My all-in-one calculus problem

[@blackpenredpen]

Learn more calculus on Brilliant: 👉brilliant.org/blackpenredpen/ (now with a 30-day free trial plus 20% off with this link!)

[@maxvangulik1988]

i like how the limits of integration are actual limits

[@isavenewspapers8890]

I've always preferred the term "bounds of integration". I mean, considering that we're already using the word "limit" for something else in calculus, doesn't it make sense to use a different word here?

[@prabhakarsingh6821]

​@@isavenewspapers8890 using the word "bound" just makes so much sense....idk why most people don't call it that

[@Elprofium]

nothings better than solving an integral on Christmas's

[@hanckNCR]

its christmas?

[@anadishrivastava4804]

Agreed

[@michalkrawczak]

​@@hanckNCRit's always Christmas if you have integrals to solve

[@Aaron_1112]

​@@michalkrawczak😔

[@aninditabasak7694]

And Christmas also happens to be the birthday of Newton, who invented calculus.

[@trelosyiaellinika]

I've graduated from a mathematical school and even went to Mathematics faculty at the university for a year before changing my mind and becoming a general surgeon... It was a very tough decision as there was no scientific material that didn't interest me at the time... But maths has always remained my love and mania and I've always benefited from the knowledge while creating various complex macros for my work... However, I had almost forgotten most of its juicy parts... It's been more than 36 years after all! Now, I am retired and very much enjoy your videos, remembering and solving them in parallel... It charges my batteries and gives me a sense of satisfaction like winning a chess match! Thank you very much! You are doing a great job!

[@blackpenredpen]

Thank you so much for the comment!

[@chilli881]

Imagine checking your socks at early morning and finding a paper with this integral written and a message from Santa : "Integrate the above to receive gift"

[@mchanc]

well worry no longer my friend.

[@satyam-isical]

Every scary problem is not necessarily tough & Every tough problem isn't scary😊

[@EyeSooGuy]

😱(lol)

[@the_llaw]

Only thing scary is his face in the thumbnail 😂😂 but fr tho great video

[@AdityaMishra-nd7cq]

Is this KZbinr from China if yes then the china is my favorite country 😂

[@d3generate804]

@@AdityaMishra-nd7cq he is a Taiwanese living in america

[@lunam7249]

chuck norris says ..."hold my beer"

[@cheerio662]

Been watching you for 2-3 years now as a highschool student and could finally solve on of your all-in-one questions by myself! Feels great to go from knowing nothing and just liking the magic numbers to solving something that looks scary (but really wasnt) all by my lonesome. Thank you for the content you provide!

[@juxx9628]

Ok. Trying first before seeing the video. Step 1: Evaluate limits. On the bottom one, use L'Hopital rule and get (1/x)/(1/2√x). Simplify and get 0. The top one use L'Hopital rule to get (1/2√x)/(1/x). Simplify and it diverges. Step 2: Derivative. Just use the chain rule twice. f(y)= y² y(t)= sint t(x)= t² df/dx = df/dy • dy/dt • dt/dx = 2y • cost • 2t Recall the definitions of the variables: 2•2x•sinx•cosx Step 3: Power series. Recall the Maclaurin series for e^x, then put x² as the input. That easy. e^x². Step 4: The monster. The integral looks like 0-inf∫ 2•2x•sinx•cosx• e^-x² dx. Use substitution j=x², dj=2xdx (bounds of integration stays the same and we already have dj in the integral) =0-inf∫ 2•sinx•cosx•e^-j dj Recall doble angle formula for sinx and name the integral I: 0-inf∫ sin(2j)•e^-j dj = I Use IBP or DI method, just the same: D: + sin(2j) - 2cos(2j) + -4sin(2j) I: e^-j -e^-j e^-j After the setup, this ends like: I = (sin(2j)•e^-j)]0-inf + (2cos(2j)•e^-j)]inf-0 - 4I Notice that first term goes to 0 and in the second term I changed the bounds thanks to the minus sign. Now, in the second term, the limit as j approaches 0 is 2 and when j approaches infinity is just 0 thanks to the exponential and the squeeze theorem. So, finally: I = 2 - 4I 5I = 2 I = 2/5 Thanks for reading, love you.

[@samueljehanno]

Interesting

[@cemsaglam9241]

i've just realized by reading your comment that IBP is short for "integration by parts"

[@juxx9628]

@@cemsaglam9241 Yeah, it's a confusing way to write it. I first got confused because in spanish it is just simply despicted as integration by parts or "the cow" (la vaca) because of some mnemotecnic to remember IBP.

[@valentinvanhees8690]

i really liked this!! my first really hard integral that i solved first try! would love to see more power series-integrals

[@o_s-24]

All of calculus 2 summarized in 11mins. Awsome!

