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Flammy why not juet rewrite x as (x^2)^1/2 and solve it thst way..Hopenyou can PLEASE sjare why yiu did it this convoluted way. But thanks for sharing.
I'm not familiar with this kind of mathematics, but here's my approach. Let y=x^2, making the integral y^(1/2) dy. This equals 2/3*y^(3/2). Substituting back yields 2/3*y^3. Plug in the bound to get 2/3*(1-0) = 2/3. So, I got the same result. But I'm not sure how well this idea generalizes or if there are any ways in which it breaks.
That's exactly what I did. I put u = x^2, so dx^2 becomes du, and x = u^(1/2). Then integral is u^(1/2) du which gives U^(3/2) / 3/2 => 1/(3/2) i.e. 2/3.
Once you defined the terms of the original integral it all made sense. I never thought about integrating across a different function instead of just dx.
@williamnathanael412 In statistics, every random variable admits a cumulative distribution function F_X, but not every random variable admits a density/mass function. So, in general, we can define the expected value of a random variable X as E(X)=\int x dF_X(x) (integrating w.r.t. the cdf).
From the product rule, d(x*x^2) = x d(x^2) + dx*x^2. So x^3|(0->1) = int(x d(x^2))|(0->1) + int(x^2 dx)|(0->1). 1 = int(x d(x^2))|(0->1) + ⅓. So the answer is 1 - ⅓ = ⅔
yea that's basically what i thought (to let x^2=u). But this video wasn't just about getting the right answer. This vid provided a good insight on what it means to integrate w.r.t g(x)...
Is really similar to the easier way he uses later in the video, but notice that x ≠ sqrt(x²), in fact, sqrt(x²) = |x|. Still, in this case, it is true, since we are working in [0,1], but the correct way to conclude what you said I think it would be Int x dx² Int x 2x dx Using t = x² and dt = 2x dx Int sqrt(t) dt
Personaly, I'd just use some differential geometry result dx² is a volume form onto the segment [0,1], it can simply be calculated as : dx² = 2xdx (as it is the exterior derivative of the scalar field x->x²) Then your integral is simply 2 \int_0^1 x²dx=2/3, that's all folks
this was exactly the kind of video I needed right now. I've been out of university for a few years and getting back into calculus, and your explanation at the beginning could not have been better for me. I had no idea you could integrate by different functions like that, and I was amazing that the result of xdx^2 was 2/3 haha. Absolute banger and I will definitely be revisiting this multiple times
19:08 = integrate [cos(x) d(sin(x))] from 0 to 1. 1. Apply previous rule, yielding: integral {cos(x)[sin(x)]' dx} from 0 to 1 2. Since derivative for sin(x) = cos(x), that yields: integral [cos²(x) dx] from 0 to 1 3. Just solve 4. = (1/2) + [sin(2)/4] 😁 Edit: plus sign rectified 😅
nice display of Riemann/Stieltjes methods people just do the last one in practice though, now i wonder if there would be a generalization if g was not continuous
There is; in fact, the definition of the Riemann-Stieltjes integral does not require g(x) to be continuous. It must, however, be of bounded variation, which means it can't be too 'wiggly'.For example, sin(1/x) would be a bad choice in the vicinity of x=0, even if you explicitly exclude x=0. A full explanation is outside the scope of what I can write in a TH-cam comment.
It is dx² = 2xdx, so xdx² = x(2xdx), thus integral(0,1)(xdx²) = integral(0,1)(2x²dx) = 2integral(0,1)(x²dx) = 2(⅓1³ - ⅓0³) = ⅔ Note that in identifying x(2xdx) with 2x²dx I'm using that we have a module operation from the ring of smooth functions on all differential-k-forms defined by left-multiplication.
Although it might not be obvious at first the point of this integral modification, one of its strengths shines in A(x) being a partial sum function (A(x)=A(floor(x))=sum of terms
couldnt you switch the variable? if you consider x^2=t that means x=sqrt t since x is possitive so the integral just becomres integral from 0 to1 of sqrt t dt which is 2/3
At 11:53 he says “2i - i” but means, I think, “2i - 1”. I found this very confusing at first because the “i”s look almost the same as the “1”s on the blackboard.
