Actually, this is closer to 'solving an integral' than most cases. In most instances we use the Fundamental Theorem of Calculus to do it indirectly. In this case, ok it's not from first principles, but it's at least calculating the area under the graph as the integral is defined. Though even then you could say it's not really because the way you derive the formula for the area of the circle in the first place precisely to use the FTC to integrate sqrt(R^2-x^2). So it's closer but it's not lol.
TLDR; breaking up the integral by parts you get two integrals. The left hand function is an ODD function, so ANY definite integral with limits that are symmetrical around zero will be zero because the negative and positive areas will cancel out. So this is really just an integral of a quadratic, which amounts to the area of a semicircle, and therefore works out to pi.
@@enverko He/she probably means that there is a difference between a decimal point and minutes with seconds, so the video is more like 6.5 minutes long.
I really like the area approach to the second integral, it makes it so easy to see. I was immediately doing a trigonometric substitution in my head, which works fine, but this is so much easier. Cool!
As soon as one student has the answer everybody will be asking them what the password is, and if they doesn't share it, they'd be bullied and/or excluded. That's what would happen in Latin America.
The former integral is harder to solve without using the fact that it is an odd function But the second one has a formula derived from "by parts" method of integration §sqrt(a²-x²)dx = (x/2)(sqrt(a²-x²) + (a²/2)(arcsin(x/a)) + C
@@anuraagrapaka2385 you can solve it with trings too just let the x = 2 sin θ and its pretty much done just do some work with supstitution and there you go
I solved it without using any of those , by the stuff that I learned in srilankan a/l maths syllabus the trick is you have to use the [ a to b f(x) = a to b f(a+b-x) ] transformation for the one with cos x/2 and for the root 4-x^2 thing use x = 2sinu substitution. Both of which are covered in calculus 2 we had to learn them separately though its just a matter of connecting dots.
Hi blackpenredpen!! Two questions? 1) How should be solved an indefinite version of that integral? 2) Would it be possible to define a total order on the complex plane using a hilbert curve?
x³×√4-x²≡√4x⁶-x¹² therefore the first integral becomes ∫√4x⁶- x¹²×⟨cos(x/2)⟩dx which can be easily be solved using formula FIRST F(X)×∫SECOND F(X)dx-∫⟨d/dx(FIRST F(x)×∫SECOND F(x)dx⟩dx and in this case first function should be √4x⁶-x¹² and second should be cos(x/2)
As an additional note, there was a sign posted at Xinyuan Library Restaurant to name notes in a line of music to access the Wi-Fi. The notes made up Twinkle Twinkle Little Star. Now I'm guessing that's slightly easier than ^^^...
Be careful. The next time the Wi-Fi password becomes an answer to an integral in a restaurant, someone will have already devoured it (it is a restaurant). Be grateful for your integrals as always, folks!
@@greogryhouse8341 You gotta start from the circle equation, r^2=x^2+y^2 (if it's centered in the origin as in the video). Then you separate y and square root everything to obtain a function of x (y=sqrt(r^2-x^2))Since a function can have only one value of y for every value of x the function can only describe the upper or lower semicircle. Lower has a -sign in front of the square root and upper has not. If you're as familiar as he is with maths you don't need demos to visualize the approx "shape" of functions. I didn't recognize it at first but when he mentiomed it I realized. Though I agree, a demo for us would not have hurt.
I recommend using trig substitution, it's usually way faster than integration by parts for things like this. If you haven't heard of it, I recommend looking into. Here you would basically just let x = 2sin(theta)
I guessed it would be pi, but I think I would try to make my password some form of k*n where k is an integer constant and n is some transcendental number.
Yes. And by the way, the circle equation is the Pythagorean theorem a² + b² = c². The reason why they're related is because of how the theorem works: The square of the two catheti's length (short sides, here a and b) of a rectangular triangle equals the square of the hypothenuse's length (long side, c). That means, you can draw a rectangular triangle inside the circle whose hypothenuse goes from the center to the edge of said circle and the catheti are parallel to the x- and y-axis, respectively.
