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?
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)...
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
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
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
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 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
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
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
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.
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.
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 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.
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. ..
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.
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 😅
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.
@@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 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)
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.
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). 🤷♂
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
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 :)
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
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)
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.
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
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.
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 😅
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.
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.
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!
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.
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)
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.
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).
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!
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
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 🤪
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.
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.
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.
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!
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
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.
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.)
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
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(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
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 🙂
for question at the end by applying LMVT for (0,x) { [f(x)-f(0)]/ [x-0]} = f'(x) then substituting from given relation and using x=1 we will get f(x)/x
for the problem you put at the end, i think the solution is 1- 1/e. To solve, set f(x)+f'(x) = 1 y = f(x) y + dy/dx = 1 dy/dx = 1-y dy/(1-y) = dx -ln(1-y) = x ln(1-y) = -x 1-y = e^-x y = 1 - 1/e^x f(x) = 1 - 1/e^x f(1) = 1 - 1/e
I think this way is the easiest.... Integral of f(x) between 0 and 1 is = 1 Integral of f(x) between 0 and 2 is = 1 + A Since f(2x) = 3f(x)... f(x) = f(2x)/3 Therefore the Integral of f(2x)/3 between 0 and 1 is also = 1 (since f(x) = f(2x)/3) Therefore the Integral of f(2x) between 0 and 1 is = 3 (multiply both sides by 3) If the Integral of f(x) between 0 and 2 is = 1 + A Then Integral of f(2x) between 0 and 1 is = (1 + A)/2 (half the limits because 2 squashes the function and (1 + A) has been squashed so half that aswell) Therefore (1+A)/2 = 3 rearrange this and... A = 5
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.
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.
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.
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)
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.
The answer is 1-1/e. Just imagine that since f(0) = 0 the biggest f'(0) can be is 1. Then when f increases a bit f' can only be 1-f and so on created an increasing concave down function. Getting the exact answer of 1-1/e comes from solving f' + f = 1
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.
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.
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.
And an alternate way of writing that coefficient is log2(6) log2(3) + 1 log2(3) + log2(2) log2(3*2) log2(6) So, f(x) = lb(6)x^lb(3) Where lb is the binary logarithm (log base 2)
Never mind, there are infinitely many continuous (even differentiable) functions that satisfy this. You can define any continuous function f(x) on the interval [1,2] satisfying f(2)=3f(1), and then extend it with the functional equation to all positive values of x. Then you can scale the result to get a function that satisfies the integral condition too.
The solution I found is so much more simpler bro. It's just u substitution fr x=0-->x=1 integral 3f(x) dx = 3 x=0-->x=1 integral f(2x) dx = 3 x=0-->x=1 integral f(2x) d(2x) = 6 u=0-->u=2 integral f(u) du = 6 And then u get that the answer is 6 - 1 =5
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
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 :))
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)
I think I found something very simple, could anyone tell me whether this is fine? If f(2x)=3f(x) we can conclude for an antiderivative that F that F(2x)=6F(x) which is immediately seen be deriving both sides Note that for x=0 er get F(0)=6F(0) which means F(0)=0 The first integral hence yields: F(1)=1 The second integral: F(2)-F(1)=6F(1)-F(1)=5F(1)=5
let F be an antiderivative of f. f(2x) = 3f(x) integral from 0 to 1 of f(2x) = that of 3f(x) integral from 0 to 1 of f(2x) = 3 * that of f(x) integral from 0 to 1 of f(2x) = 3 F(2*1)/2-F(2*0)/2=3 F(2)-F(0)=6 F(2)-F(1)+F(1)-F(0)=6 integral from 1 to 2 of f(x) + that from 0 to 1 = 6 integral from 1 to 2 of f(x) + 1 = 6 integral from 1 to 2 of f(x) = 5
@@skylardeslypere9909 yes they employ the same methods but the solutions are structured differently, mine is a bit more friendly with all the heavy lifting done in the first step itself. Btw Nice job staying to notice the second solution
For all real a, F(x)=a*x^2 gives you F(2x)=2^2*F(x), F(x)=a*x^3 gives you F(2x)=2^3*F(x), By assuming that the question is well posed, there exists a unique real a such that f(x)=a*x^(log base 2 of 3) because we can solve for a knowing the integral from 0 to 1. Once you have a (it is [log base 2 of 3] +1 ), integrate your function [ f(x)=(log_2(3)+1)*x^(log_2(3) ] from 1 to 2 and get 5.
