The MVP of Integrals
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- Опубліковано 8 гру 2023
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Today integrate (x^69-1)/ln(x) from 0 to 1 using the Leibniz Rule for integration aka. Feynman's Technique. Enjoy! =D
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As a physics boi, I saw the thumbnail and was like “fuck yea we getting into the Feynman shit today”
I said wtf immediately I saw it 😂
You can get the same result by transforming it into a double integral of 1/xlnx over the region 0 < x < 1, x < y < x^70. You can the flip the order of integration to get the new boundaries 0 < y < 1, y^(1/70) < x < y. By integrating the first integral using a u-sub, the second integral will simplify to an integral from 0 to 1 of ln(70), which is equal to ln(70).
Incorrect
Same integral with power 68 would be better because you actually get 69 lol
Actually you’d get ln(69)
But no succ'ing it
Then how you'll create thumbnail
What helps to see how to apply the Feynman trick is the ln x in the denominator. One of the cases where the Feynman trick simplifies the integrand is when it multiplies it by a factor that helps you integrate.
So, in the case where what's annoying you is something in the denominator one idea is to use the Feynman trick to cancel it out.
So nice to see Papa using the trick that Papa Feynman taught to Papa Leibniz.
Lol the "don't be a d³/dt³(x->)"
it means "don't be a jerk"
Why does it mean that? can you explain pls this is not the first time seeing this and I still got no clue
@@mohamedjaballa6811
The derivatives of position vector are :
1th:velocity
2th : acceleration
3th :named jerk
And so on
You can see it in matt parker's video
its called "an unexciting video about distance derivatives"
I love how he eventually don't write 70
You killed me with log(succ(69)). Legendary integral
Feynman trick is always fun!
The 69 was clearly an arbitrary number. Also, using e^u=x you it is non elementary. That makes it a clever Feynman, easy enough for me to do in my head. It's fun.
Nice!
I found the same result with another (yet longer) method:
1) Substitute u=log(x) => I=integral (exp(69*u)-1)*exp(u)/u from -inf to 0
2) Expand (exp(69u)-1)/u with taylor serie and switch sum and integral
3) By another substitution you get the gamma function of k on the numerator which is divided by k!, so you find the serie of log(1--69)
OK it's more complicated, but it"s so satisfaying when everything simplify and you recognize a famous serie!
🎼It's beginning 🎶to look a lot like Christmas 🎵
We can use definite int property intf(x)dx from 0 to a intf(a-x)dx
I did it by noticing that the integrand is the integral from 0 to 69 of x^t with respect to t. After swapping the order of integration, it becomes trivial.
I love leibnitz
Me too, they make some nice biscuits/crackers!
Nice one papa
Yeah, you should also show that you can change order of derivative over t and integral over x. It's not always true you can do it. Either mention a theorem you're using or prove it.
Here we can do it because for f(x,t)=x^t/ln(x) we have d/d^2 t (f(x,t)) = ln(x) *x^t and it is bounded on sex (0,1)x(t-epsilon(t),t+epsilon(t)) for each t and epsilon taken as small we want for each t>0. It's enough to prove that certain integral in proof is as close to zero as we want. For t=0 the direct proof doesn't work, but it's not needed. Continuity of I_t in 0 is enough.
Feynman op
Where did u get the chalkboard from? Im considering getting one :)
6:35 Aren't we supposed to integrate I'(t) instead of I(t)? That should give us I(t) at upper bound-I(t) at lower bound, shouldn't it?
Of course! My bad, small typo, thanks for pointing it out! =)
@@PapaFlammy69 Really trivial mistake, not a big deal at all. Keep going!
You no want beef jerky?
nice
lover bound ;))
-68
Integrate n times with respect to x, f(x)=x^x^x^x^x^x^x^x^...
Papa Flammy. It is cold. Why you dressed so lightly?
You expect us to follow along when you're distracting us with your guns??
Bruh 😂