This Oxford Integral Question STUMPED Students

Hope everyone enjoyed! Shorter more quick style today but stil a nice, fun, simple problem that stumped a lot of oxford students - please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated.
~ Thanks for watching!
Comment answers to the challenge down below.
@blackpenredpen … I've left you a challenge at the end my last few videos - if you see this give it a go!
Thanks so much for all the recent support - hitting 1k subscribers has been amazing and so please keep sharing and interacting so I can make even more content!
Thanks again for all the recent support and please keep sharing and interacting!!
#maths #mathematics #integrals #entrance #university #Oxford #Cambridge #JEE #problemsolving#taylor #maclaurin #gaussian #gauss #statistics #whoknew #fascinating #functions #euler #funproblems #proofs #functions #physics #sums #series #limits #whiteboard #math505 #blackpenredpen #integral #trig #trigonometry

[@Tosi31415]

i subbed the result back in the first statement, then subbed x with -x in the input to get two equations I found f(x) to be equal to f(-x), which is -12x²+6

[@OscgrMaths]

Perfect! Nicely done.

[@princeyadav1233]

There is 1 more solution. F(x)=6-9x^2. This. Satisfies the original identity and integral in this. Case is equal to 3.

[@martijntervelde5600]

@@princeyadav1233 Interesting solution! It follows the structure you'd expect from f(x): f(x) = - 3 I x² + 6. But I don't think it actually solves the original equation. If you integrate your f(x) from -1 to 1, you actually get I = 6. But you have to assume I = 3 to get your solution.

[@mathmachine4266]

​I believe that is the only solution. The problem is to find f(x) such that -f(x)+2f(-x)=12x²-6, and such that ∫ [-1,1] f(x) dx = 4. So, we take your particular solution, 6-12x², and try to find a complementary solution, g(x), we get that f(x)=6-12x²+g(x) where g(x) is a function such that -g(x)+2g(-x)=0, and ∫[-1,1] g(x)dx = 0. The first statement says that g(x)=2g(-x). That means g(-x)=2g(x), and so g(x)=4g(x), which means g(x) can only be 0.

[@leif1075]

​@@OscgrMathswouldn't you agree it is confusing to have tje barisble t because evem though it's just a symbol if t is different from x pr has some relation to x then the integral of f(t)dt is ABSOLUTELY NOT THE SAME AS the integral of f(×)dx because t may ewual x squared or t may ewual 2x or whatever else..so wouodnt you agree its not totally fair or smart or at least not mathematically accurate to just swap t for x as you did?

[@jga5821]

The first thing to try with a functional equation involving both f(x) and f(-x) is computing the even and odd parts of f(x), this is, f(x)+f(-x) / 2 (even part) and f(x) - f(-x) / 2 (odd part). Doing this one sees that your f(x) is even, and your equation becomes f(x) = 6 - 3*x^2 * Int. Integrating both sides yields your result Int = 4.

[@OscgrMaths]

@@jga5821 That's great! Definitely another good approach.

[@vylon1075]

This was also my solution.

[@shreesh.06]

Bro can you make me understand this? please.

[@iddonaiss4541]

But it's wrong...

[@pneujai]

I dont understand why f is even. Here is my work. 6 + f(x) = 2f(-x) + 3x² Int Rearranging, we have f(x) = 2f(-x) - 6 + 3x² Int Replace x by -x, we have f(-x) = 2f(x) - 6 + 3x² Int Consider e(x) := [f(x) + f(-x)] / 2 = (3f(x) - 6 + 3x² Int) / 2 o(x) := [f(x) - f(-x)] / 2 = (-f(x) + 6 - 3x² Int) / 2 Since o(x) does not vanish I do not believe f is even. 2f(-x) - f(x) = 6 - 3x² Int Assume f is even then we have f(x) = 6 - 3x² Int Int = [6x - x³ Int] from -1 to 1 Int = (6 - Int) - (-6 + Int) Int = 12 - 2Int 3Int = 12 Int = 4

[@yahola4030]

man thank you for the simple explination

[@OscgrMaths]

@@yahola4030 no problem! Thanks for the comment.

[@MadScientyst]

The high level of your Math skill & insight alone was worth subscribing to....good work indeed!!

[@OscgrMaths]

@@MadScientyst thank you!

[@rosettaroberts8053]

How i did this is as follows: 1. Rewrite to 6 + f(x) = 2f(-x) + 3x^2I 2. Take the antiderivative 6x + F(x) = -2F(-x) + x^3I 3. Substitute x = -1, x = 1 to get the equations 6 + F(1) = -2F(-1) + I -6 + F(-1) = -2F(1) - I 4. Use the fundamental theorem of calculus to get the equation: I = F(1) - F(-1) 5. These three equations form a system of linear equations, which is trivial to solve with a bit of rearranging.

[@OscgrMaths]

@@rosettaroberts8053 Nice!

[@Latronibus]

Interesting that they want the integral rather than the function. In that case your approach is easier: integrating both sides gives 12+I=2I+2I and then you're basically done. It is nice of them to use 3x^2 to nudge you in this direction. Since I saw the thumbnail that didn't say what the goal was, my first response was to try to find f, and that's not too bad by just taking it to be a quadratic. If you're a little clever about it, you can rule out a nonzero linear coefficient right away as well (since the equation implies that f is even). I see other folks in the comments have gone about this by deriving the form of f from the equation instead of just guessing.

