Hey fellas! =D Spring is offering a 20% discount off EVERYTHING at my Merch shop. Just go over to papaflammy.myteespring.co/ and use code HEATWAVE20 :)
@kentlouisoctaviano44662 ай бұрын
BRO LOOK AT THE DATE ON YOUR OLD VIDEO "OH NO DADDY" BRO THE DATE IS OLDER THAN KZbin REALEASE💀💀
@leif10752 ай бұрын
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.
@YellowBunny2 ай бұрын
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.
@shaurryabaheti2 ай бұрын
thats what I did xD
@Qaptyl2 ай бұрын
dx^2 is dx^2/dx * dx = d/dx[x^2] * dx = 2xdx
@ianmathwiz72 ай бұрын
They are equivalent as long as g is continuously differentiable and strictly increasing.
@mrpsychodeliasmith2 ай бұрын
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.
@dank.2 ай бұрын
@@Qaptyl Leibniz notation is the best lol
@alexander83112 ай бұрын
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
@PapaFlammy692 ай бұрын
fucc, u got me D:
@KinuTheDragon2 ай бұрын
But this isn't (dx)^2, this is d(x^2).
@alexander83112 ай бұрын
@@KinuTheDragon for dx = x = 0 it holds
@diobrando76422 ай бұрын
@alexander8311 wait, so dx = sin(x)
@othila99022 ай бұрын
@@diobrando7642 wrong. dx=dsin(x)
@jppagetoo2 ай бұрын
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.
@PapaFlammy692 ай бұрын
:)
@phonon12 ай бұрын
Very useful integral in probability.
@williamnathanael4122 ай бұрын
Help explain
@phonon12 ай бұрын
@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).
@phonon12 ай бұрын
@@williamnathanael412 see properties section in this wiki article as a starting point: en.wikipedia.org/wiki/Expected_value?wprov=sfla1
@diobrando76422 ай бұрын
@@phonon1isn't the generalized case the integral over Omega of X in dP?
@ekxo11262 ай бұрын
yes he should also show for a discrete probability how the integral works out to be the usual sum
@FaerieDragonZook2 ай бұрын
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
@PapaFlammy692 ай бұрын
Very nice approach! :)
@siwygameplay2 ай бұрын
This is just integration by parts btw
@renanwilliamprado53802 ай бұрын
@@siwygameplay Not exactly, it depends on your interpretation.
@tfg6012 ай бұрын
@@renanwilliamprado5380 and what interpretation is that
@vincentproplayer2 ай бұрын
Isn't this solvable by writing x as sqrt(x^2) and then treating it as int{sqrt(t)dt} with t= x^2
@hassanniaz75832 ай бұрын
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)...
@ZipplyZane2 ай бұрын
@@hassanniaz7583Sure, but I thought he'd use substitution at the end to show how to relate the two different types of integrals.
@Brandon-be2uw2 ай бұрын
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
@digbycrankshaft75722 ай бұрын
Yes. Exactly how I approached it.
@neboskryobchannel53032 ай бұрын
Why using change of variables, isn't it just Int 2x^2 dx = 2/3 * x^3 ?
@mitch5232 ай бұрын
No one gonna talk about the farmers tan?? Goes crazyyyyy
@eigenchannel-1372 ай бұрын
Papa flammy has been engulfed in flamen!
@ricardoparada53752 ай бұрын
Oh wow, I wasn’t expecting to see a Riemann-Stieltjes integral today
@flamurtarinegjakyt37452 ай бұрын
Never seen a definite integral done by definition. Very exciting video
@PapaFlammy692 ай бұрын
thx :)
@ChrisRossaroDidatticaDigitale2 ай бұрын
12:00, 2i-1.
@ofridaniel21272 ай бұрын
Yeah i thought i was crazy for a second
@adgalad252 ай бұрын
It yields the same result. But yeah.. it's 2i-1. I had to do the sum like 5 times to convince me 😂
@Legion22Cl2172 ай бұрын
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
@dank.2 ай бұрын
Exterior derivative go brrrrr
@tomfan58632 ай бұрын
that was my immediate thought. dx^2 is 2 x dx, and then do the integral.
@GeodesicBruh2 ай бұрын
Yuuuup this is the way to do it quickly.
