Math for fun: integral of ln(x) from 1 to ? is equal to 2

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I want the area under the curve y=ln(x) from 1 to some number t to be 2, but how can we achieve this? Not only do we have to use calculus integration by parts, but we also need to use the Lambert W function to solve the resulting equation for us.
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#blackpenredpen #math #calculus #apcalculus

[@blackpenredpen]

Get started with a 30-day free trial on Brilliant: 👉brilliant.org/blackpenredpen/ ( 20% off with this link!)

[@IbrahimaBah-hs8do]

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[@salehsattouf2320]

In Finding Nemo they could have just used the Lambert-W function and then they would easily find Nemo

[@maximilianarold]

They need to heve Nemo in the exponent first and I doubt they had that

[@bodesshorts8640]

Yay the fish is back

[@shophaune2298]

Fun fact, the general form with an area of N under the curve is t = (N-1)/W((N-1)/e)

[@ekut1922]

the fish function is kinda bad imo, you give away a medium fish and a small fish for a big fish but I personally find that since you have to eat the fish in one go or it'll go bad the m fish and s fish serve a better purpose for consumption and if you are collecting fish, I doubt you want to decrease the number of fish you have unless your tank is running out of space, but bigger fish generally need more space to be healthy so I wouldn't use this as a solution even then

[@mitochondria1674]

Ahh isnt that fishy

[@gibson2623]

@@mitochondria1674 LoL

[@romanbykov5922]

that's what I thought right away when I saw the question :)

[@acetylsalicylicacid]

What an interesting problem! I could see things like this being used for many applications, like how long you need to drive to drive a certain displacement.

[@mhm6421]

For anyone wondering you can simplify the result like this: = e^(W(1/e) + 1) = e * e^W(1/e) = e * W(1/e) e^W(1/e) / W(1/e) = 1 / W(1/e)

[@riccardomarino9786]

I didn't understand why e*W(1/e)e^W(1/e)=1

[@vinko8237]

I got the last result first ;)

[@munkhjinbuyandelger]

@@riccardomarino9786 because, by definition, W(x)*e^W(x)=x. It is the true property of Lambert’s W function.

[@sandyjr5225]

Even I got the last result first.

[@kindacringengl]

Hey bprp, I just wanna say a massive thank you for your videos, not only are they super entertaining for me but when I started watching them a couple years ago I had no idea what was going on, not about calculus, complex numbers or anything. But slowly I started picking things up and I started to basically get a online class from you, whilst being entertained with great content. Now that my curriculum is more lined up with your content and european style (IB by the way), I actually use the stuff you use a lot and it helps me get really good grades, so yeah sorry for the paragraph but thanks :)

[@therealist9052]

Wow first real practical application of the Lambert W function I've ever seen that could come up in everyday calculus! Thanks so much!!

[@simonteo6414]

Thanks to your videos I was actually able to solve this one on my own, never thought I would like math this much. Thank you for reigniting my interest in math :)

[@sovietwizard1620]

It's crazy how after watching so many of your videos, I can solve problems like these without any problem at all in a matter of seconds. Lambert W fuction really is so good afterall.

[@blackpenredpen]

That’s awesome, cheers!

[@sumedh-girish]

Paused the video, tried it first, comes out approx 3.591. 1 / (Fish 1/e) Edit : Fixed typo

[@sumedh-girish]

yay a heart!

[@Trust_the_brain]

Fun fact, I actually did this question and t can also be write as the 1/LamberW(1/e), i feel like the method you used was harder than mine though but its fun to see that e^( LampbertW(1/e) + 1) is the same as 1/LamberW(1/e) seems like it wouldn't be true😀

[@romanbykov5922]

perhaps it's just not nice to have W in the denominator :)

[@XJWill1]

It is actually a simple consequence of the product log function being an inverse of f(x) = x * exp(x) In this channel, the most common usage is W( f(x) ) = x but you should not forget the other usage which is f( W(x) ) = x W(x) * exp( W(x) ) = x then you simply divide by W(x) to get exp( W(x) ) = x / W(x) and that identity can be applied to this problem to show the equality that you mention: exp( W(1/e) + 1) = e * exp( W(1/e) ) exp( W(1/e) + 1) = e * (1/e) / W(1/e) exp( W(1/e) + 1) = 1 / W(1/e)

