the last question on a Harvard-MIT Math Tournament!

[@blackpenredpen]

Learn more #calculus 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@cicik57]

my i ask, why following is not working: if f(2x) = 3f(x) then integral from a to b (f2x) = F(2b)- F(2a) = 3(F(b) - F(a)) so integral from 0 to 2 f(x) = F(2) - F(0) = 3(F(1) - F(0))=3*1 = 3 something must be wrong in this allroach, can you point to it, pleace?

[@ivandebiasi6657]

The integral from a to b of f(2x) is not F(2b) - F(2a). Try just with f(x) = 1: the integral is b-a but F(2b) - F(2a) is 2b-2a = 2(b-a). The correct formula is 1/2 (F(2b) - F(2a)). Also f(2x) = 3f(x) not F(2x) = 3F(x)...

[@emmanuelbalogun3404]

This looks wrong. You stated that u = 2x, so how can you then leapfrog into stating that u is equivalent to x? not convinced

[@samuelmoss2480]

there’s actually an easier solution: just integrate the identity they give you from 0 to 1 and substitute to find an expression for integral from 0 to 2 then subtract

[@hmiscoordinator2751]

Apologies for horrible shorthand, using my phone. Int(f,x,1:2) = Int(f,x,0:2) - Int(f,x,0:1) = Int(f,2x,0:1) - Int(f,x,0:1) = 3*Int(f,x,0:1) - Int(f,x,0:1) = 3*1 -1 = 2. #

[@drpeyam]

Your videos are an integral part of my day 😁

[@chsh9686]

🤣🤣

[@naturelover4148]

Hello Sir!

[@muratkaradag3703]

Yours and BlackPenRedPen, I love to watch at the end of the day 🤩🥰

[@bryantwiltrout5492]

I see what you did there sir

[@amirmahdypayrovi9316]

@@muratkaradag3703 also me♥

[@Macieks300]

the first solution is like if someone forgot that 1+1=2 and instead used the equation 1+1/2+1/4+...=2

[@blackpenredpen]

😂

[@RB-ew6lo]

perfect summary :-)

[@MrMusic55123]

too easy for Harvard-MIT

[@siobhangraham7280]

1+1 may equal 2, but f(1)+f(1) doesn't necessarily equal 2f(1) :P

[@devyanshatharv1861]

@@siobhangraham7280 it doesn't?

[@jessehammer123]

Wow, both solutions are so neat! I appreciate how yours doesn’t require anything infinite.

[@xinpingdonohoe3978]

Except the geometric series.

[@lih3391]

@@xinpingdonohoe3978 what do you mean?

[@xinpingdonohoe3978]

@@lih3391 it says "doesn't require anything infinite" but there's a geometric series, which is an infinite series.

[@gamerdio2503]

@@xinpingdonohoe3978 His solution doesn't use a geometric series though

[@stephenbeck7222]

Xinping Donohoe watch to the end of the video, he has an alternate solution.

[@vladimirkhazinski3725]

When I first saw the question, I thought it was ridiculously easy. When I saw the official solution I thought it was ridiculously hard lol

[@blackpenredpen]

And then ridiculously easy again, right? 😆

[@saruarahemadrakib5187]

@@blackpenredpen 😏

[@nanamacapagal8342]

@@roxannemackinnon2213 it turns out the function IS cx^r (it's (log_2(3)+1)x^(log_2(3)))

[@Willtedwards]

Did you know how to solve it when you first thought it was easy, or just assume it was easier to solve?

[@Atlas_Enderium]

The official solution was so needlessly complicated 💀 substitution is such a great tool

[@vitalsbat2310]

skull substitution

[@miantony6493]

kzbin.info/www/bejne/sGOyqol_f9-hftk This question is super difficult

[@btwtempest]

your solution made this look ridiculously easy

[@noahtaul]

To solve the question at the end: move the 1 to the other side and multiply by e^x. You can pull the whole thing together to get (e^x*f(x)-e^x)’

[@omaromy8722]

That is brilliant 👏 👌

[@DRoo95]

Brilliant solution, I found a more extensive/less neat way to prove it (first proving that f(x) + f'(x) = 1 is the optimal solution, then solving the ODE), but I found the same answer: Question: how did you go from e^x*f(x) + e^x*f'(x) - e^x

[@omaromy8722]

@@DRoo95 He didn't go anywhere, it's still the same, it's just written in the derivative form

[@DRoo95]

@@omaromy8722 I must say I haven't heard the derivative form in my uni years. Enlighten me. Why are we allowed to remove the e^x * f'(x) term by taking the derivative over the remaining terms? EDIT: Ah never mind. Reverse multiplication rule. Whoops 😂 brilliant way of finding it

[@manizarin2233]

wow I solved it with ODEs but this one is way to better and more elegant👌👌

[@kobethebeefinmathworld953]

I think there's a faster way to do the integral: by FToC, I = I_1 - I_2 where I_1 is from 0 to 2 and I_2 is from 0 to 1 which is already given to be 1, then for I_1 we can do u-sub by letting u = x/2 to change the bounds into 0 to 1 and the expression inside of the integral becomes f(2u) * 2du to apply the given formula f(2x) = 3f(x) and obtain f(2u)* 2du = 6 f(u) du. So it yields to I_1 = 6*1 = 6 and we get the answer I = 6 - 1 = 5

[@kobethebeefinmathworld953]

Oh, never mind. I didn't finish the video and just realized you used the same method.