You are right, but whatever comes out of it will be of size n and thus will be ruled out by the n^-3 factor. That might be the reason he did not correct the error if he became aware of it
Nice! Little piece of new info. I tried it before watching and got the same answer. What I did was I set x^2 as y. Solved for x to get sqrt(y). Didn’t change the bounds cuz it’s just 0 to 1 and those don’t think would really change given this scenario. So now it’s the integral from 0 to 1 of sqrt(y) with respect to y. This gives the answer of 2/3 as well. Playing for a bit, changing the bounds to that function is accurate. If it were 0 to 2, it’d be the integral from 0 to 4 since (0)^2 is 0 and (2)^2 is 4. Comparing that to the form of solving this for continuous functions. Integral from a to b of f times g’ dx. Gets the same answer for the 0 to 2 situation. Which is 16/3.
For context At 2:40 and just a senior at highschool Isn't this the same as Integral of √xdx from 0 to 1 Can't we just extrapolate the answer from that Or let x²=u then it just becomes a normal integral With the last formula, Int cosx d(sinx) = Int cosx × cosx dx = Int cos²x dx = Int (cos2x+1)/2 dx = (1/2) Int (cos2x +1) dx = (1/2) (sin2x/2 + x + C1) = Sin2x/4 + x/2 + C Sorry if I did anything incorrect But is that really the only way to do that, i feel.we can maybe play around with the fact that cosx and sinx are derivates/integrals of each other
We got just use the substitution, x^2=t. x= root(t) assuming x is positive for this integral. So we got root(t)dt, = 2/3t^3/2= 2/3x^3, and applying the limits, we get 2/3
This is how I did it: Let x² = t, Such that dx²/dt = 1, Hence dx² = dt. As a result the integral §xdx² becomes §√tdt which equals 2/3 √t³ + c Of which you can re-sub to obtain 2x³/3 + c. Substituting bounds, you then get 2/3.
It's simple, you can Wright the x to ((x^)1/2)^2. Then you simply integration it and you will get (2x^3)/3. Then you input the limit and the ans is 2/3.
Commenting from just the thumbnail to say that it makes perfect sense if you just say y=x², implying dy = 2x dx, so dx² = 2x dx. In total: int_0^1 x dx² = int_0^1 x 2x dx = int_0^1 2x² dx = [2/3 x³]_0^1 = ⅔
I did it is a much simpler way Integral [x d(x^2)] x^2 going from 0 to 1 = Integral [x du] u going from 0 to 1, where u = x^2 du = 2 x dx and when u = 0, x = 0 and when u = 1, x = 1 for the limits of the definite integral So, Integral [ x du ] u going from 0 to 1 = Integral [ x (2 x dx) ] x going from 0 to 1 = 2 * Integral [ x^2 dx ] x going from 0 to 1 = 2 * [ x^3 / 3 ] from 0 to 1 = (2/3) * [ x^3 ] from 0 to 1 = (2/3) * (1^3 - 0^3) = (2/3) * (1 - 0) = (2/3) * (1) = 2/3
So there are two branches of x for the equation y=x^2, hence I would expect that there are two possible solutions, namely +2/3 or -2/3, corresponding to each branch. Can you explain why we disregarded the other branch?
Yet another way: d(x^2) may be multiplied by 1, where 1 = dx/dx. Next d(x^2) * (1/dx *dx) makes d(x^2)/dx *dx, and this yields to 2x*dx, guessed immediately, waiting impatiently till 18:00. No formal prove, just "an engineer" version, but when our mathematical language uses some basic "grammar" rules it should be created in that way - our convention way. The 2x multiplier makes non-linear expansion on x axis of a regular Riemann integral, showed quickly somewhere in the mid part of the movie.
Information-Based Unification of Forces: a) Central Idea: All fundamental forces (gravity, electromagnetism, strong, weak) emerge from a single information field. b) Unified Force Equation: F = -∇(ℏc/l_P² · log(I/I₀)) Where I is the local information density and I₀ is a reference density. c) Implications: - Potential resolution of incompatibilities between quantum mechanics and general relativity - New approach to grand unification theories - Prediction of new particles or forces at extreme energy scales
In the spirit of the intro meme 0:03 I like to think about it like this: dx² = 2xdx ∴ ∫xdx² = ∫x(2xdx) = 2∫x²dx = ⅔x³ + C. hence when the integral is evaluated from 0 to 1 it equals ⅔. 19:19 sin(x) is differentiable everywhere on the real number line so it's differentiable on that interval. dsin(x) = cos(x)dx ∴ ∫cos(x)dsin(x) = ∫cos(x)[cos(x)dx] = ∫cos²(x)dx = ½x + ¼sin(2x) + C. hence when the integral is evaluated from 0 to 1 it equals ½ + ¼sin(2) ≈ 0.7273243567.