[@OscgrMaths]

Thanks for the comment, and definitely agree that it's nice of them to give you 3x^2!

[@luccabrunner8106]

I think showing the integral with f(-x) is also I with substitution is fun, but I think a symmetry approach would have been more elegant. ✨

[@drynshock1]

6+f(x) = 2f(-x) + 3Ix^2 (denote the integral as I). Let x = -x, 6+f(-x) = 2f(x) + 3Ix^2. Substracting both equations: f(x)-f(-x) = 2f(-x)-2f(x) => f(x) = f(-x) thus f is even. Then, 6 = f(x) + 3Ix^2 => f(x) = 6 - 3Ix^2. Now int_0^1 f(t) dt = int_0^1 6-3It^2 dt => [6t - t^3.I]_-1^1 = 6 - I - (-6 + I) = 6 - I + 6 - I = 12 - 2I. Therefore, I = 12 - 2I => I = 4

[@BederikStorm]

the last step: as you know I=4, you can put it into f(x) = 6 - 3Ix^2 and get final form of the function f(x) = 6 - 12 x^2

[@ishantkhanchi5539]

Wow you explained it really well

[@OscgrMaths]

Thank you!

[@FernandoRuiz-wc9eu]

Amazing explanation by an amazing teacher, hat off!!

[@OscgrMaths]

@@FernandoRuiz-wc9eu Thank you!!

[@yplayergames7934]

Wow, I like liked this one, new sub

[@OscgrMaths]

@@yplayergames7934 Thanks a lot! Appreciate the comment.

[@martijntervelde5600]

Love the solution path you took! My first thought was drawn to functional equations to find f(x) (done too many of those recently). But your approach just demonstrably simpler. Just to share the alternative path: we can first substitute -x for x to get f(-x) = 2 f(x) + 3 I x² - 6. Substituting this into the original equation: 6 + f(x) = 2 (2 f(x) + 3 I x^2 - 6) + 3 I x² 6 + f(x) = 4 f(x) + 9 I x² - 12 -3 f(x) = 9 I x² - 18 f(x) = -3 I x² + 6. And integrating this result yields: ∫ f(x) dx = - I x³ + 6x I = -2 I + 12 3 I = 12 I = 4.

[@OscgrMaths]

@@martijntervelde5600 Thats great! That method you used at the start was exactly how I solved for f(x) after having found I. Thanks for commenting!

[@ffc1a28c7]

You can solve directly for f by noting that the integral (and x^2) is symmetric (integral of f(x)=integral of f(-x)), so taking I as the integral, you have 6+f(-x)=2f(x)+3x^2I. Subbing this back in, you end up with 6+f=2(2f-6+3x^2)+3x^2I or f=-3x^2I+6. Then, you can plug in f, solve for I, and get both in one fell swoop.

[@J4UMZ4UM]

You can also derivate both sides three times, and you get f’’’(x) = -2f’’’(-x) = 4f’’’(x). Where you conclude that f’’’ = 0 and f(x) is ax^2 + bx + c. Now just plug in

[@theupson]

only if f''' exists for all x. how do you propose to prove that?

[@rishabhhappy]

nice approach sir .... i observed fx = f (-x) which is an easy kill though and solved the equation by some manipulations...... but your soln was so simple and elegant

[@sylowlover]

Way I did it: take the derivative of both sides 3 times to find out f must be a quadratic, then directly sub it in and get that f=-12x^2+6. Integrating is a cake walk from there.

[@O_of_1]

Another well made video. Definitely subscribing you fuelling my late night math binging

[@OscgrMaths]

@@O_of_1 Thank you! Glad you enjoyed.

[@youteubakount4449]

Separate f into even and odd part p and q. You get p(x)-6+3x²I = 3q(x) ; therefore p(x)-6+3x²I=0. Also notice that the integral of f between -1 and 1 is the same as that of p. So using p(x) = 6-3x²I, you can integrate between -1 and 1 and get I=4.

[@youteubakount4449]

Your approach would have been my 2nd attempt if this hadn't worked! You have better intuition ; I think it's easier your way

[@OscgrMaths]

Both great methods!! Thanks so much for the comment.

[@mathunt1130]

Start off as he did by noting that the integral is just a constant. The make the swap x -> -x, and you have a linear simultaneous equation for f(x), and f(-x). This yields f(x)=6-4Ix^2. Simply integrate between -1 and 1 to get I.

[@OscgrMaths]

Nicely done! That's a great technique.

[@ZXD121]

Integrals suck to solve but are awesome to use.

[@ayoubabid213]

Its easy just remarq that f is even Then f(x)=6-3x^2 *C where c is a constant which is obviously the given integral Our goal is to determine C Just integrate from both side from -1 to 1 Then c=12-2c Then c=4 Then the olnly solution is : f(x)=6-12x^2

[@OscgrMaths]

Interesting! I did it a different way without assuming f(x) is even - i rearranged for f(-x), then subbed -x into the function in the place of x and rearranged for f(-x) again. I then equated my two expressions and rearranged for f(x) on its own on one side of the equation. This shows this is the only possible solution since you assume nothing about f(x) to obtain it. Thanks for the excellent comment!!