[@xum0007]

I’m only a freshman so I’m taking algebra 2 honors right now. I must say this looks way harder than what I do in class right now (which is a pretty low standard) but if you’re interested in the subject it shouldn’t be too bad.

[@matheusdossantos9252]

​@@xum0007Algebra II also called "Linear Algebra"? After the diagonalization content it can get a little more complicated depending on your teacher.

[@michellekagansbu]

@@matheusdossantos9252 I don't think he means linear algebra

[@MokshitArora.]

That e^x² at the denominator was great . I was thinking it to be some different series and was thinking to use limit as a sum (converting an infinite sum to definite integral)

[@M7RAA]

how did he get that though?

[@MokshitArora.]

@@M7RAA use tailor series expansion on e^x you will get the series or if you know series of sine and cosine then also you can get that After that replace x with x² and you will get the mentioned series We can reverse it also by finding function with series by writing it as a limit on summation and then converting into Reimann sums then integrating

[@PRIYANSH_SUTHAR]

This guy can intimidate you with full innocence.

[@sergeygaevoy6422]

And it is a Laplace transform in the end.

[@andripula8986]

to end with a repeating integral, brilliant problem!

[@TsukkiSenpai727]

So what’s the answer to 1/5 + 1/5 ? BlackPenRedPen: sooo actually

[@aubertducharmont]

When you got to the final form of the integral, I would just use contour integration to get the answer. I dont like doing that much integration by parts. And also that series in the numerator arent necesserily described by the e to -x squared formula. As you wrote only a finite number of parts, in this case four, there is an infitnite amount of formulas for these four parts of the series. One could pick that after x^2/6 would come 69 and find a formula for this, with use of the Gregory-Newton formula.

[@phillipalter6499]

My calc professor will love this, thanks

[@arvinpersaud156]

I'm going into precalc next year and I'm kind of excited to be starting calculus. I've been watching these videos for a few years now, and I feel accomplished that I can solve this by myself. Thank you for all of these videos, they give some really interesting equations, and I've learned a lot from them. I hope you keep making quality videos.

[@hsod0]

You are really awesome!!! Actually, thank you for what you are doing, I'm into mathematics even more because of your videos and I'm really having fun watching them. Please, keep it up. These videos really make my day

[@tambuwalmathsclass]

Wow, incredible. 💪 But isn't the final answer supposed to be -2/5 ?

[@ABHIGAMING-yo9my]

Bro function is always positive so answer should be positive

[@joshhh___]

@@ABHIGAMING-yo9myThe function f(x) = sin(2x)e^(-x) is not always positive on [0, inf), but ∫₀^∞ f(x)dx is still equal to 2/5.

[@stapler942]

"Two limits, a derivative, a power series, and an integral wander onto a board..."

[@myththelegendtyson]

We should have an advent of integration. Each day a new integral problem

[@TypoKnig]

Merry Calcu-mas!

[@RefreshingShamrock]

SLOW DOWN ONE HOLIDAY at a time! We haven't even made it past Thanksgiving yet!

[@blackpenredpen]

My bad 😆

[@dinokiller9186]

The numerator was easy but I couldn't guess the denominator part 👍👍

[@yencheonglee5940]

This question is simple. The limits can be found easily, next I replace t=x^2 and come out with \int e^{-t}sin(2t) dt, and then I solve lim_{s -> 1} Laplace transform of sin(2t) by subtracting s=1 in the result.

[@Siddhartha.Chatterjee]

I have not watched it yet... But please tell me it's 2/5 Edit: Ok, I messed up somewhere at plugging infinity at the last part (for some reason I forgot that even with infinity, the sin & cos function would be finite, and applied L'Hopital, somehow ended up having I=-4I, allowing me to say I=0 at x->infinity), but anyways the answer still ended up the same....

[@jakehu]

The kid who just guesses 2/5😂

[@pedri_meet]

That was great!! It's like quick revision

[@aimgaming4744]

Love these kind of questions, keep it up!

[@pekorasfuturehusband]

I’ve been wanting another all in one problem for a while now, thanks for the early present!

[@softllamaspajamas]

What a thrilling problem! I’ll give it a go myself closer to Christmas!

[@PhysicalScienceInSinhala]

It's amazing 😃❤️

[@Peter_1986]

I once saw an integral that had integrals as limits of integration, lol.

[@MichaelZankel]

It’s not Christmas without integration!

[@skywalker5653]

I love that from watching your calculus videos and using brilliant I was actually able to follow along and solve it in my head though I have done no formal cal classes 😅

[@nikko2505]

This is simply Laplace Transform

[@scottleung9587]

Yay - the answer is 2/5 for the 25th (of December)!