[@walaaaaaa3456]

@@kobethebeefinmathworld953 xD

[@learningmathswithconcepts7203]

Yeah exactly,i did same

[@sushrut4328]

I did the same, and this way is so easy. Why to complicate things? duh!

[@hongshengzhu4411]

I use the same method as yours :D

[@UberHummus]

Your insight is always phenomenal

[@opstall]

I'm terrible at all contest problems (for a math professor, anyway), and this one took me under a minute. Thanks for sharing.

[@blackpenredpen]

I was actually very confused at first when I saw their official solution. But then after I decided to try it on my own, I solved in within minutes.

[@davidebic]

The first method was mind-boggling but uselessly complicated.

[@3ckitani]

Harvard MIT do be flexing their integral skill

[@dbro1205]

Agreed. It was just fancy steps

[@georgefan2977]

Basically a race of whoever comes up with a more uselessly sophisticated solution

[@bilalmalik5002]

@@georgefan2977 For a uselessly complicated question

[@gabrielroberto5819]

when he was doing the first method I thought he would do it in an easier way, I mean, after he made the integral from 1/2 to 1 he could do the integral from 0 to 1/2 and it wold be enough to draw a conclusion by making an equation, but it wanted to be unnecessarily difficult, seem that MIT tried to use all calculus features kkkkkkkkkkk

[@Re-lx1md]

The MIT Tournament problems you gave at 0:18 are proving to be fun! Problem 3 has me stumped currently

[@Notthatkindofdr]

In Problem 3 I think you first need to solve for f'(0), g'(0), and h'(0). But Problem 7 is really bizarre. It does not even make sense the way it is written (since A(M) is a random variable and cannot just have a "limit value"), but even when I interpret the problem to mean what I think they intended, it is not a calculus problem at all (more of a number theory problem). 🤷‍♂

[@sinecurve9999]

Consider integral[f(x), {x, 0, 2}]. We can make the substitution 2u = x transforming our integral into integral[2*f(2*u), {u, 0, 1}]. Applying the rule f(2x) = 3f(x), our integral becomes integral[2*3*f(u), {u, 0, 1}] = 6. integral[f(x), {x, 1, 2}] = integral[f(x), {x, 0, 2}] - integral[f(x), {x, 0, 1}] = 6 - 1 = 5.

[@wowZhenek]

Man, I feel proud of myself for literally instantly figuring out the solution, which ended up similar to your approach

[@blackpenredpen]

👍

[@cosmicgxming8808]

I'm currently taking AP Calculus BC in high school and I understood everything up to 2:41 but I lost track about everything after that lmao. The hard part about this type of math isn't necessarily the execution process itself but about connecting all the information in the question to formulate all the steps needed to arrive at a conclusion.

[@adamschroeder3946]

Man said in the description "good challenge for calc 1 students"... umm nah calc 2 maybe? (im in BC too lol)

[@willo1345]

Yes, takes a lot of practice. There are a lot of rules to remember and you often make little mistakes that screw you in the end. Practice is the only way to really get it because you won't know how to solve a problem until you do it.

[@martinrosol7719]

​@@willo1345Actually, you should understand the rules, not "remember" them.

[@willo1345]

@@martinrosol7719 You do not have time to derive everything from scratch on a test and, even if you did, that increases the chances of human error. So yes. You need to remember the rules.

[@xninja2369]

Bro this wasn't even hard to understand , specially limit part was easiest some type of limit can't be solved by L'HÔpitals rule or infinity GP or anything and you need to solve it and it takes 2-3 pages to solve. ..

[@andrewkarsten5268]

I immediately did the first step of the first method, and realized it would repeat, but it gave me the idea for the second method, so I did the second method instead. That was much simpler in my opinion

[@ankitbasera8470]

This is how I did it f(2x) = 3 f(x) Integrating both sides from x=0 to x=1 Integration 0 to 1 of f(2x) = 3 Put 2x = t I get Integration 0 to 2 of f(t) = 6 Divide the range of integration from 0-1 and then 1-2 Integration 1 to 2 f(t) = 5

[@KewlWIS]

this is also cool

[@IdeesDePhysique]

The solution from the MIT was so convoluted. I naturally approached the problem with your solution in 1 min. lol.

[@alexdotdash7731]

Such a great teacher! Amazing video🔥

[@miantony6493]

kzbin.info/www/bejne/sGOyqol_f9-hftk

[@ayaanpatel9667]

again an amazing video from you

[@soundest8768]

That’s ridiculously satisfying. Thanks for the step-by-step.

[@VinTheFox]

When I plotted the curve on a graph to visualize the problem, I intuitively came up with the first method as the way to solve it. That might be the approach they took as well.

[@byronvega8298]

I did it waaaay differently. Take the integral from 1 to 2 of f(x) and set it equal to y Then add 1 on both sides but on the LHS as the integral from 0 to 1 of f(x) Then make the substitution x->2v Now use the identity 3f(v)=f(2v) Resulting in six times the integral of f(v) from 0 to 1 which implies 6 = y+1

[@titiyop7286]

It's the same thing that he did but the other way around

[@nqnqnq]

the second method (yours) is so neat and simple!

