Is this integral 0?
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- čas přidán 21. 08. 2024
- integral of sqrt(1-x^2) from 1 to 1
#shorts #calculus
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bprp #fast
OG fans know that he did this with √(1+x²) as well
Blindfolded
The substitute function must be bijective. Most of the books don't tell that.
Injective
I mean that’s what the jacobian tells you
What does that mean can som1 tell
@@fahimnabeel606 It just means that it's invertible.
@@davidepaolillo7218 Yeah this is more right. Generally, the jacobian is det(W^T*W) with W being a matrix of the partials and W^T being W transpose. Injective is all you really need
The substitution x=sin(t) works. Borns of integration became -pi/2 and pi/2
What you can do is use the fact that f(x) is symmetric, f(-x) = f(x), split the integral from (-1, 1) to (-1,0) plus (0, 1) and then you can see the two different integrals are equal due to symmetry, now the limits of u are from 0 to 1 and the u substitution works.
When there are no symmetric arguments, use complex calculus and residue theorem iykwim.
Fun)
For guys who does not understand why it does not work i will explain.
Substitution in integral required that on a;b exists function and it is continuous and its derivative exists too.
g(x)' there has x in the denominator but in the -1;1 exists 0 which mean function will break there.
Not differentiable on the closed interval.
how to take a derivative of polynomials
1.factor
2.differnate using the product rule
3.expand
a example should help
x²+2x+1
(x+1)(x+1)
(x+1)*1+(x+1)×1
2x+2
So that’s the answer? Weird how it works that way
@@ryanjagpal9457 It's a bit of a joke. You can just differentiate every monomial individually lol
@@BroArmyCommander For example like (3 + a)^3 how would you do that in
(3+a)(3+a)(3+a), I can do the first two not sure about the last one, it’s probably too hard for cubed
This is actually hilarious 😆
@@roygalaasen Hehe
Why, in theory, does the U-sub fail?
Upper integral and lower integral is the same (u=0) which means answer is 0 when the question is made out of a circle that has center on origin so x² + y² = 1 and if u leave y alone u get y=sqrt(1-x²) which is the function in the question
@@cyrenux right, but I’m asking why the substitution fails in theory. Meaning I know the integral doesn’t equal 0 but what about the substitution was wrong
@@Happy_Abe my bad but now that you said it what i thought according to a shorcut i saw a while ago that this function is even so when u u sub it changes nothing, just like any even function which has upper as x=a and lower as x=-a you will still get 0 so right way to solve them is by drawing the shape first but i could be incorrect, im just trying to guess
We dewen sum maths 'ere innit bruv 3Head
@3head individual I haven’t heard of that for u-sun, can you elaborate on that or point me to some further readings on this?
I’m curious of the formal way we define a substitution in integrals to be valid
You can use sine or cosine function to substitute the x.
Yes, but rather point the misteak :-)
This is on par with doing a u = e^-x^2 sub for the Gaussian to get 0, or a u = cosx sub for integrating cod between -a and a etc…
Basically for any even function, u cannot simply substitute for the integrand else the upper and lower limits will always evaluate to be the same value
The problem is that the function being integrated wouldn't be of the form f(u(x))u'(x), because u' isn't defined at endpoints.
Also if you did do a U sub, the new function wouldn't have been continues at 0 😉
what
@@nikhilnagaria2672 just needs to be continuous in the interval you're integrating, otherwise you'll need to split the integral and use limits to get the answer.
dU : *Am I a joke to you*
📠
it's in the dots
You're teaching in simplest ways.
Unbelievable person You're.
Just like calculus/deviate 👍😌
My 2 solutions using integrals:
1. Substitute x = cos(u), dx = -sin(u)du,
Get integral -S sin^2(u)du, solve using integration by parts and trig identities and get pi/2
2. Switch to polar coordinates and solve ;D
When you have. √(a^2 - x^2) can't you just let x = a*sin(u) or x = a*cos(u) ?
I don't understand anything but i like the video because it is so well explained (i just assume, because he usually explains so well)
I don't even know what integrals are, why am I watching this?
Ik integrals just help you find the area under the curve :))
😆
Love this man you will never fail in maths
I hope now you do.
@@Lser03 antiderivative is awesome as well, even better in my opinion.
The problem is that the function inside the integral can't be written as f(u(x))u'(x).