@vincentstone72722 ай бұрын
I like this way
@discontinuity75262 ай бұрын
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
@gabrieleinsiedel18492 ай бұрын
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 😅
@bobdavid012 ай бұрын
Slight correction, the sin(2)/4 has a positive sign
@nablahnjr.67282 ай бұрын
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
@tomkerruish29822 ай бұрын
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 KZbin comment.
@KinuTheDragon2 ай бұрын
@@tomkerruish2982"I have a marvelous proof of this fact that this comment section is too narrow to contain." - Fermat if he used KZbin
@nigerianprinceajani2 ай бұрын
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.
@taterpun62112 ай бұрын
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
@bingchilling47172 ай бұрын
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
@georgasatryan38762 ай бұрын
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.
@gavinh81462 ай бұрын
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.
@fritskuijk2 ай бұрын
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
@douglasstrother65842 ай бұрын
That was fun! Very "Michael Penn".
@ben_adel34372 ай бұрын
I love evil looking integrals and the fact that we can actually do stuff to calculate them
@encounteringjack56992 ай бұрын
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.
@-.-Infinity-.-Ай бұрын
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
@abhirupkundu27782 ай бұрын
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
@epicperson99612 ай бұрын
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.
@tushi102 ай бұрын
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.
@user-SK22-calc2 ай бұрын
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.
@Djenzh2 ай бұрын
YESSSS PAPA FLAMMY IS BACK WITH SOME WEIRD INTEGRAL, LET'S FUCKING GOOOO!!!!!
@threepointone4152 ай бұрын
The Rie-womann Integral
@Mrlonely3452 ай бұрын
Its becomes really easy if we write dx^2 = 2xdx . And the answer is 2/3 simple
@kappasphere2 ай бұрын
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
@firozabegum43732 ай бұрын
"Nonlinear Partition Scaling" explains it all
@dougr.23982 ай бұрын
dx ^ 2 = 2x dx
@dougr.2398Ай бұрын
Notation is unclear is dx^2 interpreted as (dx)^2 or d(x^2)??? Both in the video and my comment!!!!
@dougr.2398Ай бұрын
If the former is intended, then an additional integral sign is needed, so the latter is assumed to be the intent and correct
@cheddastacker2 ай бұрын
Looking absolutely yoked man 🔥
@7th_dwarf5422 ай бұрын
very clear and didactic 👏 thank you for this contribution
@wagsman99992 ай бұрын
Look out, the guns are out to stay!
@als2cents6792 ай бұрын
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
@laitinlok1Ай бұрын
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.
@jewgenijmoldawski33062 ай бұрын
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
@Kunal12552 ай бұрын
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?
@marekrawluk2 ай бұрын
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.
@theangledsaxon67652 ай бұрын
Why haven’t your vids been recommended in so long?? Missing papa flammy!
@koenth23592 ай бұрын
xdx^2 = x•2xdx = 2x^2dx = d(2x^3/3) So the value of the given definite integral is 2/3 - 0 = 2/3.
@ready1fire1aim12 ай бұрын
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
@wjalp2 ай бұрын
Watched until the end! Also the assignment's answer is = 1/2+(sin(2))/4 :DD
@loganhagendoorn63272 ай бұрын
love your content, thats such a smart way to solve this!
@RuthvenMurgatroyd2 ай бұрын
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.
@GodzillaFreak2 ай бұрын
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.
@thesuhasvasishta2 ай бұрын
no you are absolutely correct,
@GodzillaFreakАй бұрын
@@thesuhasvasishta Oh, that's great :D
@jebarijihed2 ай бұрын
hey great video ! Is it possble so to do the integral of 1/dx ?
@dragileinchen14852 ай бұрын
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.
@melonking97522 ай бұрын
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.
@Awdcguk2 ай бұрын
Use the absolute value for that
@melonking97522 ай бұрын
@@AwdcgukBecause of the square root? Also do you think my way makes sense?
@kostasch56862 ай бұрын
The 6 dislikes are probably coming from the very misleading thumbnail featuring Feynman and x*dsinx.
@thekingofgindio2 ай бұрын
I'm afraid to ask why your right bicep seems larger than your left one
@PapaFlammy692 ай бұрын
Masturbation I guess.
@thekingofgindio2 ай бұрын
@@PapaFlammy69 😎
@celestindupilon27732 ай бұрын
@@PapaFlammy69 Aber PappaFlammy, du solltest doch wissen: 99, 100, Handwechsel!