[@KurtColville]

Your solution is SO much better. Well done!

[@wavingbuddy3535]

Everyones saying the official solution was really complicated, but it wasn't really, it was mathematically beautiful and utilizes A level/ basic year 1 university algebra.

[@adnanhussain9906]

Maybe not necessarily “complicated” but people say it is because of the other more efficient and easier way to answer the question.

[@omerbar7518]

When I saw your solution I just flopped in my chair. I love how you take solutions to MIT test and be like: "yeah but I can do it 14 times better". Keep the good content up! (although I'm 4 months late lol)

[@dabmanplz1451]

Nobody seems to calculate f(x) and is the simplest way for me. Knowing that f(2x)=2x for the usual f(x)=x, and f(2x)=4x for the usual f(x)=x^2, then our f(x) must be something like f(x)=x^k, with k between 1 and 2. Just substituting in the condition of the problem gives us that f(x) = x^log2(3). Then you just need to do the integrals. With the first you realize our function actually needs a constant: f(x) = a * x^log2(3), with a = log2(3)+1, because that's the value that makes the integral 1. And them you just do the second integral, and its 5. I believe this method may be a little longer that the fastest one proposed, but you need to know practically nothing about integrals to solve it: no changes of variable, no nothing. And also you learn what is happening with f.

[@abhimanyus1670]

You don't get a calculator. But that's a method for sure

[@PotmosHetoimos]

Thank you, I was wondering what f was!

[@MarkMcDaniel]

You're simplified technique to solve this was great!

[@ideatronzvvvicunit9476]

Both solutions are realy great . What is wonderful in the first solution is the idea, the approach using infinite sum series

[@adamtedd2000er]

Since f is Riemann integrable on (0,2), it is continuous a.e. on (0,2) and therefore it can be shown the function agrees a.e. with the formula: f(x) = x^( log_2(3) ) * (log_2(3)+1). Interesting to note you can find an explicit formula from the vague assumptions about f(x).

[@artemisnimrod2948]

My solution: start calculating points on the line that satisfies f(2x)=3f(x) starting with (1,1). (1,1), (2,3), (3,9), (4,27), (5,81). The pattern is pretty obvious. It's just (2^n,3^n). So, the equation f(x)=3^[log{2}(x)] goes through these points. The integral of this function from 0 to 1 is less than 1, though, so we need to add a scalar, "a." a*int{0 to 1}(3^[log{2}(x)])dx ==> a = log(6)/log(2). So, [log(6)/log(2)]*int{0 to 1}(3^[log{2}(x)])dx =1. Change the bounds of integration: [log(6)/log(2)]*int{1 to 2}(3^[log{2}(x)])dx = 5. I have no idea if this is a sound way of doing this, but it's how I got the right answer.

[@furiousfajitaa2367]

This restored my passion for maths

[@laprankster3264]

I literally just guessed f(x)=cx^(n) where (c,n) are real numbers. I solved for n to get log2(3). And just integrated to get cx^(log2(6))/log2(6). I then plugged in the F(1)-F(0) to find c=log2(6) (yay for cancellation). I finally plugged in F(2)-F(1) to get 5 as the final answer.

[@hemantbhosale3060]

We can substitute f(x)=f(2x)/3 In given integral which gives integral from 0 to 2 is equal to 6 but we know integral 0 to 2 is integral 0 to 1 plus integral 1 to 2 we know integral 0 to 1 is 1 and integral 0 to 2 is 6 therefore integral 1 to 2 is 5

[@sajedtabbabi]

Some day I'll be a mathematician

[@ΑνναΜαρια-ψ7σ]

idk if its right but F’(x)=f(x) so in this case we have F(2x)/2=3F(x) . The integral from 0 to 1 for f(x) is F(1) -F(0) and the one from 1 to 2 is F(2)-F(1) . By putting 1 in the first you get that F(2)=6F(1) so the second integral is 5F(1) . Also by putting 0 you get that F(0) =0 so from the first integral F(1)= 1 so 5F(1)= 5

[@kostasl1808]

Είσαι ο μοναδική στα σχόλια που έχει ακριβώς την ίδια λύση με τη δική μου και η μοναδική που δεν το έλυσε μέσω Αλάσκας. Ή εμάς στα σχολεία μας τα κάνουν πολύ απλά ή σε όλους τους άλλους πολύ περίπλοκα.

[@F_A_F123]

F(2x)/2 = F(x) is generally wrong. Because if you increase the F by a constant, the thing you get still is the antiderivative of f(x), but it doesn't satisfy F(2x)/2 = 3F(x). (F(2x) + C)/2 = 3(F(x) + C) F(2x)/2 + C/2 = 3F(x) + 3C C/2 = 3C Which is wrong for C ≠ 0

[@zerglingsking]

That first method was pretty cool! I did something more similar to your answer: I take integral of f(2x) between 0 and 1, which is equal to 3 times the integral of f(x) between 0 and 1. But it is also 1/2 * integral of f(x) between 0 and 2. So we get 1/2 * ( integral of f(x) between 0 and 2) = 3 * (integral of f(x) between 0 and 1) The right side is equal to 3, and the integral between 0 and 2 of f(x) can be separated between 0 and 1, and 1 and 2. The integral betwene 0 and 1 is known so we have 0.5 * (integral of f(x) between 1 and 2 + 1) = 3 which gives integral of f(x) between 1 and 2 = 2 * ( 3 - 1/2) = 5!