Wait am I coping? Can't we just index u = x^2? We would get int(0-1) u^(1/2)du Which solves to (2/3)u^(3/2)](0-1) And if we put back in x^2 for you we get (2/3)x^3](0-1) Which seems identical to the expected result. That seems to simple though so please tell me where I'm just completely wrong.
I really like to see the views on this vid. Hope this shows you, what your community really wants to see. I dont think more than 1% wants exponent rules.
I think we could've just solve it by saying x²=y and x=√y and so ∫¹₀ x dx² will be ∫¹₀ √y dy and the result will be (2y^3/2)/3 |¹₀ and the result of it is 2/3.
To equalize the subdivisions of the ordinate axis, I replaced x with SQRT(y) and dx^2 with dy, and the integral value of 2/3 pops out. (limits are unchanged) I have no idea if this presents any kind of general solution. I see others did the same.
What would you do if the limits of integration don't match up? For all the examples you do dx^2 and dsin(x) the limits, 0 and 1 when plugged in result in new limits sin(1) = 1 and (1)^2 = 1 and similar for zero. For the Thumbnail integral the limits 0 to 2 don't match up and sinx never reaches 2. Since this trick operates on similar principals to U-Substitution wouldn't we need to change the limits of integration, and for the thumbnail example also split the integral into two pieces since sinx is not 1to1 from 0 to 2.
Wow, what a complicated way to use the notation. This is how I do it: // my definition of integral, as a cancellation of S and d, where d is implicit diff operator S [d f] = f - f_0, d[f_0]=0 S (x d[x^2]) = S x*2*x dx = 2 S (x^2 dx) = 2/3 S d[x^3] = 2/3 x^3 - f_0 = 2/3 1^3 - 2/3 0^3 = 2/3
Shouldn't the x^2 be in parentesis? I mean, dx^2 looks like dx*dx, while d(x^2) would look like 2x*dx. After all, in a second order derivative, where the notation used has no parentesis, we kinda mean dx*dx. I mean the "denominator" of the derivative, btw.
much much simpler would it be to just take d(x^2) = 2x*dx, then we would get 2*int(x^2 dx)|(0->1), and then basically (2*(x^3 /3))|(0->1) which is equal to 2/3...
17:35 I'm not sure and maybe I'm wrong, but isn't there a theorem that for a complex function a derivative exist? Does that mean complex functions are always integrateble that way?
Can't you just set X = x^2 (so x = √X), and then calculate the normal integral of √X d X, with X from 0 to √1? At least for this example, it seems to give the same result 2/3.
I have one question. If the intuition is that we're dividing the segment according to the function g rather than linearly, why do we still take x_i to be i times delta x?
it actually applies in all cases. it's merely that the standard case results in f'(x) being the multiplicative identity, so it's not necessary to be aware of it: if f(x) = x then f'(x) = 1 ∫ g(x) df(x) = ∫ g(x) f'(x) dx = 1 ∫ g(x) dx = ∫ g(x) dx
I'm confused about how at 10:15 you say x_i is still i/n when we've just redefined delta(x) to be g(x_i)-g(x_i-1), since x_i = 0 + idelta(x), don't we need to recalculate x_i using our new delta(x)?
9:55 When doing the new integral, you wrote the new f(x_i) as the same as the one in the previous integral, namely f(x_i) = f(i/n) = i/n. However, this defining of x_i used the previous value for Delta x (Delta x = i/n), instead of the new value for Delta x (Delta x = g(x_i) - g(x_i-1). How can you still say that f(x_i) is the same even after this new value for delta x?
Isn't there some sense in which d(x^2) is simply equal to 2x dx? It is not some kind of trick, they are literally equal and can be substituted for one another, which makes this wacky integral in fact very easy!
That is ONE solution, if dx² means d(x²). But if dx² would mean (dx)² the solution is 0, because the volume under a surface of infinitesimal thickness tends to zero.
Papa Flammy make a video on ISI (Indian Statistical Institute)entrance exam questions and on CMI (Chennai Mathematical Institute) entrance exam questions.These are pure math and statistics research institutes and the question level of these institutes are even higher than Jee advance.They are challenging problems you will surely like them.I guarantee you.😊😊😊
To those who say "why don't we use d(x^2)=2xdx or let u=x^2": surely you didn't read the title. Think twice or you will be r/wooooshed. This is not a place to show how good you are at "your calculus". Flammable Maths makes serious mathematic jokes, pay some respect.
