good at doing.
Talk about biomathematics. I want to complete a master's degree in it, but I am hesitant between it and industrial mathematics
Great sweater. Merry Christmas!
Still impressive of calculus 2 knowledge level. I am still self study calc 2 and it seems complex. Glad you solved it and good luck in your goal.
Excellent explanation. I wish you were my teacher 👍
You explained the odd and even function better than my math teacher could in a whole semester I appreciate but feel sad for my teacher being horrible at his job 🙃
That I2 looks simple but the way you solved is pure integral mathematics.
You are legend solving integral with using integral legend hats off 🥳🥳
your hand writing is too beautiful for you to be a mathematician! I also loved the explanation of the second half of the integral
aaah thank you! 🥺
Wtfdym "to be a mathematician" 😭😭😭😭😭💀🙏🙏🙏
@EllieSleightholm Thanks for sharing Ellie. I really hope you can respond to my other comment whenever you can. Happy Holidays
@@PhalitSehgal with my limited experience of just high school math i notice we never write so neatly ever. "mathematicians are lazy" as one of my teachers said, we just wanna get to the end haha
So, should we be less demanding about the handwriting of a mathematics teacher when compared to the other teachers? Why?
can u make a video on doing a STEP paper
Yes!!
Thought this would be a really tough one. But, surprisingly it's too easy.
Loved this video. ❤🎉
this was really elegant. wow
try to solve any JEE ADVANCE questions
After 1st step we can just use kings property in first and in the second one we can put x as 2 sin theta
I saw this problem as a joke before. They were asking for the last 10 digits of the solution for free wi-fi. Of course, it was a joke.
Thank you very much My teacher ❤
Took me most of the video to see the design of the jumper😂. Great video rolling back my undergraduate days !!
I definitely would have used trig substitution without thinking. Good catch about the half circle. Definitely an easier way than trig sub.
In Portugal x^2+y^2=r^2 isnt a circle but a circumference. To be a circle it has to be it has to be x^2+y^2
I am currently starting year 11 maths and this stuff has always seemed super hard to learn but you taught it well so that even I can understand thank you also wondering how hard this question is compared to other
I hope you will guide us, I am interested in Physics and Maths, so which should I do Masters in?
Great video but I really wondered what your approach had been to solve that integral of the semicircle. Obviously with these boundaries your approach was by far the simplest, but could you maybe show with which technique you would approach such an integral? I study medicine but find a lot of joy in doing math, so that would be awesome! I came across that sort of integrals a lot whilst practicing the length of a curve by using integration of line segments…
Because of your this video, I could get a mcq correct in my exam today
Good work ellie mam .
The smartest brain I have ever seen ..❤
We gettin free wifi with this one 🔥🔥🔥
can you try and solve some math olympiad questions ? I do them myself and they're so fun to do and watch
Yes!! I’ve been planning a few 😁
As a student who's preparing for JEE.... THe first thing i did was ignore the x cube cos x/2 term... odd function, integral with opp limits is 0... then it took 20 seconds.... Depresses me out that JEE turned me from a student who loved mathematics to a robot that calculates this in less than 30 secs
I just use Matlab Symbolic solver; that way the answer is quick, never any errors and my head doesn't hurt...
woah she's really good at match yoo
I saw this meme when i was like 11, obviously i had 0 clue what integration was, now I'm 17, preparing for jee, and even tho the question does look scary i considered giving it a try, and the question did trickle down after applying king's lol. Refreshing and nostalgic
Optical Illusion in your sweater 😮
Beautiful explanation, your follower is an Iraqi Arab
Pi by inspection. Odd part vanishes. Rest is 1/4 the area of a circle of radius 2 = pi.
Although I don't understand anything, but these are beautiful
Can you make a vedio on cube root of unity
Thanks professor 😊😊😊😊
UR explaining complex MATH problems simple)
I didn't watch the video, just saw the question and here's how I solved it in my head (btw dx is under square root, that's wrong, also first digits, what do you mean picture):
1. Whenever you see integral of f(x) from -a to +a, it can be transformed to integral of [f(x) + f(-x)] from 0 to +a.
2. Applying 1, you would get integral of sqrt(4-x^2) from 0 to 2.
3. You can solve this directly, or by a trig substitution, but also observe that if we set y = sqrt(4-x^2) -> x^2 + y^2 = 2^2, it's the equation of a circle.
4. In the circle, x goes from 0 to 2, and same for y, it also goes from 0 to 2. And we need to get integral of y dx.
5. That's just the area of the first quadrant of that circle. A = pi * 2 * 2 / 4 = pi
6. Alright so, pi is the answer.
7. Here's how you can remember upto 24 digits of pi: How I want a drink alcoholic of course, after the heavy lectures involving quantum mechanics. All of thy geometry, Herr Planck is fairly hard.
(just count the no. of letters in each word)
It is easy to see that the result is pi, but I still don't know what the password is. 314? 3.14? 3.1415? How many digits? With or without period?
hp-prime spits out the answer ‘not quite π’.
You do not even need a pen and paper to solve that integral. Everybody would have got free wifi😂😂
This is really funny that instead of giving free wifi, they ask the people to seek to get wifi to solve an integral
Thank you for the video.
hi ellie, can you create math formula for storage any number then recall that number also with math formula? your videos was awesome. i subscribed 😊🎉❤
I spotted the odd part straight away and realised the integral was the the area of a quadrant of a circle of radius 2. It would have been better if the password was the LAST 10 digits of the answer though.
Great video! Thank you!
Great teacher
Nice lernt now thing on how to use analysis and logic Thnk u
great video👍
Tank you so much
Very informative videos
Very unique explaination ,
In this circle example, why does the circle integral not cancel out (nvm I realized as I asked the question)
Since when angles started solving integrals
Do you add constant at the end of answer??
any can please clean me out for I1
f(-x) = -f(x)
f(x) = f(x)
so when we integrate from 2 to -2
f(2) - f(-2) = f(2) - ( -f(2) ) = f(2) + f(2) = 2f(2)
so how can I1 = 0 ???????????
Broooo….in integration we calculate area under the function i.e., area under f(x)..one area will be negative valued and other area will be positive valued…or other way is we sum the f(x) values….f(-2) +f(2)+f(1.99)+f(-1.99)….f(1)+f(-1)……(multiplied by infinitesimal x coordinate ) and the f(0) value of odd function is always zero…i think you got the ans.
the way i called it being pi from the beginning
here before watching this vid, the function could be expressed in form of an odd function and an standard integrand, The -2 to 2 would take care of the odd and we will get pi.
Now watching how she does it.
That's a great way to solve it
Can some one explain to me why did she consider integrating from -2 to 2 as (1/2 of pi r ^2 )not just ( pi r^2) . What is the reason for including 1/2?
I should do masters in math to get free wifi 😂😂
Where did you get your sweater?
Why you do not apply odd & even function concept in 2nd step, i mean for I2?
Answer plz
Nice. Thank you.
couldn't we use the formula integra(sqrt(a^2 - x^2)) formula there for i2
Well done