[@jga5821]

@@OscgrMaths Seeing that it is even is straightforward, simply find the even and odd parts of the function. Once we have shown it is even the equation becomes f(x) = 6 - 3*x^2*I, from which your result of I=4 follows immediately.

[@brickie9816]

how do you know f is even though

[@wychan7574]

Interesting

[@allansmart5358]

As someone who loves watching this type of content but goes straight over my head, where is a good place to start learning this type of math?

[@OscgrMaths]

I think very often delving into harder ideas without lots of practice on the more basic ones can often be what catches a lot of people out. 3blue1brown's course on youtube on calculus and linear algebra are excellent and would give you a really firm grip on the two. Hope this helps!

[@allansmart5358]

@@OscgrMaths I will take a look at those videos. I also feel that knowing this will help me with my passion, data analytics. Thank you :)

[@mustafaseyitt]

Integral equations are awesome. Nice

[@kikilolo6771]

<a href="#" class="seekto" data-time="305">5:05</a> I thought that was the question. so in the original equation we let x be -x. We do the difference of the two lines to get that f is even and the sum to get that f(x)=-3Ix^2+6 and by integrating we find I. (so the only function is the one associating -12x^2+6 to x).

[@CalculusIsFun1]

I did it a bit different. 6 + f(x) = 2f(-x) + 3x^2(L) 6 + f(-x) = 2f(x) + 3x^2(L) 12 + 2f(x) - 4f(-x) - 6x^2(L) 6 + f(-x) - 2f(x) - 3x^2(L) 18 - 3f(-x) = 9x^2(L) 6 - f(-x) = 3x^2(L) integrate both sides from -1 to 1 12 - L = 2L 12 = 3L L = 4 So the integral of f(x) from -1 to 1 is 4.

[@OscgrMaths]

@@CalculusIsFun1 Nice!

[@Tenebrae0o]

Holly cow. I wouldnt even know where to begin with this one. Makes one humble knowing how little I know outside of stuff I have to handle at work and uni. Though i am curious how we have never gone through integrals of functions like this in uni in any of the mechatronics courses we had. I look forward to further videos!

[@OscgrMaths]

Thanks for the comment! Glad you enjoyed.

[@redpepper74]

I felt the same way the first few times I came across these kinds of questions, but with a little practice it gets easier to see the patterns. It’s a little weird that there aren’t any courses on functional equations (afaik)

[@dir2002usable]

I apologize if somebody did the same. Was lazy to read all comments. Starting from where the video left us: 6+f(x) = 2 f(-x) + 12 x^2 thus f(x)-f(-x) = f(-x) +12x^2 -6 notice LHS is an odd function, for the RHS to be the same we need f(x) = -12 x^2 +6 +g(x), where g(x) is odd , plugging back to the initial equation, one will figure out very fast that g(x) = -2 g(x) and therefore g(x) ==0

[@kmyc89]

I know math tricks like multiplication by 1 and addition with zero, but (<a href="#" class="seekto" data-time="110">1:50</a>) supriced me.

[@slavinojunepri7648]

Well explained

[@bushwalker6214]

Solved within 3 minutes. Very simple

[@ShashwatSharan]

You can also solve for the function itself which will make it an even interesting problem. It's f(x) = -12x²+6

[@OscgrMaths]

Yea that's the challenge I left at the end! Nicely done.

[@ghtoostvilliers]

We can write 6+f(x)=2f(x)+3x²I AS f is an even function. f(x)=6-3x²integral-1 to 1 (6-3t²I)dt =>f(x) =6-3x²(12-2I) Again integrating both sides within limits -1 to 1 we have: I=6(2)-(12-2I)(2) I=12-24+4I -3I=-12 I=4(required solution) 🇮🇳 🤝🇮🇱 .

[@OscgrMaths]

@@ghtoostvilliers Nice!

[@HamzaKhan-vi2zy]

its not mention f(x) is even how can you assume somethin like that

[@ghtoostvilliers]

Question was: 6+f(x)=2f(-x)+3x²I On replacing x with -x in above equation we can have 6+f(-x)=2f(x)+3x²I Now solving these 2 equations you can have: 3(f(x)-f(-x))=0 f(x)=f(-x) That's why 'f' is an even function.

[@saath9738]

Before seeing the vid : the definite integral is just a constant, so let's call it I 6 + f(x) = 2 f(-x) + 3I x^2 replacing x with -x 6 + f(-x) = 2 f(x) + 3I x^2 adding both equations : 12 + f(x) + f(-x) = 2 f(x) + 2f(-x) + 6Ix^2 f(x) + f(-x) = 12 - 6Ix^2 subtracting both equation : f(x) - f(-x) = -2[f(x) - f(-x)] f(x) = f(-x) 2 f(x) = 12 - 6Ix^2 f(x) = 6 - 3Ix^2 now to find I we can integrate this from -1 to 1 I = 12 - 2I I = 4

[@booshkoosh7994]

Great! Thanks!