When i solve the integral of 0.5(4-x^2)^0.5 i will get 2/3x(4-x^2)^1.5 Which results in 0 between -2 and 2. (because of the 4 minus x squared, which is for both numbers 0) Why am i not getting pi?
Sorry but your accent is a bit unique ,it would be easier for students like me to understand if you added English subtitles . Thanks for the amazing video
Since 1 + 1 is such a difficult mathematical expression to evaluate, you're going to have to use taylor expansion series to find an approximation. Since ln e = 1, then 1 + 1 = ln e + ln e = ln e^2 by logarithm properties. Now use the taylor expansion for ln(1-x) and e^x to approximate the result to about 2.
You don't. f(x) = sqrt(a² - x²) is the equation for the *graph* (i.e. the outer edge) of a semi-circle. It's its *integral* which allows you to calculate the area, not the function itself.
Sometimes you can solve an integral without solving a single integral.
Actually, this is closer to 'solving an integral' than most cases. In most instances we use the Fundamental Theorem of Calculus to do it indirectly. In this case, ok it's not from first principles, but it's at least calculating the area under the graph as the integral is defined. Though even then you could say it's not really because the way you derive the formula for the area of the circle in the first place precisely to use the FTC to integrate sqrt(R^2-x^2). So it's closer but it's not lol.
A koan
Wise words from a wise person.
😂
TLDR; breaking up the integral by parts you get two integrals. The left hand function is an ODD function, so ANY definite integral with limits that are symmetrical around zero will be zero because the negative and positive areas will cancel out. So this is really just an integral of a quadratic, which amounts to the area of a semicircle, and therefore works out to pi.
This video is 2π minutes long 😎
Not really... but I get what ur saying
@@budtastic1224 ?
@@enverko He/she probably means that there is a difference between a decimal point and minutes with seconds, so the video is more like 6.5 minutes long.
@@sethgrasse9082 ruining the fun lol
EDIT: How about this, the time stamp at the end is 2π
haHAA
#yay
I will use same strategy for everyone who asks for hotspot 😂😂
How?
Nice video, blackpenredpenandsometimesblue :)
Ian Santee : )
Why you should learn maths: for wifi
Oon Han yup!!!!
I really like the area approach to the second integral, it makes it so easy to see. I was immediately doing a trigonometric substitution in my head, which works fine, but this is so much easier. Cool!
Those pen skills doe
Slaf thanks
If they did that in America, everyone would be really offended
I hope they do it in america
@@SirCobra229 I want that too. Watching people getting triggered is my enjoyment
As soon as one student has the answer everybody will be asking them what the password is, and if they doesn't share it, they'd be bullied and/or excluded. That's what would happen in Latin America.
@@johnq4841 people here in America are just really bad at it.
Anything offends people in america
That WiFi password thingy lol!
I will now use the same thing if people ask me for hotspot.
Careful, they may change the password after this
😂
Can you make a video solving this without odd functions & semicircles
The former integral is harder to solve without using the fact that it is an odd function
But the second one has a formula derived from "by parts" method of integration
§sqrt(a²-x²)dx = (x/2)(sqrt(a²-x²) + (a²/2)(arcsin(x/a)) + C
@@anuraagrapaka2385 you can solve it with trings too just let the x = 2 sin θ and its pretty much done just do some work with supstitution and there you go
I solved it without using any of those , by the stuff that I learned in srilankan a/l maths syllabus the trick is you have to use the [ a to b f(x) = a to b f(a+b-x) ] transformation for the one with cos x/2 and for the root 4-x^2 thing use x = 2sinu substitution. Both of which are covered in calculus 2 we had to learn them separately though its just a matter of connecting dots.
Length is 6:28 witch is roughly 2pi
Marcin Łukaszyk coincidence ???
Don't think so.
One does not upload a tau-long video by coïncidence.
Actually 2π >3
@@giovanniferreira4699 I hope that u get what is it
Don't you mean 6.47 mins long?
Next time I see a password locked by an integral, I'm just gonna guess pi, tau and e, in that order. (I prefer tau over pi)
I see. Thats why when I was in the cafeteria and I ordered a pie the clerk said " Enjoy free unlimited internet access "
I'm sure 97% people in the world will get triggered when they find out that the answer to this "huge" Intergral is π.