[@JBOboe720]

f(x)=log_2(6)x^log_2(3) totally works despite x

[@eja1731]

When you realize he’s holding a Pokeball the whole time

[@stefanopetrone9900]

Using the identity f(2x)=3f(x) and some logic i found the function f(x) = c * x ^ (ln(3)/ln(2)) Found c with known integral and finally calculate the other integral

[@TheDannyAwesome]

My initial reaction was that f(2x)=3f(x) looks like a functional equation for some Ax^n. Under the assumption that there is a unique correct answer to this problem, we can find one function satisfying it, and take that as f(x). Use the functional equation to find n, and then use the fact that the area between 0 and 1 is 1 to determine A.

[@adayah2933]

Nah, there is a ton of such functions.

[@ATL45]

I did this out explicitly by realizing the given information describes a function for which a vertical dilation can be equivalent to a horizontal dilation. The family of power functions, f(x)=ax^n, have this property, and for these values (2 and 3) we end up with n=log_2(3) and a=n+1=log_2(6). Evaluate F(2)-F(1) and you get 5.

[@Maxence1402a]

About the last question, let g(x) = f(x)-1+exp(-x), which gives g(x)+g'(x) = f(x)+f'(x)-1 0, g(x) > 0, then let a = max {y in [0,x), g(y)=0} (which exists since g is continuous and g(0)=0), and mean value theorem yields b in [a,x] such that g'(b)=g(x)/(x-a) > 0, therefore g(b)+g'(b) > 0 (x >= b > a so g(b) > 0) which contradicts g(x)+g'(x) = 0, g(x)

[@paxtonjk]

I literally learned about geometric series today In my ap calc bc class. So this was a fun example of what I learned today.

[@yourmom4476]

I solved it by assuming a monomial solution ax^n. Using the first information gives a(2x)^n=3ax^n -> 2^nx^n=3x^n -> 2^n=3 n=log2(3) Our function is ax^log2(3). Putting this into the integral from 0 to 1 and taking the antiderivative gives (a/log2(3))x^log2(3)+1 evaluated from 0 to 1. log2(3)+1 can be simplified to log2(6) by making 1 into log2(2) and using log rules. (a/log2(6))(x^log2(6)) evaluated from 0 to 1. 1 to any power is 1 and 0 to any power is 0 so a/log2(6) which must be equal to 1, so a=log2(6). Our final function is log2(6)x^log2(3). Putting this into the integral from 1 to 2 and taking the antiderivative gives (log2(6)/log2(6))x^log2(6) evaluated from 1 to 2. The logs cancel, 2^log2(6)=6 1 to any power is 1 6-1 is 5. It doesnt prove that any function with these requirements has an integral from 1 to 2 of 5 but it solves the problem :)

[@jaredwoolsey5692]

Dang you just beat me to it. I made the same assumptions and got: f(x) = x^(log2(3))*(log2(3)+1) Nice job!

[@Jeff-zc6rr]

Justt integrate the identity f(2x)=3f(x) from 0 to 1. Then right side is just 3. Do a change of variable u=2x and you will get an integral from 0 to 2 on the left side. but split up into two integrals from 0 to 1 and then 1 to 2. That second integral is what you need and you will get the answer 5.

[@dbro1205]

I found the second solution as more efficient

[@cwcarson]

Used infinitely fewer terms 👍

[@dbro1205]

And the second substitution for the second method was just way easier

[@farklegriffen2624]

For your challenge at the end: the answer is infinity. For instance: Assume f(x) = lim [ a→1, 1/(3x-3a) + 1/(3a) ], as 'a' gets arbitrarily close to 1, f(x) approaches infinity. This function also satisfies the requirements that f(x) + f'(x) ≤ 1 and f(0)=0.

[@MathIsFun137]

Unfortunately, your function is not continuous everywhere. From the conditions in the problem, f’(x) exists for all x, which means f(x) is continuous for all x. But your function does not exist when x = a. In fact, since your function does not take an actual value at x = 1 by what you yourself showed, you know that it’s not continuous. I’m also unsure about being able to define a single-variable function like that in the first place, as one could argue that there’s technically two variables there.

[@Biggyweezer69]

That limit evaluates to 1/3(x/(x-1)) and is undefined at 1 as the limit from the left is -infinity and the right is +infinity. If the function were to increase going from left to right it would be bounded by f(x)

[@DRoo95]

You make the mistake of trying to "move" asymptotic behaviour to infinity. -For any of the approaching functions for a -> 1, the function isn't continuous, thus not fulfilling the constraints. You can't just make a "half limit", taking one property from the real function for a=1 (continuity on [0,1) ) and take another property from the limit for a -> 1 (the value for f(1) ). In general, even a limit of continuous functions does not always lead to a continuous functions. -If you fill in a = 1, you do get an asymptote at x = 1, which is -inf when approaching from the lower side of x. -If you do do a -> 1, you're making a limit of non continuous functions, thereby not proving your function is continuous. In fact, if you're interested. You could check my reply, in which I proved 1-1/e to be the maximum answer.