[@Trust_the_brain]

@@XJWill1 I used a more elementary friendly way to get to my result here it is : We have gotten to the point x ln(x) - x = 1, we use logarithm rules and expand the x to ln(e^x) now we get ln(x^x) - ln(e^x) = 1 lograithm rules: ln(x^x/e^x) = 1 take out the ln, we get x^x / e^x = e, now take denominator to the other side and multiply: x^x=e^x * e now take the root of x on both sides: x= e^(x/x) * e^(1/x) we get x= e^1 * e^(1/x) take the e^1 to the other side: x/e = e^(1/x) , now take the x to the other side: 1/e = 1/x * e^(1/x) as you might have noticed you can take the lambert W on both sides, W(1/e) = W(1/x * e^(1/x)) simplify W(1/e) = 1/x hence x= 1/W(1/e), yours is obviously much more simpler but as a highschooler myself, I find that this way is much simpler, though I see how your idea provides for a much better and simpler, and thanks for the proof btw.

[@Trust_the_brain]

@@romanbykov5922 Hmm not sure, 1/x is usually a pretty friendly function

[@cdkw2]

Such a common problem man, thank you for the solution.

[@farhansadik5423]

t can be further simplified to just 1/W(1/e). I still have not taken full calculus classes but I do love the lamber W function

[@mjpledger]

Instead of using the lambert function and then solving numerically, you can just use recursion to solve numerically. Since t.lnt-t=1 then t=(1+t)/ln(t). Let ans equal some starting number e.g. 10 and then ans=(1+ans)/ln(ans) and repeat until it converges e.g. 4.777239, 3.694211, 3.592233,] 3.591122, 3.591121. It converges pretty quickly and easy to do on a calculator.

[@abcd-m1n7w]

Evaluate the surface integral ∬SF⋅dS for the vector field F(x,y,z)=x^2i+y^3j−xzk, where S is the part of the paraboloid 0.6+1.1x^2+y^2−z=0 that lies above the disk x^2+y^2≤1 and has upward orientation

[@lazarusisaacng]

Nice question and good answer, love it.❤

[@gribvkvadrate6233]

please try this equation: e^i^x=x

[@mhm6421]

You mean e^(ix) = x or e^(i^x) = x

[@adb012]

@@mhm6421 ... As the OP wrote it it means the second.

[@gribvkvadrate6233]

​@@mhm6421, the second one, the tower of powers

[@KingGisInDaHouse]

You can also use parametrical equations to integrate ln x y=t x=exp(t) $ydx dx=exp(t)*dt $t*exp(t)dt You don’t get out of integrating by parts but it’s a little more intuitive.

[@AubreyForever]

Can you do a clear video of the Lambert function for high school students? Thank you.

[@andrec.2935]

Gosto muito da condução dos seus vídeos, da forma como faz matematica.

[@YanbekArt]

I knew the answer was going to use Lambert W function the second I saw the problem)

[@onetwo7191]

How? I’m new to lambert W, how do you recognize it so early?

[@staticchimera44]

@@onetwo7191 It's a practice thing, kinda like how you can expect to solve a quadratic whenever powers of 2 are involved. The Lambert W function appears naturally in natural log and e problems, and since the integral of ln(x) is probably some weird product of natural logs and such, it's not a bad guess that the Lambert W function would come into play

[@sweettoy3824]

I did the integral then used Newton-Raphson iteration to approximate t as 3.59, which is correct to 3 significant figures. I think that's more satisfying than using the W.

[@the_nuwarrior]

​@@sweettoy3824W is an analitic function, sometimes is better to have one

[@eve_the_eevee_rh]

Integration of lnx, by parts: u = lnx, u' = 1/x v' = 1, v = x 2 = [uv - integration of u'v] 2 = [xlnx - integration of 1] 2 = [xlnx - x] Upper bound is t, lower bound is 1 2 = tlnt - t - ln1 + 1 2 = tlnt - t + 1 1 = tlnt - t 1 = t(lnt - 1) Iiii have no idea how to simplify and/or solve this

[@idjles]

I guarantee the answer uses Lamdba W function!!