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Flammy why not juet rewrite x as (x^2)^1/2 and solve it thst way..Hopenyou can PLEASE sjare why yiu did it this convoluted way. But thanks for sharing.
I'm not familiar with this kind of mathematics, but here's my approach. Let y=x^2, making the integral y^(1/2) dy. This equals 2/3*y^(3/2). Substituting back yields 2/3*y^3. Plug in the bound to get 2/3*(1-0) = 2/3.
So, I got the same result. But I'm not sure how well this idea generalizes or if there are any ways in which it breaks.
thats what I did xD
dx^2 is dx^2/dx * dx
= d/dx[x^2] * dx
= 2xdx
They are equivalent as long as g is continuously differentiable and strictly increasing.
That's exactly what I did. I put u = x^2, so dx^2 becomes du, and x = u^(1/2). Then integral is u^(1/2) du which gives U^(3/2) / 3/2 => 1/(3/2) i.e. 2/3.
@@Qaptyl Leibniz notation is the best lol
nice try, but we know that dx is very small, so higher orders are basically zero which yields the trivial result of zero for the integral
fucc, u got me D:
But this isn't (dx)^2, this is d(x^2).
@@KinuTheDragon for dx = x = 0 it holds
@alexander8311 wait, so dx = sin(x)
@@diobrando7642
wrong. dx=dsin(x)
Once you defined the terms of the original integral it all made sense. I never thought about integrating across a different function instead of just dx.
:)
Very useful integral in probability.
Help explain
@williamnathanael412 In statistics, every random variable admits a cumulative distribution function F_X, but not every random variable admits a density/mass function. So, in general, we can define the expected value of a random variable X as E(X)=\int x dF_X(x) (integrating w.r.t. the cdf).
@@williamnathanael412 see properties section in this wiki article as a starting point: en.wikipedia.org/wiki/Expected_value?wprov=sfla1
@@phonon1isn't the generalized case the integral over Omega of X in dP?
yes he should also show for a discrete probability how the integral works out to be the usual sum
From the product rule, d(x*x^2) = x d(x^2) + dx*x^2. So x^3|(0->1) = int(x d(x^2))|(0->1) + int(x^2 dx)|(0->1). 1 = int(x d(x^2))|(0->1) + ⅓. So the answer is 1 - ⅓ = ⅔
Very nice approach! :)
This is just integration by parts btw
@@siwygameplay Not exactly, it depends on your interpretation.
@@renanwilliamprado5380 and what interpretation is that
Isn't this solvable by writing x as sqrt(x^2) and then treating it as int{sqrt(t)dt} with t= x^2
yea that's basically what i thought (to let x^2=u). But this video wasn't just about getting the right answer. This vid provided a good insight on what it means to integrate w.r.t g(x)...
@@hassanniaz7583Sure, but I thought he'd use substitution at the end to show how to relate the two different types of integrals.
Is really similar to the easier way he uses later in the video, but notice that x ≠ sqrt(x²), in fact, sqrt(x²) = |x|.
Still, in this case, it is true, since we are working in [0,1], but the correct way to conclude what you said I think it would be
Int x dx²
Int x 2x dx
Using t = x² and dt = 2x dx
Int sqrt(t) dt
Yes. Exactly how I approached it.
Why using change of variables, isn't it just Int 2x^2 dx = 2/3 * x^3 ?
No one gonna talk about the farmers tan?? Goes crazyyyyy
Papa flammy has been engulfed in flamen!
Oh wow, I wasn’t expecting to see a Riemann-Stieltjes integral today
Never seen a definite integral done by definition. Very exciting video
thx :)
12:00, 2i-1.
Yeah i thought i was crazy for a second
It yields the same result. But yeah.. it's 2i-1. I had to do the sum like 5 times to convince me 😂
Personaly, I'd just use some differential geometry result
dx² is a volume form onto the segment [0,1], it can simply be calculated as : dx² = 2xdx (as it is the exterior derivative of the scalar field x->x²)
Then your integral is simply 2 \int_0^1 x²dx=2/3, that's all folks
Exterior derivative go brrrrr
that was my immediate thought. dx^2 is 2 x dx, and then do the integral.