[@DRoo95]

@@Biggyweezer69 if you're interested, you could look at my reply. I think I've got a prove why 1-1/e is the maximum answer.

[@DRoo95]

@@MathIsFun137 if you're interested, you could look at my reply. I think I've got a prove why 1-1/e is the maximum answer.

[@stephenbeck7222]

Yeah, your approach is what I would do. I don’t even know who would come up with splitting the original integral into sub intervals between 2^n-1 and 2^n in order to solve when there’s a much more intuitive way.

[@floydmaseda]

A real analysis professor would come up with the geometric series method. Real analysis professors are in their own weird world.

[@bjao8619]

I think the original solution was also quite nice though, yes not the most efficient but it was easy to follow and there wasn’t as much confusion with the switching variables. Idk I just started learning analysis and it was really cool seeing a lot of the things I learnt being applied in a solution like that.

[@spazticdrummer7]

OMG I did it the way you did before watching any of the solutions. Time to take the rest of the day off.

[@theeternalwanderer190]

Interesting method, thank you for sharing! I did it by looking at the pattern to get f(x) = k 3^log_2(x) for some constant k, did the given integral to find k= ln(6)/ln(2) and then integrating to get 5. Not as elegant though 😅

[@hrvojedjurdjevic2123]

I also thought that finding solution without finding f(x) would be more elegant, but I did it the same way as you did, anyway. Since f(0)=0, I assumed f(x) = c x^a for some constants c and a, from f(2x)=3f(x) it follows a=log_2(3), and from the value of the known integral it follows c=a+1. As the resulting function is same, it yields the same result when integrated from 1 to 2.

[@F_A_F123]

Did you prove that that's the only function satisfying the criteria?

[@michaelwarnecke3474]

At the end, we just want the function for which f(x) + f'(x) = 1, which is: f(x) = 1-e^(-x) f'(x) = e^(-x) f(x) + f'(x) = 1-e^(-x) + e^(-x) = 1 f(1) = 1-e^(-1) = 1-1/e Not a proof, but the way I got the solution.

[@kqnrqdtqqtttel1778]

Yeah you need to prove that requirement f(x) + f’(x) = 1 yields the maximum value in the first place

[@MrHotBagel]

So why do you set f(x)dx = f(u)du in the last portion of your problem. I am having trouble conceptualizing that since f(u)du=f(2x)*2dx, when you perform substitution back. I understand that the "x" or "u" is simply a dummy variable; I do not understand how the dummy variable "x" is similar to the x previously. This may have been why this method was not chosen; since it seems to coincidentally lead to your integral, from 1 to 2, being equal to 5.

[@pedroribeiro1536]

The sponsor of this video is Brilliant and this solution was *super brilliant*

[@CalculusIsFun1]

Did it in my head. Nice warm up for the day.

[@jakehawks1090]

That entire problem felt like a backwards knight move in chess, going backwards to go forwards later. Incredibly smart solution 👍

[@maaikevreugdemaker9210]

Loved it. Thanks.

[@dlmcnamara]

I'd start with the decomposition: \int_1^2 f(x) dx = \int_0^2 f(x) dx - \int_0^1 f(x)dx ; substitute 2u=x into first integral; second integral =1 (given); easy peasy.

[@methodiconion8523]

I got my answer in a very different way. I decided to find a specific function which satisfied the first requirement. After a little thinking, I came up with g(x)=3^log2(x)=x^(ln(3)/ln(2)). I integrated this and found something which doesn't have an integral from 0 to 1 of 1, so I multiplied it by its denominator to get F(x)=x^(ln(3)/ln(2)+1). Putting in x=2 boils down to F(2)=3*2=6. Edit: Added a single closing parenthesis. P.S. I feel like my function obviously doesn't work for the domain from minus infinity to 0. Anyone know if there's a valid solution sub zero that my answer fails to get, but which this substitution method can produce? P.P.S. Just watched both of the video's explanations. I don't believe either is able to find a solution for x=0, -((-x)^(log2(3))*(log2(3)+1))|x

[@vvthanh]

you can define f(x)=0 for any negative x, it doesn't matter what the negative part of the function looks like I like your approach, very brute force, but not really the spirit of the question 🤪

[@seansean1728]

It’s perfectly well defined as a complex function

[@Infinium]

I’ve always loved your channel, another great video! 😊

[@punstress]

LOVE THIS! Can't believe how they complicated such a simple solution ... but it was kind of cool. Darn showoffs!

[@chaosredefined3834]

I can see an approach to do the f(x) + f'(x) inf, this becomes the derivative. Step 2: Find the maximum value of f(x + 1/n) with this strategy. Step 3: Find the maximum value of f(x + 2/n) with this strategy. Step 4: Find the general pattern to get the maximum value of f(x + k/n) with this strategy. Step 5: Use induction to prove that it is the maximum. Step 6: Use this to find the maximum value of f(x + n/n). Step 7: Take the limit as n -> inf of the above, and set x = 0, and f(x) = 0. This will give you lim n-> inf f(n/n) which is f(1) This is way too much work for me to do with my day off, but if anyone wants to try it, feel free to tell me how you go.