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.
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
I was surprised by the complicated HMMT solution. I solved it by observing that integral[f(cx), {x, a, b}] = integral[f(x), {x, ca, cb}]/c for any function f (it's just horizontal stretching by a factor of c). This is analogous to your substitution of u = 2x, but eliminates that explicit step. Anyway, 3integral[f(x), {x, 0, 1}] = integral[f(x), {x, 0, 2}]/2 6 = integral[f(x), {x, 0, 1}] + integral[f(x), {x, 1, 2}] integral[f(x), {x, 1, 2}] = 5
Learn more #calculus 👉 brilliant.org/blackpenredpen/ (20% off with this link!)
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?
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)...
This looks wrong. You stated that u = 2x, so how can you then leapfrog into stating that u is equivalent to x? not convinced
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
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. #
the first solution is like if someone forgot that 1+1=2 and instead used the equation 1+1/2+1/4+...=2
😂
perfect summary :-)
too easy for Harvard-MIT
1+1 may equal 2, but f(1)+f(1) doesn't necessarily equal 2f(1) :P
@@siobhangraham7280 it doesn't?
Your videos are an integral part of my day 😁
🤣🤣
Hello Sir!
Yours and BlackPenRedPen, I love to watch at the end of the day 🤩🥰
I see what you did there sir
@@muratkaradag3703 also me♥
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
And then ridiculously easy again, right? 😆
@@blackpenredpen 😏
@@roxannemackinnon2213 it turns out the function IS cx^r (it's (log_2(3)+1)x^(log_2(3)))
Did you know how to solve it when you first thought it was easy, or just assume it was easier to solve?
The official solution was so needlessly complicated 💀 substitution is such a great tool
skull substitution
ruclips.net/video/z2OyVIJznHw/видео.html
This question is super difficult
Wow, both solutions are so neat! I appreciate how yours doesn’t require anything infinite.
Except the geometric series.
@@xinpingdonohoe3978 what do you mean?
@@lih3391 it says "doesn't require anything infinite" but there's a geometric series, which is an infinite series.
@@xinpingdonohoe3978 His solution doesn't use a geometric series though
Xinping Donohoe watch to the end of the video, he has an alternate solution.
your solution made this look ridiculously easy
The first method was mind-boggling but uselessly complicated.
Harvard MIT do be flexing their integral skill
Agreed. It was just fancy steps
Basically a race of whoever comes up with a more uselessly sophisticated solution
@@georgefan2977 For a uselessly complicated question
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
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
Oh, never mind. I didn't finish the video and just realized you used the same method.
@@kobethebeefinmathworld953 xD
Yeah exactly,i did same
I did the same, and this way is so easy. Why to complicate things? duh!
I use the same method as yours :D
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)’
That is brilliant 👏 👌
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
@@DRoo95 He didn't go anywhere, it's still the same, it's just written in the derivative form
@@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
wow I solved it with ODEs but this one is way to better and more elegant👌👌
I'm terrible at all contest problems (for a math professor, anyway), and this one took me under a minute. Thanks for sharing.
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.
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
this is also cool
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.
Your insight is always phenomenal
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
Some day I'll be a mathematician
Man, I feel proud of myself for literally instantly figuring out the solution, which ended up similar to your approach
👍
When you realize he’s holding a Pokeball the whole time
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.
You don't get a calculator. But that's a method for sure
Thank you, I was wondering what f was!
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.
Man said in the description "good challenge for calc 1 students"... umm nah calc 2 maybe? (im in BC too lol)
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.
@@willo1345Actually, you should understand the rules, not "remember" them.
@@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.
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. ..
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.
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 😅
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.
@@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
@@violintegral in fact, @noahtaul had an even neater solution.
Start with f + f'
@@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)
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.
The solution from the MIT was so convoluted. I naturally approached the problem with your solution in 1 min. lol.
The MIT Tournament problems you gave at 0:18 are proving to be fun! Problem 3 has me stumped currently
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). 🤷♂
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
It's the same thing that he did but the other way around
the second method (yours) is so neat and simple!
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 :)
Dang you just beat me to it.
I made the same assumptions and got:
f(x) = x^(log2(3))*(log2(3)+1)
Nice job!