[@DotDotEight]

you mena lambert?

[@sowndolphin5386]

you mean mean?​@@DotDotEight

[@DotDotEight]

@@sowndolphin5386 yes my bad, i didnt notice that typo

[@neilg2256]

The result can be simplified to 1/W(1/e)

[@MichaelRothwell1]

My solution was pretty similar to that in the video, but I put the answer in a neater form (as some others have in their comments) ∫₁ˣln t dt=[t(ln t-1)]₁ˣ (using integration by parts) =x(ln x-1)-1(0-1) =x(ln x-1)+1 So we need x(ln x-1)+1=2 x(ln x-1)=1 x(ln x-ln e)=1 x ln(x/e)=1 (x/e)ln(x/e)=1/e ln(x/e) e^[ln(x/e)]=1/e So (as 1/e>0) there is a unique real solution, which is ln(x/e)=W₀(1/e) x/e=e^W₀(1/e) [⇒x=e^[W₀(1/e)+1] ] But W₀(a)e^W₀(a)=a, so e^W₀(a)=a/W₀(a) So x/e=(1/e)/W₀(1/e) x=1/W₀(1/e).

[@TanmaY_TalK]

Lambert W function ❎ bprp's fish function ✅ 😊

[@joyneelrocks]

Whenever I see problems that use the Lambert's W Function, I just say, nah too lazy, and end up using the Newton-Raphson Method 😂😂

[@moktamoni1373]

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[@Samiul_007]

He's from Taiwan

[@darcash1738]

Dam I saw it from the thumbnail and thought we got to choose the lower bound too and had an answer since we could freely choose one Not my first Lamberto tho. I see ln, I know. Get integral and factor out a b to be nicer. b(lnb-1) = 1 Now set up lambo by creating an e^ term. Then have it match what’s next to it by multiplying e^-1. e^(lnb) (lnb-1) = 1 e^(lnb-1) (lnb-1) = e^-1 Lambo: lnb-1 = W(e^-1) Finish: b = e^[W(e^-1)+1]

[@dd_hd2]

Please make video on Fourier transform.

[@arashsoleimany4688]

plz make a video and explain a simple solution for Morley's theorem.🙏

[@janeknowakowski5732]

You can put the answer in an alternate form as follows: t=1/W(1/e), which looks better in my opinion.

[@MohamedBenbrahim-u9m]

Integral entre 0and1(-ln(x))^1/2

[@identify3195]

i dont know about W function cuz i am in high school . i tried and got t should be between 3 and 4

[@TerenceTao-w1i]

Try this insanely hard integral : Integral between 0 and pi/2 of tan(x)^(1/(x^x))

[@jackkalver4644]

I got the solution, but I had trouble checking it.

[@mribang9128]

hey bprp, could you try making a vid about tetralog someday? I wld appreciate if you do.

[@magnusmalmborn8665]

Could you use series expansion of ln(x) instead of W?

[@KavinKesav]

Bro plz try this d/dx(log base x of n)

[@EyeSooGuy]

Hey Steve! Show us a HARD analysis problem - at the 899 (end of PhD or even postdoc) level. 😈

[@celsonguenha3845]

Your help please Integral of ((x^2 + sin x)/(x^2 +1))dx

[@mangalesh7936]

Though I know how to solve this I came to see because to see bprp's Lambert fish function. 😅

[@Samiul_007]

Can you solve for y? x = log(base b) of [ (W(y' /lnb)) /lna ]

[@MathAdam]

Just checking in to see how you pronounce ln x.

[@blackpenredpen]

😆

[@Engy_Wuck]

W still feels like cheating. Or "here magic happens".

[@user232-sbydse]

Is there any way to expand ( a+b+c+d)^7

[@azzteke]

Troll question!

[@pkvlogs5078]

Well did it after expanding logt upto t-1 or even upto of power 3 from there we can get an approx value 1+ (2)^1/2 instead using lambert transformation

[@IlyesBenahmed-vf6gi]

Try this very difficult integral : ∫_0 ^(pi/2) tan(x)^(1/x) dx. So you want a VERY difficult question ? Solve the differential equation : ∫_0 ^pi ( ∫_y' ^y'' ) dx dt is y'''(e).

