This Integral is Nuts

Can you solve this seemingly simple Integral of x integrated with respect to x²? Let's cover the Riemann-Stieltjes Integral, shall we? :) Enjoy! =D

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0:00 Intro
1:25 Riemann Integral
7:43 Stieltjes Integral
17:31 The Cool Way

[alexander]

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

[YellowBunny]

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.

[FaerieDragonZook]

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

[jppagetoo]

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.

[vincentproplayer]

Isn't this solvable by writing x as sqrt(x^2) and then treating it as int{sqrt(t)dt} with t= x^2

[mitch]

No one gonna talk about the farmers tan?? Goes

[gabrieleinsiedel]

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 😅

[LegionCl]

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

[gavinh]

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.

[flamurtarinegjakyt]

Never seen a definite integral done by definition. Very exciting video

[ricardoparada]

Oh wow, I wasn’t expecting to see a Riemann-Stieltjes integral today

[nigerianprinceajani]

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.

[georgasatryan]

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.

[nablahnjr.]

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

[discontinuity]

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

[nikitaluzhbin]

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

[bingchilling]

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

[kappasphere]

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

[taterpun]

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 <= x). Since A is dcts when adding next term and zero else, the dA’s are nonzero only at discrete points, leading to an integral representation: sum x<=k<=y f(k)ak = integral x to y f(t)dA(t). You can then use ibp to prove Abel’s summation formula

[encounteringjack]

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.