[@redroach401]

An easier way to solve the last bit is to remember, when ever you have sin or cos with exp, you can set the trig functions equal to the Imaginary part of the exp function, meaning the problem becomes a simple exp integral. In this scenario, we would have the Imaginary part of the integral from 0 to inf of e^(i2u)*e^(-u) du. This is obviously just e^u(2i-1)/(2i-1) eval: 0 to inf. Infinity diverges so we are left with Im(1/(1-2i)). Multiply by the conjugate and separate the fraction to get the Imaginary part being 2/5.

[@AlumniQuad]

IT'S A CHRISTMAS MIRACLE!

[@Priyanshu-q7s]

we can solve it by gama function

[@DravenFNM]

i think its -2/5, you overlooked the last fraction

[@thebeardman7533]

It is to early for I still have calc lectures but when Christmas comes be assured that I will solve it

[@jonny8448]

Thanks professor!!! Christmas is coming and I have to find a crazy Christmas problem for my channel!!!🎄🧑‍🎄🤶 PS. Not as crazy as yours!!! I wouldn't be able to come up with something like this!!!🤩🤗

[@stevencarr4002]

To get the limit why not put u = ln(x), then we have e^0.5u in the denominator and u in the numerator as u goes to infinity. This is obviously zero.

[@nickfleiwer5272]

Thanks a lot for this years Christmas present 😂😂😂 but I might return it later haha

[@gameworld6740]

This is... A nightmare

[@Curiescat-f5f]

Since it's my bday, i'll take this as my bday gift

[@yoniziv]

Loved it

[@TomMarAlem1987]

My boy's giving us a surprise in the denominator.

[@Jadamhodges]

Wonderful!!!😊

[@pritamsur1926]

Please solve this integration.. integral of (32-x^5)^(1/5)🙂

[@TozzaYT]

u sub?

[@pritamsur1926]

@@TozzaYT mathematics🙂

[@benatia99]

ty much appreciated

[@brucekritt7036]

Strange.. The answer I'm getting is -(2/5). Based on (d/du)[e^(-u)*(sin(2u)+2*cos(2u))] = -5*e^(-u)*sin(2u). I checked that derivative carefully.

[@fwelling2703]

gonna come back to this video in a year to see if I understand yet.

[@codehucau5564]

all nightmare come in one

[@atommax_1676]

Actually an easy one. Thank you

[@hidden_leafy]

Best Christmas gift I've ever received lol

[@armanavagyan1876]

Thanks PROF 👍

[@namename7000]

Hello, how to solve factorial equations like this: 3x!-x^x-2=0 do you have a video about this?

[@richardfredlund8846]

0,1,2 are trivial solutions, but for different numbers that looks really hard... interesting looking problem type.

[@migueldomingos4570]

If x's domain is positive integers: You can just do some bounding. Rearrange to 3x! = x^x + 2 and notice that the RHS grows much faster than the LHS, to formalize it you can prove by induction that for x>= 3 x^x > 3x! and thus all solutions will be smaller than 3 and you can easily check that 0,1 and 2 works as richard stated

[@atishthatei8842]

make me fun as i do in cristmas . thanks bro . but quite a easy one

[@anticlashers2617]

I likes your videos ❤. Love from india🇮🇳

[@CrushOfSiel]

Ah damn, I was close. Been a while since I did calculus. I got the limits and the numerator right but I thought the denominator was cos(x) and then I was stuck, it is similar.

[@knowledge90s93]

Which of the following sequences could represent the impulse response of a stable discrete-time system? k^2 (-0.65)^k 2^k ksin(k)

[@igorhaladjian5718]

Love the Christmas T-shirt !

[@suscraft7767]

solved it quite easily! can u make an starter 3 hour pack on definite integrals!

[@diptenduchatterjee950]

Yess!! Done in the first attempt. Good question

[@umertaiyab5500]

i wanted to know how does trigonometric substitution work when you substitute sinx or cosx as they can only have the value from -1 to 1.

[@conanedojawa4538]

i think that the limit of sinx /e^x when x goes to infinity the sine function goes to a finite value 1 or -1 but e^x goes to infinity then the limit will be zero but I don't know it will be 0 plus or 0 minus

[@A_Random_Ghost]

If you're talking about the final limit. When you have a bounded numerator and a denominator that goes to infinity. You can just conclude the limit goes to zero. And the reverse goes to infinity.

[@A_Random_Ghost]

@@abcd-ug8tj Yeah, I forgot that was a thing 😅.

[@hotlatte1222]

Great work!! But i think it is more likely for Halloween, not Christmas.