[@Jeremy-jj4nj]

Your solution is the effective one. The other one is nice to flex muscles though :)

[@laxminarayanbhandari855]

I have a super easy way. Integral from 1 to 2 f(x) dx= integral from 0 to 2 f(x) dx- integral from 0 to 1 f(x) dx = 2*integral from 0 to 1 f(2x) dx-integral from 0 to 1 f(x) dx =5*integral from 0 to 1f(x) dx=5

[@ruchirkadam8510]

NICE.

[@laxminarayanbhandari855]

@@ruchirkadam8510 thanks

[@srijanagarwal964]

Try breaking the given integral into (0,2) and (2,1) and substitute x=2u in the first one....some may find that easier.

[@schizoframia4874]

That’s so cool!!

[@Alex-ud6kv]

Interesting solutions! Since f(x)=x^2 follows a pattern of f(2x)=4f(x) i realized f(x)=ax^n (actually absolute value of x since it should work for all x, but that doesn't matter for this question). Then plugging that into f(2x)=3f(x) yielded n=log_2(3). Plugging into the given integral then yielded a=1+log_2(3), and we then have f(x)=(1+log_2(3))x^(log_2(3)) which we can use to solve the final integral. *"log_2(3)" meaning "log base 2 of 3"

[@albertobernal2537]

It took me at least 30 mins to watch the video in full due to pausing and backtracking; it's been many years since my last math course at uni. But halfway through, it became very clear. I'm looking forward to watching your earlier uploads on a regular basis and refresh my calculus. Thx a lot sir!

[@ftbex9224]

I like your video,they not only train my English but also give me a demonstration to teach calculus

[@sbunny8]

The way I did it is f(2x)=3f(x) told to me it's an exponential function, i.e. f(x)= ax^n and I just had to find a and n. Since f(2x)=3f(x), we get a(2x)^n=3ax^n, hence (2^n)=3 so n=log2(3). To get a, use the integral (a/(n+1))x^(n+1) from 0 to 1, so a/(n+1)=1, hence a=n+1. Now we know the mystery function is *f(x)=(n+1)x^n where n=log2(3)* . The integral of that is x^(n+1). Evaluate from 1 to 2 and we get 2^(n+1)-1, which is 2^(log2(3)+1)-1. That evaluates as (3)(2)-1, which is 5. Took about the same length of time, and I not only found the integral but I know the function itself, so I'm ready for follow-up questions like what's f(2). I suppose my answer isn't as clever but it works.

[@richardvidil6994]

Why not just integrate both sides of f(2x) = 3f(x) from 0 to 1? Since we know f(x) integrated from 0 to 1 is 1, the right hand side is just 3. Now sub u = 2x, on the left hand side, changing the limits to from 0 to 2 and introducing the 1/2 factor. We now have integral of f(u) du from 0 to 2 = 6 (clearing the fraction). Therefore integral of f(u) du from 0 to 1 plus integral of f(u) du from 1 to 2 = 6, but we already know integral f(u) du from 0 to 1 is 1, therefore integral of f(u) du from 1 to 2 = 6 - 1 = 5.

[@DRoo95]

At the question in the end: The optimal strategy will be to always maximize f'(x). (So f'(x) = 1 - f(x)). If some other function fulfilling this constraint g is better (so g(1) > f(1) ), then by midpoint theorem, there has to be a point x1 in [0, 1) where g(x1) = f(x1) and g(x) > g(x) for all x > x1. However, since g(1) > f(1), we can again use the midpoint theorem to find that for some x2 in (x1, 1) we have g'(x2) > f'(x2). Since g(x2) > f(x2) we have g(x2) + g'(x2) > 1. Thus such function cannot exist. Now we've established that f' = 1 - f, we have to solve this differential equation. We have: df/dx = 1 - f df/(f-1) = -dx (multiply both sides by -dx/(1-f) ) Int(df/(f-1)) = -Int(dx) Ln(f-1) = c1 - x f-1 = e^c1 * e^-x f = c2 * e^-x +1 Now we need to fill in f(0)=0 to find that c2 = -1 and we find: f(x) = 1 - e^-x With f(1) = 1-1/e = 0.63... the maximum value. This one was a lot harder than the one in the video 😅

[@violintegral]

I have an alternate, albeit similar solution to this problem. Since working with a differential equation is much easier than working with a differential inequality, I set f(x) + f'(x) = a =< 1. This first order equation is separable, but I think a nicer solution involves treating it as a first order linear equation. Multiplying both sides by e^x, we have e^x·f(x) + e^x·f'(x) = a·e^x. The left hand side is clearly d/dx(e^x·f(x)), so we can integrate both sides, giving e^x·f(x) = a·e^x + C. Dividing both sides by f(x), we have found f(x) = a + Ce^-x. Using the initial condition f(0) = 0, we can see that C = -a, so f(x) = a(1 - e^-x). At this stage we can evaluate f(1), giving f(1) = a(1 - e^-1). Since 1 - e^-1 > 0, f(1) is trivially maximized when a = 1, given the restrictions on a, so f(1) = 1 - e^-1, as you showed.