Yuuuup this is the way to do it quickly.
I like this way
this was exactly the kind of video I needed right now. I've been out of university for a few years and getting back into calculus, and your explanation at the beginning could not have been better for me. I had no idea you could integrate by different functions like that, and I was amazing that the result of xdx^2 was 2/3 haha. Absolute banger and I will definitely be revisiting this multiple times
19:08 = integrate [cos(x) d(sin(x))] from 0 to 1.
1. Apply previous rule, yielding: integral {cos(x)[sin(x)]' dx} from 0 to 1
2. Since derivative for sin(x) = cos(x), that yields: integral [cos²(x) dx] from 0 to 1
3. Just solve
4. = (1/2) + [sin(2)/4] 😁
Edit: plus sign rectified 😅
Slight correction, the sin(2)/4 has a positive sign
nice display of Riemann/Stieltjes methods
people just do the last one in practice though, now i wonder if there would be a generalization if g was not continuous
There is; in fact, the definition of the Riemann-Stieltjes integral does not require g(x) to be continuous. It must, however, be of bounded variation, which means it can't be too 'wiggly'.For example, sin(1/x) would be a bad choice in the vicinity of x=0, even if you explicitly exclude x=0. A full explanation is outside the scope of what I can write in a TH-cam comment.
@@tomkerruish2982"I have a marvelous proof of this fact that this comment section is too narrow to contain."
- Fermat if he used TH-cam
It is dx² = 2xdx, so xdx² = x(2xdx), thus integral(0,1)(xdx²) = integral(0,1)(2x²dx) = 2integral(0,1)(x²dx) = 2(⅓1³ - ⅓0³) = ⅔
Note that in identifying x(2xdx) with 2x²dx I'm using that we have a module operation from the ring of smooth functions on all differential-k-forms defined by left-multiplication.
Although it might not be obvious at first the point of this integral modification, one of its strengths shines in A(x) being a partial sum function (A(x)=A(floor(x))=sum of terms
couldnt you switch the variable? if you consider x^2=t that means x=sqrt t since x is possitive so the integral just becomres integral from 0 to1 of sqrt t dt which is 2/3
just say u= x^2 and you'll have integral(sqrt(u))du what equals ⅔*u^3/2 equals ⅔*(u^½)^3 and since u = x^2, it means sqrt(u) = x so: ⅔*x^3.
At 11:53 he says “2i - i” but means, I think, “2i - 1”. I found this very confusing at first because the “i”s look almost the same as the “1”s on the blackboard.
You are right, but whatever comes out of it will be of size n and thus will be ruled out by the n^-3 factor. That might be the reason he did not correct the error if he became aware of it
That was fun! Very "Michael Penn".
I love evil looking integrals and the fact that we can actually do stuff to calculate them
Nice!
Little piece of new info.
I tried it before watching and got the same answer.
What I did was I set x^2 as y. Solved for x to get sqrt(y). Didn’t change the bounds cuz it’s just 0 to 1 and those don’t think would really change given this scenario.
So now it’s the integral from 0 to 1 of sqrt(y) with respect to y.
This gives the answer of 2/3 as well.
Playing for a bit, changing the bounds to that function is accurate. If it were 0 to 2, it’d be the integral from 0 to 4 since (0)^2 is 0 and (2)^2 is 4. Comparing that to the form of solving this for continuous functions.
Integral from a to b of f times g’ dx. Gets the same answer for the 0 to 2 situation. Which is 16/3.
For context At 2:40 and just a senior at highschool
Isn't this the same as Integral of √xdx from 0 to 1
Can't we just extrapolate the answer from that
Or let x²=u then it just becomes a normal integral
With the last formula,
Int cosx d(sinx)
= Int cosx × cosx dx
= Int cos²x dx
= Int (cos2x+1)/2 dx
= (1/2) Int (cos2x +1) dx
= (1/2) (sin2x/2 + x + C1)
= Sin2x/4 + x/2 + C
Sorry if I did anything incorrect
But is that really the only way to do that, i feel.we can maybe play around with the fact that cosx and sinx are derivates/integrals of each other
We got just use the substitution, x^2=t. x= root(t) assuming x is positive for this integral. So we got root(t)dt, = 2/3t^3/2= 2/3x^3, and applying the limits, we get 2/3
This is how I did it:
Let x² = t,
Such that dx²/dt = 1,
Hence dx² = dt.