[@sarv1494]

X = e^(W(3/e) + 1)

[@mhm6421]

no

[@sarv1494]

@mhm6421 i saw end of the video, i got it wrong sorry

[@martinb.3997]

I've even asked WolframAlpha this question, but it seems to bug out and ask me to reask. When I formulate it another way, it just shows me a numeral approximation but nothing else. Could you try and solve it, please? Thank you! x^x^x = 2

[@scottleung9587]

Nice!

[@Relativity-d2]

B = (84 - A)/(1 + A) (A and B both are positive integers. A, B > 1 ) then find the value of AB.

[@archangecamilien1879]

I have to say, I don't remember what the antiderivative of ln(x) was...I suppose if it was that straightforward, lol, there wouldn't be a video about this...maybe it doesn't have a straightforward derivative?...I don't know, lol...

[@usptact]

What if I want that area to be i?

[@Micheal2009-u4o]

can you please come to Iran someday? pleaaaaaaaaase

[@josephnaylor]

i tried it; my brain was hurting i was so angry but i got 3.5911223...

[@josephnaylor]

I don't know the relationship between this number and e or ln

[@lorenzopattaro3529]

I don't think I've actually understood what is the W function

[@Earthzooka]

It's called Lambert W function, also known as the Product Log function. I'm not an expert on the topic myself, but I think of it as an inverse function of f(x) = xe^x

[@Samir-zb3xk]

Its just a function we define as the inverse function of y=xe^x; simalar to how if we have y=e^x, its functional inverse is y=ln(x) This function allows us to solve equations that can't be solved using normal algebra techniques. Such as equations where we have a variable outside and inside an exponent or logarithm.

[@Bayerwaldler]

bprp has a couple of videos about the Lambert W function on his channel.

[@lawrencejelsma8118]

​@@Bayerwaldler... He is incomplete. It is wrong to never evaluate W(x)s by calculator or computational for higher level Numerical Analysis Senior level University computer math courses. How did he find W(1/e) to get his approximate value of 3.59112? There are no lookup tables that will equal a log(x), ln(x), or trigonometry type equivalent table for x values. From Wikipedia the Lambert Function is a summation formula. W(x) = summation of n=1 to infinity [((-n)^(n-1))/(n!)](x)^n for |x| < 1/e. Wikipedia states outside this range the solution for 1/e

[@bunthornpgoneshop6603]

Can you prove that why (-)(+)=- (-)(-)=+

[@GurmesaAmeni]

Please Make a video with a youngest an and famous mathematician and physician subbrno isaac bari

[@azzteke]

Troll, go home

[@tommasotiberi5666]

Math for fun fact: integral of ln(x) from 0 to 1 equals -1

[@mrpineapple7666]

I am the first one commenting on your video

[@white_145]

I dont understand the W function, can someone explain? It doesnt seem to give any information at all. Why not just invent some function K so that t = K(2)?

[@prakashhuilgol6445]

Please make a detailed video on fields medal

[@leonardobarrera2816]

Stop making easy exercices, I feel that I am good at math

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The first one to comment on! 🕶

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I am the 300th viewer

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first

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I think i'm first ?

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@@jackwoods7070 So what? Quoting from a book that is more than 2000 years old and was written by ignorant people proves nothing.

[@artsmith1347]

@@bjornfeuerbacher5514 Why do you say they were ignorant? Have you achieved omniscience, so no one will ever be able to say you were ignorant?

[@alexandermorozov2248]

Нашёл t≈3.5911215, решал через функцию Ламберта W(x).

[@TheBoeingCompany-h9z]

normal person : W(xe^x) = x bprp : W(🐟e^🐟) = 🐟

[@TheFastProgrammer]

the lambert w function is bprp's favorite math concept

[@a.b3203]

Lambert what?

[@General12th]

The W function is my favorite function I never learned about in a proper math class!

[@alexandermorozov2248]

Если использовать формулы для функции Ламберта, то ответ можно упростить: t=e^(1+W(1/e))=e*e^W(1/e)=e*(1/e)*1/W(1/e)=1/W(1/e) Получаем: t=1/W(1/e) 😜

[@alexandermorozov2248]

Здесь использовали формулу: e^W(x)=x/W(x)