[@OscgrMaths]

@@booshkoosh7994 Thanks so much for the comment!

[@SilentArc]

I’m a Columbia graduate student and don’t worry I don’t know either. Mainly because I don’t study this for my major XD

[@viktor-kolyadenko]

In <a href="#" class="seekto" data-time="101">1:41</a> we can set x -> -x end give f(x) as function with parameter I.

[@OscgrMaths]

@@viktor-kolyadenko true, that's how I found f(x) once I'd finished the problem.

[@joakkim2642]

I did it differently. Set x-->-x and subtracting the 2 equations we get: f(x)=f(-x). Now the original equation becomes: f(x)=6-3Ix². Then: I=integral(6-3It²)dt... Solving we get: I=4

[@OscgrMaths]

@@joakkim2642 Great solution! Thanks for the comment.

[@vincentzevecke4578]

You are very intelligent young man

[@0519ryan]

Yes

[@OscgrMaths]

@@vincentzevecke4578 Thank you!! Very kind of you to comment.

[@wojciechswiderski1532]

clever trick!

[@OscgrMaths]

@@wojciechswiderski1532 Thank you!

[@vector8310]

I never knew McCauley Culkin was that proficient in higher math.

[@sethlupo4736]

Hi, nice video 👍. I think I was able to prove that f(x)=-12x^2+6 always. Note when you know I, you can rearrange the eq to f(-x) = -6x^2+3+(1/2)f(x). You can Apply this twice, saying f(-(-x))= -6x^2+3+(1/2)f(-x) = -6x^2+3+(1/2)[-6x^2+3+(1/2)f(x)], and algebraically solve for f(x).

[@OscgrMaths]

@@sethlupo4736 That's how I did it!! It's a very neat solution. Thanks for the comment!

[@takyc7883]

genius

[@OscgrMaths]

@@takyc7883 Glad you enjoyed!!!

[@marcdavies7046]

Short videos are good too!

[@OscgrMaths]

@@marcdavies7046 Good to know! This has been my biggest video so I think people prefer these.

[@abhinavyadav8408]

SIR currently im in grade 10 and i'm thrilled to see what more there is to learn

[@OscgrMaths]

Glad you're looking forward to it!

[@redpepper74]

Calculus and discrete math are pretty dang cool if I say so myself, you got some neat stuff to look forward to 😎

[@EvergrowingInfinity]

Subscribed! Your channel is great.

[@OscgrMaths]

@@EvergrowingInfinity Thank you!

[@legendsplayground7017]

Simple and elegant 👍💪 looking forward for more content, Jesus bless.

[@OscgrMaths]

Thank you!

[@ChandrasegaranNarasimhan]

Hey, there might be a mistake in solution. When finding integration bound by substitution, why f(-1)=1? Sorry my bad. So you may be right.

[@hafez591]

f(x)=6-12x^2

[@shivx3295]

i think we could also find the function easily by just replacing x with -x but i think your approach is more cool

[@OscgrMaths]

@@shivx3295 Definitely a good other approach! Thanks for the comment.

[@mab9316]

Nice problem.

[@OscgrMaths]

@@mab9316 Thanks!

[@МаксимАндреев-щ7б]

C = integral from -1 to 1 of f(t)dt 6+f(x) = 2f(-x)+3Cx^2 6+f(-x) = 2f(x)+3C(-x)^2 Subtract: f(x)-f(-x) = 2f(-x)-2f(x), 3f(x) = 3f(-x), f(x) = f(-x) So 6+f(x) = 2f(x)+3Cx^2, f(x) = 6-3Cx^2 C = integral from -1 to 1 of (6-3Ct^2)dt = 12-2C C = 4 f(x) = 6-12x^2

[@jos3perez98]

I arrived at the same solution (I = 4) by using a reference function. I isolated f(x) and then proposed f(x) = ax^2 + bx + c. Substituting into the equation, I found a = -3I, b = 0, and c = 6. Therefore, f(x) = -3I x^2 + 6. Then I used the definition of I as the integral of the function between -1 and 1 and got: I = -2I + 12, which gives I = 4. That would make f(x) = -12x^2 + 6. Would that be a correct way of solving the problem? Thanks in advance, and keep up the great work!

[@OscgrMaths]

Absolutely gives you the right solution! Doesn't necessarily show it's the only one, but it seems good to me!! Thanks for the comment it's great to see lots of different solutions.

[@yaronbracha4923]

Loved it

[@OscgrMaths]

@@yaronbracha4923 Thanks for the kind comment, really appreciate it!

[@nanamacapagal8342]

I subbed in -x and got that f(x) must be even, swapped f(-x) for f(x), then solved for the integral Then checked. Then proved it's the only function that satisfies this.