As a result the integral §xdx² becomes §√tdt which equals 2/3 √t³ + c
Of which you can re-sub to obtain 2x³/3 + c.
Substituting bounds, you then get 2/3.
It's simple, you can Wright the x to ((x^)1/2)^2. Then you simply integration it and you will get (2x^3)/3. Then you input the limit and the ans is 2/3.
we can compute dx^2 using the chain rule:
dx^2=dx^2/dx*dx=(x^2)'*dx=2xdx
int x dx^2=int 2x^2 dx=2/3x^3. after evaluation we get 2/3.
YESSSS PAPA FLAMMY IS BACK WITH SOME WEIRD INTEGRAL, LET'S FUCKING GOOOO!!!!!
The Rie-womann Integral
Its becomes really easy if we write dx^2 = 2xdx . And the answer is 2/3 simple
Commenting from just the thumbnail to say that it makes perfect sense if you just say y=x², implying dy = 2x dx, so dx² = 2x dx. In total:
int_0^1 x dx²
= int_0^1 x 2x dx
= int_0^1 2x² dx
= [2/3 x³]_0^1
= ⅔
"Nonlinear Partition Scaling" explains it all
dx ^ 2 = 2x dx
Notation is unclear is dx^2 interpreted as (dx)^2 or d(x^2)??? Both in the video and my comment!!!!
If the former is intended, then an additional integral sign is needed, so the latter is assumed to be the intent and correct
Looking absolutely yoked man 🔥
very clear and didactic 👏 thank you for this contribution
Look out, the guns are out to stay!
I did it is a much simpler way
Integral [x d(x^2)] x^2 going from 0 to 1
= Integral [x du] u going from 0 to 1, where u = x^2
du = 2 x dx
and when u = 0, x = 0 and when u = 1, x = 1 for the limits of the definite integral
So,
Integral [ x du ] u going from 0 to 1
= Integral [ x (2 x dx) ] x going from 0 to 1
= 2 * Integral [ x^2 dx ] x going from 0 to 1
= 2 * [ x^3 / 3 ] from 0 to 1
= (2/3) * [ x^3 ] from 0 to 1
= (2/3) * (1^3 - 0^3)
= (2/3) * (1 - 0)
= (2/3) * (1)
= 2/3
Let u=x^2, du/dx= 2x, du= 2x dx , so the integral becomes x 2x dx, when u=0, x=0, when u=1, x=+-1.
My first guess was: let y=x^2 and therefore x=y^1/2. Then integrate y^1/2 with respect dy. The result is of course also 2/3
So there are two branches of x for the equation y=x^2, hence I would expect that there are two possible solutions, namely +2/3 or -2/3, corresponding to each branch. Can you explain why we disregarded the other branch?
Yet another way: d(x^2) may be multiplied by 1, where 1 = dx/dx. Next d(x^2) * (1/dx *dx) makes d(x^2)/dx *dx, and this yields to 2x*dx, guessed immediately, waiting impatiently till 18:00.
No formal prove, just "an engineer" version, but when our mathematical language uses some basic "grammar" rules it should be created in that way - our convention way.
The 2x multiplier makes non-linear expansion on x axis of a regular Riemann integral, showed quickly somewhere in the mid part of the movie.
Why haven’t your vids been recommended in so long?? Missing papa flammy!
xdx^2 = x•2xdx = 2x^2dx = d(2x^3/3)
So the value of the given definite integral is 2/3 - 0 = 2/3.
Information-Based Unification of Forces:
a) Central Idea:
All fundamental forces (gravity, electromagnetism, strong, weak) emerge from a single information field.
b) Unified Force Equation:
F = -∇(ℏc/l_P² · log(I/I₀))
Where I is the local information density and I₀ is a reference density.
c) Implications:
- Potential resolution of incompatibilities between quantum mechanics and general relativity
- New approach to grand unification theories
- Prediction of new particles or forces at extreme energy scales
Watched until the end! Also the assignment's answer is = 1/2+(sin(2))/4 :DD
love your content, thats such a smart way to solve this!
In the spirit of the intro meme 0:03 I like to think about it like this:
dx² = 2xdx
∴ ∫xdx² = ∫x(2xdx) = 2∫x²dx = ⅔x³ + C.
hence when the integral is evaluated from 0 to 1 it equals ⅔.