I thought he was gonna explain why this doesn't work.
czcams.com/video/waqlinTaKO8/video.html
I'm not equivalent to a formula! *cries in matter*
You can only substitute a function as u if it doest not have any maxima or minima in the interver in which we are integrating
The rule says that we can let a function as u only if it is strictly monotonous and it must be differentiable in the interver
Where can I find a “theory of substitutions “? , In which book?
Break the limits in two parts... from -1 to 0
Then 0 to 1
Then integrate...
1-x ^2 is an even function
then the integral will be from 0 to 1
To me you're not supposed to change the limits of the integral from x to u. Since you made the substitution for u you could have solved the integral with your substitution by changing back to x instead of u. From there find the value of the integral.
Is'nt it using trigonometry sub ?
Yes by putting x= sinu perhaps
what if you want integral of √(1+4x²) from 0 to 2 dx
Use trigonometric substitution. Let x=(1/2)*tan(u), and proceed further.
I faced the same issue recently
Can you do a video on why sometimes it doesn't work
try graphing
Substitution has to be injective
Because for even functions, like the one here area of graph is same from x = -1 to X = 0 and x =0 to x = 1. So this integral can simplified to twice the integral from x= 0 to x = 1 U sub will work after that
@@deadmayday6702 injective?
@@pedrosso0 it should not map two values to one here (-1)^2 = 1^2 so u=x^2 is not injective.
YOU HAVE MADE US DIFFICULTY WITH THIS BEARD THAT KEEPS THESE LAST TIMES, WHEN EXPLAINING THE EXERCISES! ...
Нужно разбить на 2 интеграла, от 0 до 1 и от 1 до 0.
Half of the circle is above the x axis and half is below. The sum of the positive half and the negative half is zero, so the first solution is OK. %D
It's not below man. It's the positive part of a circle
Does it only fail when both the limits become equal or does it fail in other circumstances too?
It is WRONG whenever the substitution is not injective.
I do not think that this is correct way to substitute limits because as x varies from -1 to 1 , u varies from 0 to 1 , not 0 to 0 , which can be proved by considering the inequality that since
-1
-1
That's not true. The inequality is not preserved when squaring because y=x² is not an increasing function over the interval [-1,1]. For example, x=0: -1
Me when I'm asked to integrate an odd function with limits -a to +a
I lost him at x=1
When you graph this curve, notice the area under the curve in [-1, 0] is the same as the area under the curve in [0, 1]. Thus, you can rewrite the integral to 2 times the integral of the function from 0 to 1.
Ok...
That is called an “even function”, and your observation is correct.
You can do some trig sub though
Couldn’t you set the lower limit to 0, upper limit to 1, and place a 2 out front due to it being symmetrical?
Yes we can
Collab with #MathRocks pls
But...what if it was a indefinite integral with no upper limit and lower limit? Then could we use the *u* sub?
Yes but the function y = sqrt(1-x^2) has y-axis symmetry so the integral of y from -a to a will be the same as 2 * the integral of y from 0 to a. If you do a u sub there it should work. Also you can sub x = sin(u), dx = cos(u)du to get the integral of cos(u)^2 from -π/2 to π/2 and that will work for 100%
This is just the half area of the unit circle 🙏
Please prove or explain what actually happened
me who already knows rhe integral: 👁👄👁
Let X be cos(u)^2 or sin(u)^2
Goat.
U-sub didn't get jt wrong, U subbed wrong haha
You have to let x either be sin or cos theta
For this only, I got subbed to your new channel.
You would not believe, the moment you discussed this amazing error,
My thumb subbed you mistakenly.
Your sub gone wrong as well 🤣.
Why doesn't it work? When does it not work?
It doesn't work because the integrand isn't bijective over the domain bounded by the two limits of integration. In other words, it isn't a 1-to-1 function.
You also can't just make u equal to anything you want, you have to be able to reconcile it such that you can replace the dx with du, in order for u-substitution to work.
Why not just let x=cos theta or sin theta
I think the point here is that pi/2 = 0
Isn’t that just finding the area under the top half of the unit circle, so the answer is pi/2 ?
Edit: Oh I felt so happy I got one right
This realisation is why I didn’t post a comment about this being an even function, hence you can integrate from 0 to 1 and multiply by 2.
You changed the variable but you didn't change the derivative dx 😂
If it's wrong why did the question give us like this?
