Bruh dx² is basically 2x dx

Was this a 3 hour test bwcause if it is 2 hours it is IMPOSSIBLE to get full credit

what e=2?

Say the line squirreljak

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?

23 minutes wasted, while you can solve it in 30 secs just by substituting x^2 to y, then x = y^0.5

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

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?

Ingenieurs be like 😐 with 😐 = 😂

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

How did you get [ 2*i - i ] in the left bottom corner of the first page? Thx

İs 0.0000000000000000000000000000000001 mecosecond

How is x_i=i/n for the new integral? Doesnt it assume a linear scaling?

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

Just use the fact that d(x^2)=2xdx

Me using the /execute command in minecraft:

i need the sliding set up at home so i donr have to pop a squat to write on the bottom

Well i suck at math so even if his work is wrong i wouldn't know

This isn’t quantum calculus. Bad title.

That was fun! Very "Michael Penn".

Is it possible to learn this power?

wow

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)?

Work in your Tanktop next time, fu** T-Shirts!

Hello

(a+b)²=a²+b²+2ab Like the C never forget the 2ab

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.

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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.

This also proves division is the inverse of multiplication! :DD

nice riemann-stieltjes integral👍🏻👍🏻

Funny apple joke HAHAHA

Watched until the end! Also the assignment's answer is = 1/2+(sin(2))/4 :DD

Certainly a hmmmmmm moment

ln(cos(x))ln(sin(x))/sqrt(tan(x)) dx please

Product rule for derivative ?? d(x^2) = 2xdx then your integral is solved ? You redemonstrated the antiderivative of a monomial with the definition of the integral >:[ Nice exercise though

Should have used a double integral.

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

I really enjoyed this video! Keep it up! :DD

isn't d(x²) just 2xdx?

I'm jealous...

xdx^2 = x•2xdx = 2x^2dx = d(2x^3/3) So the value of the given definite integral is 2/3 - 0 = 2/3.

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

I didn't get the 3rd and the 1st from the bottom

At pi² = g it was already over

here’s what i do i just design a binary full subtractor digital circuit and convert it to a boolean function with two inputs, then i plug those inputs in and work out the solution

This is sad to see

Hey man they’re just trying to use scare tactics… if anything you can counter sue and you’ll win

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.

The story of the Grothendieck prime 57 (see it on Wikipedia y'all) more than makes up for the shirt. I would have made a new shirt