19:19
sin(x) is differentiable everywhere on the real number line so it's differentiable on that interval.
dsin(x) = cos(x)dx
∴ ∫cos(x)dsin(x) = ∫cos(x)[cos(x)dx] = ∫cos²(x)dx = ½x + ¼sin(2x) + C.
hence when the integral is evaluated from 0 to 1 it equals ½ + ¼sin(2) ≈ 0.7273243567.
Wait am I coping?
Can't we just index u = x^2?
We would get int(0-1) u^(1/2)du
Which solves to (2/3)u^(3/2)](0-1)
And if we put back in x^2 for you we get
(2/3)x^3](0-1)
Which seems identical to the expected result.
That seems to simple though so please tell me where I'm just completely wrong.
no you are absolutely correct,
@@thesuhasvasishta Oh, that's great :D
hey great video ! Is it possble so to do the integral of 1/dx ?
I really like to see the views on this vid. Hope this shows you, what your community really wants to see. I dont think more than 1% wants exponent rules.
I think we could've just solve it by saying x²=y and x=√y and so ∫¹₀ x dx² will be ∫¹₀ √y dy and the result will be (2y^3/2)/3 |¹₀ and the result of it is 2/3.
Use the absolute value for that
@@AwdcgukBecause of the square root? Also do you think my way makes sense?
The 6 dislikes are probably coming from the very misleading thumbnail featuring Feynman and x*dsinx.
I'm afraid to ask why your right bicep seems larger than your left one
Masturbation I guess.
@@PapaFlammy69 😎
@@PapaFlammy69 Aber PappaFlammy, du solltest doch wissen: 99, 100, Handwechsel!
@@PapaFlammy69what?
nice riemann-stieltjes integral👍🏻👍🏻
How did you get [ 2*i - i ] in the left bottom corner of the first page? Thx
To equalize the subdivisions of the ordinate axis, I replaced x with SQRT(y) and dx^2 with dy, and the integral value of 2/3 pops out. (limits are unchanged) I have no idea if this presents any kind of general solution. I see others did the same.
What would you do if the limits of integration don't match up?
For all the examples you do dx^2 and dsin(x) the limits, 0 and 1 when plugged in result in new limits sin(1) = 1 and (1)^2 = 1 and similar for zero.
For the Thumbnail integral the limits 0 to 2 don't match up and sinx never reaches 2.
Since this trick operates on similar principals to U-Substitution wouldn't we need to change the limits of integration, and for the thumbnail example also split the integral into two pieces since sinx is not 1to1 from 0 to 2.
I have a tip for you for better videos
* better lighting
🙏
Before watching it, my answer is 2/3
Wow, what a complicated way to use the notation.
This is how I do it:
// my definition of integral, as a cancellation of S and d, where d is implicit diff operator
S [d f] = f - f_0, d[f_0]=0
S (x d[x^2])
= S x*2*x dx
= 2 S (x^2 dx)
= 2/3 S d[x^3]
= 2/3 x^3 - f_0
= 2/3 1^3 - 2/3 0^3
= 2/3
Shouldn't the x^2 be in parentesis? I mean, dx^2 looks like dx*dx, while d(x^2) would look like 2x*dx. After all, in a second order derivative, where the notation used has no parentesis, we kinda mean dx*dx. I mean the "denominator" of the derivative, btw.
Well this seems interesting.
much much simpler would it be to just take d(x^2) = 2x*dx, then we would get 2*int(x^2 dx)|(0->1), and then basically (2*(x^3 /3))|(0->1) which is equal to 2/3...
Imagine saying this only works for real functions and then going crazy with the i's.
Okay, this makes sense. I thought you were going to integrate over (dx)^2, not d(x^2), which would really be crazy
Or just substitute u=x^2.
sqrt(u)=x
x=0 -> u=0
x=1 -> u=1
Integral(0->1)(sqrt(u)du) = (2/3)u^(3/2) | (0 -> 1)
= (2/3)(1)^(3/2) - (2/3)(0)^(3/2)
= 2/3
We should get the same value for the integral if it's just a different partition of the x-axis.
just write x=(x2)^1/2 and it becoomes a pretty simple integral
Let's do Lebesgue-Stiltjes now
Really?
Just a clickbate!
d(x^2) = 2xd(x)
S[0,1] x * 2xd(x) = S[0,1] 2x^2d(x) = [0, 1] | 2/3x^ = 2/3 * 1^3 - 2/3 * 0^3 = 2/3
17:35 I'm not sure and maybe I'm wrong, but isn't there a theorem that for a complex function a derivative exist? Does that mean complex functions are always integrateble that way?
just substitute x² with y, then you get
Integral from 0 to 1 of √y dy, which results in 2/3. Done.