[@miraj2264]

I probably should've watched the intro rather than just working off the thumbnail. I assumed the goal was to find f(x), which is easy enough if you assume f has a taylor series. If you differentiate 3 times, you'll get f'''(x) = -2f'''(-x). Plugging in x = 0 gives us f'''(0) = -2f'''(0) ==> f'''(0) is 0. If you continue to differentiate and plug in 0, the same idea holds. In other words, this thing has a taylor series that dies after the first 3 terms i.e. it's a quadratic (I'll use ax^2 + bx + c below): Plugging 0 into the original equation: 6 + f(0) = 2f(0) + 3*0*I ==> f(0) = 6 ==> a*0^2 + b*0 + c = 6 i.e. c = 6 Differentiating the original equation and plugging in 0: f'(0) = -2f'(0) + 6*0*I ==> f'(0) = 0 ==> 2a*0 + b = 0 i.e. b = 0 Differentiating the original equation twice and plugging in 0: f''(0) = 2f''(0) + 6I ==> f''(0) = -6I ==> 2a = -6I i.e. a = -3I I = Integral(ax^2 + 6) from -1 to 1 ==> I = (1/3)*ax^3 + 6x evaluated from -1 to 1 = 2a/3 + 12 ==> 3I = 2a + 36 ==> -a = 2a + 36 ==> a = -12 So f(x) = -12x^2 + 6. I = -a/3 = 4

[@OscgrMaths]

@@miraj2264 Nicely done!!! Haven't seen anybody do it this way yet. Thanks for the comment!

[@subramanyakarthik5843]

Simplest form integral function I = F[x] from -1 to 1

[@SlimThrull]

Can I? Maybe. Am I going to? Naw.

[@Lolwutdesu9000]

I can understand that doing so for the integral of f(t)dt, since it's a constant and is not a function of x, but I'm not convinced by the statement that the integral of f(x)dx from -1 to 1 is also equal to I. It could be equal to something else entirely. For example we could be looking at an equation regarding displacement (for x) and there is also a function of time (for t) inside. Saying that f(x)dx is equal to I simply because f(t)dt is I is a bit arbitrary, unless I'm missing something?

[@OscgrMaths]

@@Lolwutdesu9000 Thanks for the question! No matter what the variable is the function is identical. We could choose to label our x axis with x or t or u - whatever we want - and the area underneath will be exactly the same. This is because it's just the label of the axis that's changing, nothing about the actual height of the function for a given input.

[@Lolwutdesu9000]

@@OscgrMaths I'm not sure I follow. It could be that f(t) = e^t, which when integrated between said bounds is 2.35 (2 dp), whereas f(x) could be x^2, or an odd function where it equals 0. We have no way of knowing this so we can't just arbitrarily say they're all the same because the letter doesn't matter.

[@L3o_o.]

​@@Lolwutdesu9000u missing the point, f is a unique function here we are trying to find

[@حسينالقطري-ب8ص]

Interesting problem, and nice explanation. Thanks for it. It actually reminded me another interesting problem, I forgot the source: Suppose that f is a function on the interval [1,3] such that -1 ≤ f(x) ≤ 1 for all x and integral[1,3] f(x) dx = 0. How large can integral[1,3] f(x)/x dx be? I have a the solution, if someone is interested. ----- For your challenge question, here is my attempt: (not sure if I am correct). We figured out I=4, so we have 6+f(x)=2f(-x)+12x^2 If f(x) is even, then f(x)=f(-x) So 6+f(x) = 2f(x)+12x^2 Or f(x) = 6-12x^2. When f(x) is odd, then I must be 0, (a property of definite integrals of odd functions from -a to a is 0). This contradicts our reault that I = 4, so I am thinking (not sure) f(x)=6-12x^2 is the only form. --- Hope to know the answer for your question Thanks for the video

[@OscgrMaths]

Thanks so much for the great comment!! I'll give the other question a shot - it seems really interesting. Your answer for f(x) is absolutely right - nice work. Your thought that the function could not be even is great and simplifies things. I rearranged for f(-x) twice (once with the original expression, and a second time swapping -x for x) and then from there solved for f(x) but your method yielded the correct answer and was simpler too so looks good!

[@حسينالقطري-ب8ص]

@@OscgrMaths Thanks a lot. Kindly see your email. Thank you.

[@gregoriousmaths266]

what if f(x) isnt odd or even

[@حسينالقطري-ب8ص]

@@gregoriousmaths266 Any function f(x) can be expressed as the sum of two parts (even part fe(x)) and (odd part fo(x)) By substituting these fe(x)+fo(x), you will see that fo(x) will become zero, which contradicts the result I=4

[@naml3833]

My beautiful goat

[@eofirdavid]

For just finding the integral your solution is good. However, some of it might look like magic which just somehow works out, and it is not clear what to do in general. For me, when approaching such questions I always try separating the "different" elements of the problem. In a sense this is the basic of everything in linear algebra, and this is just an extension of this idea. The different parts are (1) f(x), (2) f(-x) and (3) the integral 3x^2 I . One way to get rid of the integral, is by noting that it is constant (given f). Taking the 3rd derivative takes it out of the equation, and you end up with f'''(x)=-2f'''(-x). As for the f(-x), the operation of predecomposing with the minus sign is very simple, in the sense that if we apply it twice we get back f(x) (namely, it is periodic with period 2). This type of phenomena happens all the time, and basically everywhere, so it is always good to look for such properties. In particular, you get that f'''(x) = -2 f'''(-x) = -2 * (-2 f'''(x) ) = 4f'''(x). Now we have a simple equation with just f, we conclude that its 3rd derivative is zero, and therefore f is quadratic ax^2+b+c. You can go back to the original equation, and plug it in, to get a linear system of equations in a,b,c. This is not the shortest solution but, once you have it you can look for shortcuts. Also the ideas are applicable to other problems and give you a better understanding where to look for a solution. Even here, you can decide that maybe you want to get rid of the f(-x) before the integral. Finally, it this is a much harder question, and not something specifically constructed to have a simple solution, then this methods might not solve it completely, but will give some intuition about where to go (and hopefully at least simplify the problem)

[@OscgrMaths]

@@eofirdavid Wow that's great! Thanks for the comment.