[@DRoo95]

@@violintegral nice solution! And a neater way of solving the DE. But if I'm not mistaken, you're only proving the maximum for functions with f(x)+f'(x) = a with a being a constant. Via this proof, there could technically be some other function with f(x)+f'(x) non constant but smaller than 1 that still produces better results, right? I know it seems trivial that a function with f + f' < 1 shouldn't score better than one which always had f + f' = 1, but doing that via the analytical way was kinda the first part of my proof

[@DRoo95]

@@violintegral in fact, @noahtaul had an even neater solution. Start with f + f'

[@carstenmeyer7786]

@@DRoo95 If you assume *f(x) + f'(x)* to be integrable, you can set *f(x) + f'(x) =: u(x) = 0* In the above *H(t)* is Heavyside's step-function that jumps from zero to one at *t = 0* . We use the inequality *u(t)

[@gustavoespinoza7940]

The second solution was literally the first approach that popped into my head

[@misterrivs9457]

Nice video! I did the question in a matter of a couple of minutes in a few simple steps: I let F'(x)=f(x) so integrating f(x) would give you F(x). I then used the first identity (integrating f(x) from 0 to 1 equals 1-----------call this identity {1}) to obtain F(1)-F(0)=1. I then subbed f(x)=f(2x)/3 into the first identity as well Iintegrating f(2x)/3 from 0 to 1 equals 1) to obtain 1/6[F(2x)] bounded by 0 to 1 is equal to 1. Eventually expanding out you get F(2)-F(0)=6-------call this identity {2}). Subtracting identity {2} by identity {1}: F(2)-F(0) - [F(1) - F(0)] = 6 - 1 ---> F(2)-F(1)=5 Which is the same as the integral of f(x) from 1 to 2 :))

[@adayah2933]

But you don't know if there exists F satisfying F'(x) = f(x).

[@mathboy8188]

Easier way still: You can do this in your head. On an exam, you can bypass the implicit math requirement to justify that the answer exists, and just ASSUME that there IS an answer. So ASSUMING this question has an answer, it's the same answer for ANY function that satisfies f(2x) = 3 f(x) and INT{ x=0 to x = 1 of f(x) dx } = 1. So you only need to find ANY function that satisfies that, integrate it, and you're done. For the students I tutored for the SATs, I called this this the *"Make it Concrete"* trick. Work: First, it ain't gonna be linear ( f(2x) = 3 f(x) ), and have 2 constraints, so will need 2 unknowns on a general form of function. The easiest choice to try is f(x) = a x^n, and it works: Constraint #1: f(2x) = a 2^n x^n = 3 f(x) = 3 a x^n implies 2^n = 3 which gives n ( n = ln(3)/ln(2), but don't even need to do that!). Constraint #2: 1 = INT{ x=0 to x = 1 of f(x) dx } = INT{ x=0 to x = 1 of a x^n dx } = a x^(n+1)/(n+1) ]_{ x = 0 to x = 1} = a/(n+1), which gives a ( a = n+1 = 1 + ln(3)/ln(2), but again, don't even need that!). Thus INT{ x=1 to x = 2 of f(x) dx } = INT{ x=1 to x = 2 of a x^n dx } = a x^(n+1)/(n+1) ]_{ x = 1 to x = 2} = ( a/(n+1) ) { 2^(n+1) - 1^(n+1) } = (1) { (2)(2^n) - 1 } = 2(3) - 1 = 5.

[@mehrdadmohajer3847]

thx. alot for the posting. well done🙂

[@mayhemistic6019]

I also solved this originally in a similar fashion. Took me two minutes to figure out. Integral 0->2 f(x) = Integral f(x) 0->1 + Integral f(x) 1->2 (Let's call this M). Now we do te substituiton x = 2t in the integral. So 2f(2t)dt from 0->1 which is 6f(t)dt from 0->1. So the final equation becomes 6 = 1 + M. So M = 5 🙂

[@raja2850]

Yes bro.

[@leickrobinson5186]

Yep, that’s how I did it! Easy enough to do in your head. :-D (Their solution WAY overcomplicated the problem, like walking around the block to go next door.)

[@gaspardh.7643]

There's a easier solution : Take the integral between 0 and 1 of f(x)dx =1, with u=2x and f(x) = f(2x)/3 you get 1 = 1/6 times the integral from 0 to 2 of f(u)du, and with Chasles relation you get 1= 1/6*(1+S) with S the integral to find, and that makes S=5

[@Nathan-cz8uk]

Your way is definitely cleaner than theirs. Another, more hack-y way that happens to work: if we assume the function f(x) is of the form f(x)=Ax^n (motivated by the fact that f(ax)=bf(x) works for functions of that form), then the two given conditions give n = log base 2 of 3, and A = log base 2 of 6. Then you can directly compute the desired integral with the power rule.

[@li__on6403]

I recently did something similar. The integral of t^a*(1 mod t) dt from 0 to 1 = (1/a+1)-(riemann-zeta-function(a+2)/a+2). The way you do it is similar. (Ik that x mod y doesn't really make sense for real variables, but I just used what desmos used for that expression)

[@UNIverso.Matematica]

The solution of the question of the end: f+f' f'=af => af+f=0 => f(a+1)=0 => suppose f ≠ 0, a=-1. Part. associated: f(x)=1 Complete: f(x)= Ae^(-x)+1 f(0)=0 => Ae^(0)+1=0 => A+1=0 => A=-1 => f(x)=1-e^(-x) => f(1)=1-e^(-1) If f(hom.) =0, f(hom.)(1)=0. So f(0)=1, but it is not possible, because f(0)=0 So the sol is 1-(1/e) Ansatz technique is so cool!