Your solution is the effective one. The other one is nice to flex muscles though :)
f(x)=log_2(6)x^log_2(3) totally works despite x
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
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)
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.
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
NICE.
@@ruchirkadam8510 thanks
0:35 relatable af
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.
Maybe not necessarily “complicated” but people say it is because of the other more efficient and easier way to answer the question.
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 😅
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.
Did you prove that that's the only function satisfying the criteria?
again an amazing video from you
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.
This restored my passion for maths
That’s ridiculously satisfying. Thanks for the step-by-step.
Both solutions are realy great . What is wonderful in the first solution is the idea, the approach using infinite sum series
Such a great teacher! Amazing video🔥
ruclips.net/video/z2OyVIJznHw/видео.html
I found the second solution as more efficient
Used infinitely fewer terms 👍
And the second substitution for the second method was just way easier
I literally learned about geometric series today In my ap calc bc class. So this was a fun example of what I learned today.
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!
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.
Yeah you need to prove that requirement f(x) + f’(x) = 1 yields the maximum value in the first place
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)
LOVE THIS! Can't believe how they complicated such a simple solution ... but it was kind of cool. Darn showoffs!
OMG I did it the way you did before watching any of the solutions. Time to take the rest of the day off.
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.
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).
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!
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
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 🤪
It’s perfectly well defined as a complex function
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.
Nah, there is a ton of such functions.
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.
A real analysis professor would come up with the geometric series method. Real analysis professors are in their own weird world.
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.
You're simplified technique to solve this was great!
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!
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
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.
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.)
Your solution is SO much better. Well done!
That entire problem felt like a backwards knight move in chess, going backwards to go forwards later. Incredibly smart solution 👍
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
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(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
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 🙂
Yes bro.
for question at the end by applying LMVT for (0,x) { [f(x)-f(0)]/ [x-0]} = f'(x) then substituting from given relation and using x=1 we will get f(x)/x
i legit laughed out loud as soon as i saw your solution (or rather the idea behind it)
for the problem you put at the end, i think the solution is 1- 1/e.
To solve, set f(x)+f'(x) = 1
y = f(x)
y + dy/dx = 1
dy/dx = 1-y
dy/(1-y) = dx
-ln(1-y) = x
ln(1-y) = -x
1-y = e^-x
y = 1 - 1/e^x
f(x) = 1 - 1/e^x
f(1) = 1 - 1/e
I think this way is the easiest....
Integral of f(x) between 0 and 1 is = 1
Integral of f(x) between 0 and 2 is = 1 + A
Since f(2x) = 3f(x)... f(x) = f(2x)/3
Therefore the Integral of f(2x)/3 between 0 and 1 is also = 1 (since f(x) = f(2x)/3)
Therefore the Integral of f(2x) between 0 and 1 is = 3 (multiply both sides by 3)
If the Integral of f(x) between 0 and 2 is = 1 + A
Then Integral of f(2x) between 0 and 1 is = (1 + A)/2 (half the limits because 2 squashes the function and (1 + A) has been squashed so half that aswell)
Therefore (1+A)/2 = 3
rearrange this and...
A = 5
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.
Try breaking the given integral into (0,2) and (2,1) and substitute x=2u in the first one....some may find that easier.
First method was really creative.
The MIT given solution is what my Math teacher from high school would define as Complicating Easy Problems Office
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.
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.
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)
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.
The answer is 1-1/e.
Just imagine that since f(0) = 0 the biggest f'(0) can be is 1. Then when f increases a bit f' can only be 1-f and so on created an increasing concave down function.
Getting the exact answer of 1-1/e comes from solving f' + f = 1
@@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.
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.
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.
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.
I think I even found the function that satisfies this:
(Log2(3)+1) *x^log2(3)
And an alternate way of writing that coefficient is log2(6)
log2(3) + 1
log2(3) + log2(2)
log2(3*2)
log2(6)
So, f(x) = lb(6)x^lb(3)
Where lb is the binary logarithm (log base 2)
Yea but i think its correct though
@@pieper7936 yes, definitely correct! I was just suggesting a slightly different form of that same correct answer.
I thought of this function too, but I don't see how to prove that this is the only continuous function that satisfies the conditions.
Never mind, there are infinitely many continuous (even differentiable) functions that satisfy this. You can define any continuous function f(x) on the interval [1,2] satisfying f(2)=3f(1), and then extend it with the functional equation to all positive values of x. Then you can scale the result to get a function that satisfies the integral condition too.