@lithunoisan2 ай бұрын
@@PapaFlammy69what?
@marcosmaldonado78902 ай бұрын
nice riemann-stieltjes integral👍🏻👍🏻
@juliank.35222 ай бұрын
How did you get [ 2*i - i ] in the left bottom corner of the first page? Thx
@crypticcrazy3672Ай бұрын
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.
@appleducky52342 ай бұрын
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.
@rajdeepsingh262 ай бұрын
I have a tip for you for better videos * better lighting
@BikeArea2 ай бұрын
🙏
@melonking97522 ай бұрын
Before watching it, my answer is 2/3
@robfielding85662 ай бұрын
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
@jorgeperezmolina2235Ай бұрын
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.
@mr.inhuman79322 ай бұрын
Well this seems interesting.
@alali28852 ай бұрын
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...
@evanwilliams73762 ай бұрын
Imagine saying this only works for real functions and then going crazy with the i's.
@txikitofandango2 ай бұрын
Okay, this makes sense. I thought you were going to integrate over (dx)^2, not d(x^2), which would really be crazy
We should get the same value for the integral if it's just a different partition of the x-axis.
@hellohello-tf9vc2 ай бұрын
just write x=(x2)^1/2 and it becoomes a pretty simple integral
@Phaust942 ай бұрын
Let's do Lebesgue-Stiltjes now
@nikitaluzhbin89822 ай бұрын
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
@emilleonardelli40472 ай бұрын
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?
@m.h.64702 ай бұрын
just substitute x² with y, then you get Integral from 0 to 1 of √y dy, which results in 2/3. Done.
@tmlen8452 ай бұрын
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.
@ViewtifulSam2 ай бұрын
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?
@flutterwind76862 ай бұрын
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
@sobhhi2 ай бұрын
Gnarly farmer’s tan bro
@topquark222 ай бұрын
In the Riemann integral, the partitions are equidistant. The point here is, they need not be.
@Hussain-px3fc2 ай бұрын
Before continuing the video the shirt caught my attention, why is 57 the best prime?
@PapaFlammy692 ай бұрын
Grothendieck prime
@Hussain-px3fc2 ай бұрын
Oh I see, I didn’t even notice that it wasn’t actually a prime until now 😅 and great video btw
@sumdumbmick2 ай бұрын
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
@brouchoАй бұрын
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
@bastianfrom772 ай бұрын
That kind of stuff was actually "high school" stuff in germany around mid 90s - it is called integration by substitution.
@PapaFlammy692 ай бұрын
Nope, Stieltjes has nothing to do with substitution.
@bastianfrom772 ай бұрын
@@PapaFlammy69 das sieht zumindest sehr ähnlich aus. Was ist der konkrete Unterschied?
@APaleDot2 ай бұрын
Is this the motivation for the 'd' operator for the exterior derivative?
@wjalp2 ай бұрын
I really enjoyed this video! Keep it up! :DD
@nalydify2 ай бұрын
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)?
@thomastd52 ай бұрын
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?
@bantix99022 ай бұрын
No he can't. That's what i thought as well.
@poyrazpekcan66352 ай бұрын
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)
@MathematicFanatic2 ай бұрын
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!
@ninireak73252 ай бұрын
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.
@siwygameplay2 ай бұрын
Or you can just think about this as integration by parts: x dg(x)=d(xg(x))-g(x)dx and integrate both sides
@nerdygeek89472 ай бұрын
Flammy try integral(0 to 1) (x*d[x]) ;[x] is the greatest integral function
@elcolicous2 ай бұрын
int[xd(x^2)] = int[sqrt(p)d(p)] = 2/3 p^(3/2), for p from 0 to 1: 2/3
@boranxiii2 ай бұрын
isn't d(x²) just 2xdx?
@bahaloicperrial89642 ай бұрын
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
@AmlanSarkar-wr2pr2 ай бұрын
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.😊😊😊
@MarcusPereiraRJ2 ай бұрын
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?
@thaianotran33152 ай бұрын
isn't d x^2 = 2x dx? so the integral becomes the integral from 0 to 1 of 2x^2 dx
@Nerdwithoutglasses2 ай бұрын
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.
@fahimnabeel6062 ай бұрын
What about the product integral, where the dx is in power, does this apply there as well, or is that completely different