[@chaulamyu7328]

Consider int(f(u)*du) from 0 to 2. If we let u = 2x, then we have int(f(2x)*d2x) = int(f(2x)*2*dx) from 0 to 1. Since f(2x) = 3(fx), so it also equals to int(6f(x)*dx) from 0 to1 = 6*1 = 6. Note that int(f(u)*du) from 0 to 2 = int(f(u)*du) from 0 to 1 + int(f(u)*du) from 1 to 2. Therefore, we have 6 = 1+ int(f(u)*du) from 1 to 2. Therefore, int(f(u)*du) from 1 to 2 = 6-1 = 5

[@chaulamyu7328]

Haha, just notice the second half of the video after typing the comment

[@milanh1349]

This is great. Thank you :)

[@keinemaschine3038]

My approach was simliar to yours, except I got rid of the integral equation by using the antiderivative. Then I got an easy functional equation to solve. Nicely selected Question!

[@maelhostettler1004]

Nice solution but there is a way simpler approach : f is continuous so let : F(x) = integral from 0 to x of f(y)dy thus F(0) = 0 and by definition F(1) = 1 then take the substitution : y = 2t thus dy = 2dt and F(x) = 2 * (integral from 0 to x/2 of f(2t) dt) then by definition F(x) = 2 * (integral from 0 to x/2 of 3*f(t)dt) so F(x) = 6 * (integral from 0 to x/2 of f(t)dt) so F(2) = 6 * F(1) = 6 and integral from 1 to 2 of f(x)dx = F(2) - F(1) = 6 - 1 = 5

[@ajbiffl4695]

I think finding the first, complicated solution comes down to which integral you start u-substituting in first. From 1 to 2, you go to 1/2 to 1, and so on, but if you start with 0 to 1, you get immediately to 0 to 2

[@rata4005]

I intregrated f(2x)=3*f(x) which gave me F(2x)/2=3*F(x) +C F(2x)=6*F(x)+C and also noticed that the integral 0 to 1 of f(x)dx=F(1)-F(0)=1 F(1)=1+F(0). F(2*0)=6*F(0)+C => F(0)=-C/5. Then is used the informations to determine the value of the integral 1 to 2 f(x)dx=F(2)-F(1)=6*F(1)+C-F(1)=5*F(1)+C=5*(1+F(0))+C=5*(1-C/5)+C=5-C+C=5.

[@saroshadenwalla398]

I did something similar but integrated the identity between 0 and x so I got 1/2( F(2x)-F(0))=3(F(x)-F(0)) which gives F(2x)=6F(x)-5F(0). Setting x=1 gives F(2)=6F(1)-5F(0) and so F(2)-F(1)=5F(1)-5F(0)=5(F(1)-F(0))=5

[@chimero3806]

very good question! I enjoyed it!

[@selimders5473]

Realy nice question keep it going !!! :))

[@OnionKing-cm4qh]

I love it when 2 methods (one which has a limit) are used to get the same result.

[@danielontheedge]

Integral from a to b is the same as sum of integral from a to c and c to b. Is there a mathematical theorem that supports the case where we add an infinitely countable set of intervals to form the integral from a to b? This question also reminds me of the Riemann sum of area of rectangles that approaches the area under curve as n goes to infinity. What supports the leap from adding area of infinitely countable number of rectangles that allows us to make the conclusion that it becomes equal to the area under the curve

[@mathcanbeeasy]

I=Int from 0 to 2 of f(x) dx x=2u then dx= 2du I=2*integral from 0 to 1 f(2u)du =6* integral from 0 to 1 of f(u)du=6 So I tegral from 1 to 2...=6-integral from 0 to 1...=6-1=5. Took me 1 min and 2 lines. Very nice problem to se how students think substitutions, variable changes and new limits of integration.

[@antonioprovolone2815]

The MIT given solution is what my Math teacher from high school would define as Complicating Easy Problems Office

[@johncalculusmathsclass5998]

Wonderful step, brilliant

[@mathevengers1131]

First method was really creative.

[@marcelohenrique5101]

Amazing, professor. Greetings from Brazil

[@chazzbunn7811]

I solved it using nothing more than the identity given and substitution, but a little differently. First, I did this: Int_0to1_{f(x)}dx = int_0-(1/2)_{f(x)}dx + int_(1/2)to1_{f(x)}dx = 1 int_(1/2)to1_{f(x)} = 1 - int_0to(1/2)_{f(x)}dx This will be important later. Next, I took the integral to be solved for and substituted 2u = x and then used the given identity: Int_1to2_{f(x)}dx = 2*int_(1/2)to1_{f(2u)}du = 6*int_(1/2)to1_{f(u)}du Next, I substitute the identity I listed above: 6*int_(1/2)to1_{f(u)}du = 6*[1-int_0to(1/2)_{f(u)}du] I use another substitution, u = 1/2w, and using the given identity once more: 6*[1-int_0to(1/2)_{f(u)}du] = 6*[1-(1/2)*int_0to1_{f(w/2)}dw] = 6*[1-(1/6)*int_0to1_{f(w)}dw] = 6*[1-(1/6)] = 6 - 1 = 5