I like your video,they not only train my English but also give me a demonstration to teach calculus
The sponsor of this video is Brilliant and this solution was *super brilliant*
The solution I found is so much more simpler bro. It's just u substitution fr
x=0-->x=1 integral 3f(x) dx = 3
x=0-->x=1 integral f(2x) dx = 3
x=0-->x=1 integral f(2x) d(2x) = 6
u=0-->u=2 integral f(u) du = 6
And then u get that the answer is 6 - 1 =5
So easy, int f(x) from 1 to 2 = int f(x) from 0 to 2 - int f(x) from 0 to 1 = (int f(2u) d(2u) from 0 to 1) - 1 = (3*2 int f(x) from 0 to 1) - 1 = 5.
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
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 :))
But you don't know if there exists F satisfying F'(x) = f(x).
Loved it. Thanks.
The second solution was literally the first approach that popped into my head
My kids laughed every time the video and audio sped up. One also liked the Pokemon ball.
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)
I think I found something very simple, could anyone tell me whether this is fine?
If f(2x)=3f(x) we can conclude for an antiderivative that F that F(2x)=6F(x) which is immediately seen be deriving both sides
Note that for x=0 er get F(0)=6F(0) which means F(0)=0
The first integral hence yields: F(1)=1
The second integral: F(2)-F(1)=6F(1)-F(1)=5F(1)=5
That’s actually better, nice
integral of f(x) from 0 to 1 = 1
=> F(1) - F(0) = 1
f(2x)=3*f(x)
=> integral of f(2x) from 0 to 1 = integral of 3f(x) from 0 to 1
=> (1/2)[F(2*1) - F(0)] = 3
=>F(2) - F(0) = 6
=> -F(0) = 6 - F(2) = 1 - F(1)
=> F(0) = F(2) - 6 = F(1) - 1
=> F(2) - F(1) = 6 - 1
=> F(2) - F(1) = 5
=> integral of f(x) from 1 to 2 = 5
the first solution is like von Neumann's claimed method for solving the fly-between-the-trains problem
let F be an antiderivative of f.
f(2x) = 3f(x)
integral from 0 to 1 of f(2x) = that of 3f(x)
integral from 0 to 1 of f(2x) = 3 * that of f(x)
integral from 0 to 1 of f(2x) = 3
F(2*1)/2-F(2*0)/2=3
F(2)-F(0)=6
F(2)-F(1)+F(1)-F(0)=6
integral from 1 to 2 of f(x) + that from 0 to 1 = 6
integral from 1 to 2 of f(x) + 1 = 6
integral from 1 to 2 of f(x) = 5
There's an easier method to it, just break the integral from (1 to 2) to (0 to 2) - (0 to 1) , then let x = 2t for the first part and done
I can't like this solution even more.. nice
I like this minimalist solution.
That was his second solution :) it's after the brilliant ad.
@@skylardeslypere9909 yes they employ the same methods but the solutions are structured differently, mine is a bit more friendly with all the heavy lifting done in the first step itself.
Btw Nice job staying to notice the second solution
For all real a,
F(x)=a*x^2 gives you F(2x)=2^2*F(x),
F(x)=a*x^3 gives you F(2x)=2^3*F(x),
By assuming that the question is well posed, there exists a unique real a such that f(x)=a*x^(log base 2 of 3) because we can solve for a knowing the integral from 0 to 1. Once you have a (it is [log base 2 of 3] +1 ), integrate your function [ f(x)=(log_2(3)+1)*x^(log_2(3) ] from 1 to 2 and get 5.
I got 2 in the end. I found the value of f(x) and then solved the integral.
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.
I figured it own my own using your method!
When the people making the problems go to so many math tournaments they forget about u-substitution
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
I was surprised by the complicated HMMT solution. I solved it by observing that integral[f(cx), {x, a, b}] = integral[f(x), {x, ca, cb}]/c for any function f (it's just horizontal stretching by a factor of c). This is analogous to your substitution of u = 2x, but eliminates that explicit step. Anyway,
3integral[f(x), {x, 0, 1}] = integral[f(x), {x, 0, 2}]/2
6 = integral[f(x), {x, 0, 1}] + integral[f(x), {x, 1, 2}]
integral[f(x), {x, 1, 2}] = 5