[@RashmiRay-c1y]

Given 6 +f(x) = 2f(-x) +3Ix^2, f(-x) = 2f(x) +3Ix^2-6. Therefore 6+f(x) = 4f(x) +9Ix^2-12 > f(x) = 6-3Ix^2. Integrating both sides from -1 to 1, we get I = 12 -3I (2/3) > 3I=12 > I=4 and f(x) = 6(1-2x^2).

[@dakcom-mk6mp]

Nice

[@OscgrMaths]

Thank you!

[@KorlinAng-bs7rh]

👍

[@OscgrMaths]

Thank you!

[@crochou8173]

I did it diffing two and three times to find the integral

[@edgargonzalezgerman1271]

Making functionals really changes the way you see math and the idea of having f(x) in terms of f(-x) is natural , it was interesting tho

[@OscgrMaths]

Definitely! Thanks for the comment.

[@Roksarr228]

i found that f(0)=6 f'(0)=0 f''(0)=-24 and the nth derivative of f(0) is zero for n>=3 then i used teylor series and got that f(x)=6-12x^2

[@OscgrMaths]

@@Roksarr228 Nice!

[@yasharora6772]

u can also put x=0 ... then f(0) =6

[@dannyfilmss]

this is very interesting, I love how you explain everything and i’m starting to learn calculus any tips? i’m 11 y/o

[@OscgrMaths]

I think youtube can be a great place to start so you're in the right place - good job!

[@gregoriousmaths266]

I got f(x)=-12x^2+6 but idk why it's the only solution tbh also <a href="#" class="seekto" data-time="244">4:04</a> lol ill leave it as an exercise for the reader.

[@OscgrMaths]

@@gregoriousmaths266 Yeah that's right. Lmao yeah I really didn't want to go off on a tangent in the video so I did the unthinkable...

[@bucsredsoxredwings]

The good old times when I would pull out a question and students would say "I studied at Oxford or Cambridge", just to see them sweat while my students just attack and solve.

[@sohampine7304]

Brother you are so good at this, can you please suggest some good books regarding calculus and maths in general.

[@OscgrMaths]

@@sohampine7304 Thanks so much! I'll see if I can find any to suggest you.

[@OscgrMaths]

I was recommended this fantastic text by another commenter about a month ago. Take a look at this: galoisian.wordpress.com/wp-content/uploads/2018/11/undergraduate-lecture-notes-in-physics-paul-j-nahin-inside-interesting-integrals-2015-springer-1.pdf

[@VengerVideoGamer]

Strange to hear an English person use American nomenclature when referring to negative quantities. I'm really not used to hearing someone say "negative 1" when it's usually phrased "minus 1" in the UK. It reminds me a bit of when I was in Cardiff University in lectures with Chandra Wickramasinghe in Mathematical Methods in the early 1990's. He always referred to 1 as "unity".

[@OscgrMaths]

@@VengerVideoGamer Interesting comment! I've been taught in British schools my whole life and always been told negative is more proper to use since it's clearer, but everyone I know uses the two interchangeably. Thanks for sharing!

[@VengerVideoGamer]

@@OscgrMaths it may well be a generational thing and I know we've adopted certain Americanisms over the years. I'm pushing 50 now and I finished my maths degree around the time Andrew Wiles was putting the final touches to his proof of Fermat's Last Theorem, so I may well be showing my age 😂. Anyway, nice video and thanks for the reply.

[@OscgrMaths]

@@VengerVideoGamer No problem! Glad you enjoyed it.

[@ChandrasegaranNarasimhan]

So, you can try two scenarios f(x) is odd or even. Then try.

[@rodbennett]

What happens when f(x) is neither?

[@ChandrasegaranNarasimhan]

@@rodbennett I do not know. I think it becomes the answer that has been solved.

[@kaan7866]

When we Integrate both part why we didn't take the integral of I. And thanks forr question.

[@OscgrMaths]

@@kaan7866 I is a constant since the integral is definite so it can just be taken out just like any other constant and ignored. Thanks for the question hope this helps!

[@kaan7866]

@@OscgrMaths Thanks a lot have a nice day :)

[@OscgrMaths]

@@kaan7866 You too :)!

[@adityarai3239]

Nerd, but a smart one

[@ellis-home2373]

I'm confused, why can we assume the definite integral with respect to t is the same as with respect to x? Couldn't f(t) be a completely different function from f(x) leading to the definite integrals not being equal?

[@poodook]

I = 4 on inspection

[@OscgrMaths]

Nice!

[@learnenglishwithash5383]

But I didn't get stumped!

[@OscgrMaths]

Well done!