That doraemon's music at the start tho.
This wasn't hard at all since the limits were so convenient, solving the indefinite integral of this would be difficult
Hi blackpenredpen!! Two questions? 1) How should be solved an indefinite version of that integral?
2) Would it be possible to define a total order on the complex plane using a hilbert curve?
The first part would be zero, the second will just be area of circle /2.
x³×√4-x²≡√4x⁶-x¹²
therefore the first integral becomes
∫√4x⁶- x¹²×⟨cos(x/2)⟩dx
which can be easily be solved using formula
FIRST F(X)×∫SECOND F(X)dx-∫⟨d/dx(FIRST F(x)×∫SECOND F(x)dx⟩dx
and in this case first function should be √4x⁶-x¹² and second should be cos(x/2)
Love the midnight post vids 😄😄☺☺
3:40 Wow, I totally forgot that :D
: )
And then they don't even get to access the actual internet.
That's the evilest thing I can imagine 😆😂
Ricardo Rodríguez yea well that’s how it is for everyone in China
This video was fun and the integral was brilliantly solved
No trig sub, just pure and simple maths
Thank-you
Splendid video!! Thanks!!
The second part of the integral would be 4π ×1/2
Because the limits are from -2 to 2
I think it would be faster to just input the first 6 digits of pi when asked such a question by a university :)
I admire your math skills!
Students be like : No thanks, I will use my mobile data 😂
This is just good use of logic... i wudv been losttt and just do actual integrals which wudv taken x5 longer lol
Looks so tough, but you made it, sir. Good one. 👍 #YAY
I was intriged by that cool pen change
Forgot the odd function! I was doing lots of things and i forgot that simple theorem 😵
As an additional note, there was a sign posted at Xinyuan Library Restaurant to name notes in a line of music to access the Wi-Fi. The notes made up Twinkle Twinkle Little Star. Now I'm guessing that's slightly easier than ^^^...
I love this...it's awesome!!!
Me tryna act smart: Oh yeah the answer is obviously pi
University: Congratulations!
Me: ....
What if I forgot the sixth digit of pi ?
Except the problem is....you need WiFi for some users to access this video
YT is banned in China
#YAY you actually did this?Well I can use the wifi now
Yale NG yea!!!! Lol
00:03 Doraemon starting song music😀😀
i really like your "hand font" :D
Looks nice when you write on paper instead of board
1:35am love the dedication.
why is second integral equals zero if you try to solve it by using triangle method?
Imagine if you worked all this out but only knew pi to two decimal places. Just Google it... oh.
Be careful. The next time the Wi-Fi password becomes an answer to an integral in a restaurant, someone will have already devoured it (it is a restaurant). Be grateful for your integrals as always, folks!
All of that for pi! I love it.
that's just such a great idea!
#yay good work. thank you.
I love how he used the doraemon bgm
im watching this video at 1:31
A motivated hacker would be simply bruteforce the password instead of solving the integral
advertiser survey music lol
I recognize the intro music: Doraemon opening :D
I had this melody as a ringtone/alarm clock in an old phone. Still hate it.
i did not understand how did we deduct that the second part is a semi circle.
I would think it is a full circle and not a semi-circle. Would have liked him to explain that.
@@greogryhouse8341
You gotta start from the circle equation, r^2=x^2+y^2 (if it's centered in the origin as in the video). Then you separate y and square root everything to obtain a function of x (y=sqrt(r^2-x^2))Since a function can have only one value of y for every value of x the function can only describe the upper or lower semicircle. Lower has a -sign in front of the square root and upper has not.
If you're as familiar as he is with maths you don't need demos to visualize the approx "shape" of functions. I didn't recognize it at first but when he mentiomed it I realized. Though I agree, a demo for us would not have hurt.
I solved the second part with integration by parts, and the answer is the same :D #yay
I recommend using trig substitution, it's usually way faster than integration by parts for things like this. If you haven't heard of it, I recommend looking into. Here you would basically just let x = 2sin(theta)
Now they will have to change the wifi password
A bit easy actually. I think asking for the integer solutions of some equation is harder
I guessed it would be pi, but I think I would try to make my password some form of k*n where k is an integer constant and n is some transcendental number.