How about we just use sin inv(x/a)
Pikachu might have suggested this substitution .Very good. DrRahul Rohtak Haryana India
I like the issue you're discussing here. It may seem like a conundrum, but it is not. It made me think about this issue, so I integrated it into my notes. For those interested, I posted a video response: czcams.com/video/waqlinTaKO8/video.html
But why does this not work?
Change the limits of integration
Sorry, make a trigonometric sub
@@dylandejonge5069 that's not the question.
To change variables, you need a differentiable bijection between the two domains of integration. This substitution gives the same value for x and -x, so it's not a bijection. If you split the domain into two halfs, [-1,0] and [0,-1], even u=1-x^2 substitution will work, but you have to treat the two integrals differently, (and this u sub won't really help to evaluate the integral).
@@zoltankurti thank you
where i can buy your merch?
Im in 8th grade. Why am I obsessed with these videos?
Trig sub!
Brotha please!
Sqrt (1-x^2) is an even function of x..
So, obviously there'd be symmetry within the graph of y = f(x)..
Thus, the area under the graph from x= -1 to x=1 is just twice the area under the graph from x=0 to x=1..
Id est, there'd be no problem transforming respective values
But what's the mistake! That's the point
ok but why does it not work?
The answer is correct, but if you plug in a calculator, the calculator will give you an other answer
Is there seriously not a mapping from x to u that accurately defines the solution as π/2?
The function needs to be bijective, for changing the limits of the integral to the u-world to work. Most books don't cover this condition. If the function isn't bijective, you have to separate the integral until you can add up multiple integrals, each of which is bijective.
If you don't know what a bijective function is, it means a function that the inputs map uniquely onto the outputs, with no overlap. In other words, a 1-to-1 function, or a function whose inverse isn't multivalued.
That is not how i u-sub
His English makes me forget mine
Cool, but, why?
Me it's an odd function 😀😀 I am great
On my opinion, the U-sub fail because you cannot simply set u equals anything you want. You must do something with the "dx".
You should make sure that you can find du
For example ∫x² from - 1 to 1
If we assume u=x²
We can not find du
I'm very curious, where in my life am I gonna use this 🤔🤔
Look up "Applications of Integration"
nooo i think you missed one step.... hear me out ok 😁 ---> just after the "u" sub, we cannot sub the limits just like that.... this is because the initial x was varying from -1 to 1 directly..... like -1, -0.9, .....,-0.1, 0, +0.1, .....,+0.9, +1 .K but here if we look at the limits of u which depend on the x values.... value of "u first goes from 0 to 1 and THEN only it goes from 1 to 0 so if we break the integral limits from 0 to 1 and from 1 to zero WE WILL GET CORRECT ANSWER..
Okay so you always have to break integrals like that? Say you have ∫dx from 0 to 100. Will we have to break it from 0 to 1, 1 to 2, 2 to 3, and so on??
@@nikhilnagaria2672 Yes we can but since the limit 0-->100 is all positive and increasing, like there isn't any monkey business going on when going from 0 to 100 so we can integrate without splitting limits
:)
Although if you try splitting limits you will always get correct answer
@@aryanbansal624 so what BPRP did is wrong because we should split the limits right?
@@nikhilnagaria2672 Ok so bprp didnt do anything wrong as he did mention that the u substitution was the wrong method and then he solved it graphically.
Although the u substitution WILL work if we split the limits at appropriate locations so basically there are multiple ways of solving this.
Just keep in mind, whenever we do a substitution we need to correctly replace the limits. We shouldn't just blindly plug in the values, in such cases a rough graphical sketch gives a lot of insight where we should break the limits.
@@aryanbansal624 I think you missed the key point here: the u-sub fails because the u-sub should be injective (bijective) in the domain
I dont understand but i watching '?D
You're looking more weird with your goatee 😮😢
It's just video from 2020
You can't make that u substitution because there is no x in the problem to cancel out the x left over when finding du.
Maths ki maa behen 1 kr di
Bhai substitution karo to pyar se warna mat karo
Bhai 1-x^2 is not monotonous in [-1,1] jo substitute kaise kar diye
Jara ruko sabar karo
Dekho pahle tab aage badhna
bruh
😍😍😍😍
So pi /2 = 0
😁
U need to learn before teaching 😂
It's -pi/2
∀x ∈ [-a,a], f(x) = f(-x) => ∫{-a,a}f(x)dx = 0