[@Tkcb2799]

why do you substitute I for both f(x) and f(t)? Aren't those seperate functions?

[@OscgrMaths]

@@Tkcb2799 Thanks for the comment! They're the same function just like how y=t^2 is the same as y=x^2 since t and x in this case are just labels we've chosen for our axes.

[@Tkcb2799]

@@OscgrMaths Ohhhh I didn't know that hahah thanks for the reply! In physics, t is time so I was confused why x is the same as t 😅

[@pythontron8710]

Let "Int" be the integral from -1 to 1 of f(t) 6 + f(x) = 2f(-x) + 3x^2 * Int (6 + f(x) - 2f(-x)) / (3x^2) = Int Since the integral is constant, that means the expression on the left is a constant. Since the denominator has an x^2 term, that means in order for the L.H.S to be a constant, f(x) must be a quadratic. The integral of ax^2 + bx + c from -1 to 1 is (2/3)a + 2c (6 + f(x) - 2f(-x)) / (3x^2) = (2/3)a + 2c (6 + ax^2 + bx + c - 2ax^2 + 2bx - 2c ) / (3x^2) = (2/3)a + 2c (6 - ax^2 + 3bx - c) / (3x^2) = (2/3)a + 2c Since only an x^2 term can remain in the numerator for the L.H.S to be a constant for any value of x, that means b = 0 and c=6. (-ax^2) / (3x^2) = (2/3)a + 12 -a/3 = (2/3)a + 12 -a=12 --> a = -12 Therefore, f(x) = -12x^2 + 6

[@attica7980]

The video is a clickbait in that you need to click on it to find out what the question is.

[@koennako2195]

The way I got f(x) is by noticing that the definite integral has to evaluate to some number. If isolate the integral, I = [6+f(x)-2f(-x)]/[3x^2]. Seeing that that entire thing has to be a number with all xs canceling out, we know that the top has to be quadratic. Substituting f(x)=ax^2+bx+c, we easily get -12x²+6. I also could've immediately noticed that the top also can't have any x terms, so I could've concluded that b=0 immediately.

[@OscgrMaths]

Nicely done! Thanks for the comment.

[@r.i.p.volodya]

Oscar, are you part of the student body or faculty?

[@OscgrMaths]

Hey, I'm a student at the moment. Thanks for the comment!

[@theunfightable4513]

Spotting x^2 as highest order term, we can assume f(x)= a+bx+cx^2. Plugging that into the equation and equating coefficient yields an system of linear equations which can be easily solved by hand.

[@peterm8808]

Assuming f(x) is odd, gives I=0. Then with the assumption f(-x)=-f(x) one can then solve the equation for f(x). This gives f(x)=-6/3, which is a constant,(contradiction) and hence it follows f(x) cannot be odd. The case how it can be solved when f(x) is even was already explained here in the comments.. Another idea to solve this issue is to assume that f(x) has a Taylor series \sum_{n} a_n x^{n}. This I can plug into f(x)-2f(-x)=-6+ 3I x^2. Then I compare the coefficients on the left and right side.

[@OscgrMaths]

Nicely done! Thanks for the comment.

[@julianbruns7459]

How do you prove that f has a convergent taylor series?

[@OscgrMaths]

@@julianbruns7459 I think my approach might be a good one - it involves rearranging for f(-x), then subbing in -x for x and rearranging for f(-x) again and then equating the two and rearranging for f(x) since it assumes nothing about the function I think it should hopefully mean it's the only solution for f(x) - but let me know if there's a mistake in mine! Thanks for commenting.

[@peterm8808]

Yeah thats true, but to be fair there were not many conditions given. But assuming that there is a Taylor expansion gives you at least a solution and uniquness in the space of analytical functions.

[@OscgrMaths]

@@peterm8808 Definitely

[@MrBeen992]

Next time try to put the camera directly in front of the whiteboard and not to the side. Really hard to see

[@srmpenedo]

Nice explanation, but you just wrote u -> -1, x -> 1 twice

[@OscgrMaths]

Hey! I wrote u -> -1, x -> 1 then u -> 1, x -> -1 as this is what happens at the upper and then lower bounds respectively. Thanks!

[@onerelic1337]

ok i understand, but dosent the u substitution only work if the function f(x) is and even function?

[@OscgrMaths]

In this case we're lucky with the bounds - were they different it may not have behaved so nicely. Thanks for the comment!

[@TimoDeStefanis]

I've tried this problem multiple time and keep making the same error. I know where my error is coming from and am wondering if anyone can help explain why my technique is false. before integrating 2f(-x) I take the negative out and make it -2f(x), then I integrate from -1 to 1 with the -2 outside and get -2 I which not the value in the video. from what I understand you can simplify f(-x) to equal -f(x) but in this problem it doesn't work. I originally tried this problem because I thought my technique would save time by not having to do a U sub and change the bounds but clearly this misses a sign change and gives the wrong answer. If someone could explain why my technique is wrong that would be amazing. I can never quiet sleep right knowing I don't understand why I am wrong in a concept that I thought I understood.

[@OscgrMaths]

@@TimoDeStefanis You can only make that substitution for an odd function (like sine) but cosine is an example of a function where that is not true! For cosine f(-x)=f(x). Always have to go about the u sub method. Thanks for the comment and feel free to ask any more questions!