[@blackpenredpen]

lol, it should really be for Thanksgiving since it's just next week! haha

[@hotlatte1222]

@@blackpenredpen Maybe this question fits all 3 festivals. When seeing it in the beginning, it is so horrible for Halloween. When solving it, it is like the games of finding eggs in Thanksgiving. And finally you reveal the solution with clear steps; which is just a Christmas gift. So cool.

[@mickelsantiagoquispenamuch4961]

Happy X-mass

[@neilgerace355]

7:44 Shouldn't that be minus minus 4?

[@TristanPopken]

Its minus minus minus 4 because of the double integration by parts, so it does become minus 4 in the end. This is what the +-+ row stands for on the left column od the D I method

[@mathmachine4266]

Your thumbnail makes it look like you're being held against your will

[@AlejandroGD17]

First time I'm actually able to solve one of these!!

[@PowerUpStudio_]

i solved it before watching and got the exact same solution

[@maxmccann5323]

If you close your eyes and squint your ears, you can almost hear Arnold Schwarzenegger teaching you maths

[@KesterPembroke]

Hey blackpenredpen is there in the complex numbers a function thats inverse equals it's derivative? Thank you

[@samueljehanno]

Interesting

[@MichaelZankel]

Isn’t it -2/5?? Because it was (-sin2u + 2cos2u )/(5e^u), so (-) ALL of that is (-2*1)/5 at the end!! No?

[@saadansari1757]

Even I think the same

[@MichaelZankel]

@@saadansari1757yeah, Idk why he didn’t put a (-) on the cos at the end.

[@Anmol_Sinha]

It is actually -(sin2u + 2cos2u)/(5e^u) , here -ve is in the outside. During the application or the upper and lower limit of integral, we got -(-(2/5)). I don't think in any part of the video it showed the -ve only on sin(as your comment suggests)

[@Anmol_Sinha]

​​@@MichaelZankelthe minus never got distributed in the expression. Look at the brackets carefully

[@saadansari1757]

@@Anmol_Sinha okay thanks

[@Asterisk_766]

This video exceeds the limits of my Brain

[@akgamer4215]

Solve this without denominator

[@user_08410]

9:23

[@darcash1738]

Very nice. Now let’s see Paul Allen’s integral… Nah I’m just joking Paul Allen couldn’t top this one 😂

[@tabommenor]

Bro just made calculus final boss 💀💀

[@herbie_the_hillbillie_goat]

Tis the season.

[@PV10008]

When evaluating the numberator for u=inf, you say it's finite so its precise value doesn't matter. However, how do you account for the fact that sin(2u)+2cos(2u) can sometimes equal 0? Why is it okay to assume it's non-zero in the limit?

[@carultch]

Sine and cosine are both functions of exponential order. This means that an exponential decay function as its input goes to infinity, will shrink to zero either faster than these functions, or as fast as these functions. This is one of the criteria for a Laplace transform to exist, is that the function has to be of exponential order, which is why sine and cosine have Laplace transforms, but secant and tangent do not.

[@aimlessideas1165]

2/5 for the 25th👀

[@christianvanderstap6257]

What just happened? Got my ass run over by a math truck.

[@longlong10203]

i thought you are gonna talk about the Gaussian Integral when i saw e^x^2, it's almost, phew

[@Morbius787]

Imagine getting this on you calc two test💀

[@Jee25-]

This was easy, as a 12th grader.

[@Passione2507]

​@@Jee25-yeah. Revises the basics

[@istvanszabo3275]

Merry Christmas 2u 😃

[@doug2855]

Can you explain the math behind cos, sin , tan etc. Like how did cos(45°)=1/sqrt(2).

[@doug2855]

More generically how would you hand calculate the value of cos(x). X being a random value

[@aryanjoshi3342]

@@doug2855 power series

[@Passersby98]

I'm expecting that Mr Tsao could demonstrate how to solve ODE

[@rufusmafija8674]

hey there i have an incredibly hard question for you: try to find the integral of sqrt(3x²+x) do you know to solve that?

[@Ashaiksameer]

Can u make a roadmap of mathematics and concepts in it😢

[@DC_EDITS]

Great christmas present

[@mauriziomorales5303]

Qué EJERCIZASO!!!! I LIKE IT, THANK YOU!!!!!

[@cristofer6806]

do you have any plans on doing calc 3 stuff, would love to see more of that

[@JSSTyger]

Oh what a wonderful Christmas gift!! (Murmur murmur...)

[@evansaschow]

I hate doing IBP, so I’d much rather decompose sin(2u) into its exponential form

[@hearteyedgirl]

Although I do know 1=0!,1=1!, 2=2!,6=3! and know the intention of the question but the intention itself remains ambiguous There's no way to know if the series really is x^n/n!

[@WhiteDMaxwell]

Why sin(2*infinity) and cos(2*infinity) isn't undefined?