Can't you just set X = x^2 (so x = √X), and then calculate the normal integral of √X d X, with X from 0 to √1? At least for this example, it seems to give the same result 2/3.
I have one question. If the intuition is that we're dividing the segment according to the function g rather than linearly, why do we still take x_i to be i times delta x?
d g(x) looks eerily similar to
d g(x) / dx so the result seems kinda obvious in that way.
d g(x) / dx = g'(x) can be rearranged to d g(x) = g'(x) dx
Gnarly farmer’s tan bro
In the Riemann integral, the partitions are equidistant. The point here is, they need not be.
Before continuing the video the shirt caught my attention, why is 57 the best prime?
Grothendieck prime
Oh I see, I didn’t even notice that it wasn’t actually a prime until now 😅 and great video btw
it actually applies in all cases. it's merely that the standard case results in f'(x) being the multiplicative identity, so it's not necessary to be aware of it:
if f(x) = x then f'(x) = 1
∫ g(x) df(x) = ∫ g(x) f'(x) dx = 1 ∫ g(x) dx = ∫ g(x) dx
I don't get what you do at 11:52, in the parenthesis i think i squared disappear and i get 2i -1 but you write i(2i -i), and of course over n3
That kind of stuff was actually "high school" stuff in germany around mid 90s - it is called integration by substitution.
Nope, Stieltjes has nothing to do with substitution.
@@PapaFlammy69 das sieht zumindest sehr ähnlich aus. Was ist der konkrete Unterschied?
Is this the motivation for the 'd' operator for the exterior derivative?
I really enjoyed this video! Keep it up! :DD
I'm confused about how at 10:15 you say x_i is still i/n when we've just redefined delta(x) to be g(x_i)-g(x_i-1), since x_i = 0 + idelta(x), don't we need to recalculate x_i using our new delta(x)?
9:55 When doing the new integral, you wrote the new f(x_i) as the same as the one in the previous integral, namely f(x_i) = f(i/n) = i/n. However, this defining of x_i used the previous value for Delta x (Delta x = i/n), instead of the new value for Delta x (Delta x = g(x_i) - g(x_i-1). How can you still say that f(x_i) is the same even after this new value for delta x?
No he can't. That's what i thought as well.
I tried by saying dx^2 = du then integrated both sides however that gives me a + c (tho it conveniently gives the right answer for c = 0)
Isn't there some sense in which d(x^2) is simply equal to 2x dx? It is not some kind of trick, they are literally equal and can be substituted for one another, which makes this wacky integral in fact very easy!
That is ONE solution, if dx² means d(x²). But if dx² would mean (dx)² the solution is 0, because the volume under a surface of infinitesimal thickness tends to zero.
Or you can just think about this as integration by parts: x dg(x)=d(xg(x))-g(x)dx and integrate both sides
Flammy try integral(0 to 1) (x*d[x]) ;[x] is the greatest integral function
int[xd(x^2)] = int[sqrt(p)d(p)] = 2/3 p^(3/2), for p from 0 to 1: 2/3
isn't d(x²) just 2xdx?
I didn't use this method. I instead use u=x**0.5, and i differiented both sides, and I applied it. It gave me 2/3
Papa Flammy make a video on ISI (Indian Statistical Institute)entrance exam questions and on CMI (Chennai Mathematical Institute) entrance exam questions.These are pure math and statistics research institutes and the question level of these institutes are even higher than Jee advance.They are challenging problems you will surely like them.I guarantee you.😊😊😊
But if g'(x) is a notation for dg(x)/dx, isn't it obviously deductible that dg(x) = g'(x).dx? Not trolling, really: what is so spectacular about that?
isn't d x^2 = 2x dx? so the integral becomes the integral from 0 to 1 of 2x^2 dx
To those who say "why don't we use d(x^2)=2xdx or let u=x^2": surely you didn't read the title. Think twice or you will be r/wooooshed. This is not a place to show how good you are at "your calculus". Flammable Maths makes serious mathematic jokes, pay some respect.
What about the product integral, where the dx is in power, does this apply there as well, or is that completely different
That doesn't sound rigorous
@@diobrando7642 wdym?
Say the line squirreljak