Would like to see solution of first part without the assumption of it being 0
I have not studied calculus
Is this what he means?
x²+y²=a² forms a circle and 'a' is the radius
Yes.
And by the way, the circle equation is the Pythagorean theorem a² + b² = c². The reason why they're related is because of how the theorem works: The square of the two catheti's length (short sides, here a and b) of a rectangular triangle equals the square of the hypothenuse's length (long side, c).
That means, you can draw a rectangular triangle inside the circle whose hypothenuse goes from the center to the edge of said circle and the catheti are parallel to the x- and y-axis, respectively.
Why didn't you evaluated the second integral?
I'd have far greater problems with the rest of the message than with the integral...
Yy the f(x) is considered hasf(-x) ...how the 2nd part is calculated has area graph explain please
But then you realise that the root was over dx in the original pic 💀
I use Maclaurin series. Nice problem.
I've got desmos on my phone and it runs offline, so... yeah.
Man your teachers really cared ours just stuck us in a closet USA
When i solve the integral of 0.5(4-x^2)^0.5 i will get 2/3x(4-x^2)^1.5
Which results in 0 between -2 and 2. (because of the 4 minus x squared, which is for both numbers 0)
Why am i not getting pi?
You integrated it wrong.
If I don't know the exact value of pi, how can I get WiFi even I solve the integral.
6 first digits...
314159
fun fact : no one really know the exact value of pi , 3.14 or 22/7 is just approximation , pi is just endless number going on and on (irrational)
You're @@That_One_Guy... huh....
i know 97, 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211
What if the integral was between let's say zero and three?
I just used my TI-84 calculator and got the answer in less than 5 seconds
But y equals 4-x^2 is a full circle not a semicircle
All damn Wi-Fi it was I needed was 3.14
I would record a handshake and then crack it. 6 digit password is done in nothing.
May you please do a video on the limit of n when n goes to infinity for the equation nsin(180°/n)/(90°-180°/n)
The answer is always pi, ALWAYS.
It’s pi
I think that has become a meme
Can you do a video for deriving sin^2 but with a power series all squared instead of just using derivatives of sin^2?
Sorry but your accent is a bit unique ,it would be easier for students like me to understand if you added English subtitles . Thanks for the amazing video
This is violent..
Could we even find the antiderivative of the odd function?
Sherlock Holmie Blitz Yes, if the function is continuous, then the answer is zero for odd functions.
@@justabunga1 no, that's the area not the antiderivative
我忽然不懂為什麼 Cos 是偶函數
-X 明明是 X 軸上原點左邊的負數
-X 除以斜邊應該是負的啊 ?
為什麼說 Cos(-X) = CosX ?
Y'see... uhm quq this is beautiful
Trick question , like its just Pi, an infinite sum of pi would be more fun.
I would quickly make a script to solve it numerically
Was trying to solve the leftside for at least 30min
Please make the hardest way to solve 1+1
Since 1 + 1 is such a difficult mathematical expression to evaluate, you're going to have to use taylor expansion series to find an approximation. Since ln e = 1, then 1 + 1 = ln e + ln e = ln e^2 by logarithm properties. Now use the taylor expansion for ln(1-x) and e^x to approximate the result to about 2.
This question is so easy
why can i interpret (4-x^2)^1/2 as an area?
You don't. f(x) = sqrt(a² - x²) is the equation for the *graph* (i.e. the outer edge) of a semi-circle. It's its *integral* which allows you to calculate the area, not the function itself.
Why is x² + y² = 4 a semicircle and not a circle ?
Cause it's under a square root without a plus-minus sign, so it is just the positive half.
Yay i solved it the same way as the based math god.
That problem isn't even that hard. It's just a bit lengthy
İngilizce bilmiyordum ama matematik dili evrenseldir.
Pretty much.
Bravo
subscribed
Part of it is odd. Remove that cos part
Damn, if I knew that odd function stuff I'd have gotten this right 😒
What about photomath?