[@riccardofroz]

A = Integ x=-1 to 1 f(t)dt 6+f(x)=2f(-x)+3Ax^2 x=0 f(0)=6 f(x)-2f(-x)=3Ax^2-6 IF f(x)=-f(-x) 3f(x)=3Ax^2-6 f(x)=Ax^2-2 as f(0)=-2 instead of 6 NO! IF f(x)=f(-x) -f(x)=3Ax^2-6 f(x)=6-3Ax^2 Integ t=-1 to 1 6-3At^2 = A 6-A+6-A=A A=4 f(x)=6-12x^2

[@OscgrMaths]

@@riccardofroz Nicely done!

[@jacquesmouton428]

Well done chap, great video, but there was something that caught my eye. It is that you, through-out the entire video, forgot the Integration constant. I know a constant combined with another constant still stays a constant, therefore we add it only at the end, however you unfortunately did not. So, the answer for I is not necessarily 4, but rather I = 4 + C. The constant, C, is very important because it can completely change the original function, f(x). From f(x) = -12x^2 + 6, to f(x) = -3(4 + C)x^2 + 6. Not a big problem in total, but I hope this helps. Thank you and keep up the good work.

[@OscgrMaths]

Thanks for the comment! Since it's a definite integral there is no constant of integration.

[@Jacob.Peyser]

Assuming f(x) is differentiable, I computed the 0th, 1st, and 2nd derivatives of f(x) evaluated at x=0 using the original functional equation. The results were f(0)=6, f'(0)=0, and f''(0) being equal to -6C, where "C" is the integral of f(t) from t=-1 to t=1 (with respect to t). Then, for all higher order derivatives of f(x) (with respect to x) evaluated at x=0, it is conveniently equal to 0. Assuming f(x) to be analytic, I Taylor expanded f(x) centered at x=0 using what was just derived. At that point I had f(x)=6-3Cx^2. Then, after integrating that equation from x=-1 to x=1 (with respect to x), it was be obtained that C=12-2C. Thus, C=4 and f(x)=6-12x^2. Therefore, IF f(x) is analytic, THEN f(x)=6-12x^2.

[@OscgrMaths]

Nice solution! I did it differently but got to the same result. Thanks for the comment.

[@deweiter]

You can also assume that f(x) is analytic at the start of the solution, and after some moving you'll end up with Σ(a_n)(1-2(-1)^n)*x^n = 3Cx^2-6. Therefore, a_0=6, a_2=-3C and all other terms are zero. After subsstituting that back it the series, you'll and up with f(x)=6-3Cx^2

[@ZaWarudo_TokiWoTomare]

I kinda dont get how the integral of f(t) dt is the same as the integral of f(x) dx.

[@OscgrMaths]

@@ZaWarudo_TokiWoTomare Try working through both of them... you'll see it gives you the same thing both times! It's because the variable you use to represent what you're integrating isn't special - we just choose x for convention most of the time. In some cases there could be differences (like in double integrals, dx would indicate along the d axis and dy along the y axis) but here it's absolutely fine to swap freely. Thanks for the comment and feel free to ask anything more!

[@zzzman9287]

Let y = integral of f(x) dx Substitute t = x then dt = dx Now substitute t for x into the original integral: y = integral of f(t) dt

[@OscgrMaths]

@@zzzman9287 Nicee good job!! This is great.

[@ZaWarudo_TokiWoTomare]

@@zzzman9287 I didn't think it would be that simple. 😕

[@ZaWarudo_TokiWoTomare]

@OscgrMaths Thanks, this helped. Can you please answer the multiple choice questions on the older MAT papers from 1996-2006. They don't have free worked through solutions online, so I was hoping you could make the solutions with the explanations in video format.

[@susanario-l4q]

erm whats the application of this irl...?

[@kewalmer7225]

yeah it was simple got it within 30 seconds but interesting nonetheless thankyou.

[@eliibi4009]

What level of maths is this? A-level?

[@OscgrMaths]

@@eliibi4009 Very simple level maths - A level at most, just involves some problem solving.

[@cosimo7770]

O-level for any normal average 14-year old

[@MrLink-dk7yn]

I = -12

[@rafjeevarafjeeva5952]

6+f(x)=2f(-x)+3x² int(f(t) dt) 6+f(-x)=2f(x)+3x² int(f(t) dt) (because of the borns) f(x)-f(-x)=2f(-x)-2f(x) 3(f(x)-f(-x)=0 So f(-x)=f(x) So: 6+f(x)=2f(x)+3x²int(f(t)dt) f(x)+3x²int(f(t)dt)=6 The intégral is constant so we have f(x)=ax²+6 And int(f(t)dt)=2a/3+12 ax²+6+3x²(2a/3+12)=6 x²(3a+36)=0 For all x So 3a+36=0 => a=-12 So f(x)=-12x²+6

[@rafjeevarafjeeva5952]

I am in highschool btw

[@albajasadur2694]

@@rafjeevarafjeeva5952 I appreciate your approach of finding a scaled integral I rather than solving for the integral I and substituting it back into the equation.