Another LOVELY Integral Problem From Oxford Uni
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- Опубліковано 5 жов 2024
- Hope everyone enjoyed! Another shorter Oxford video - hope people enjoy this as much as the last one! 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.
Thanks so much for all the recent support - hitting 2k subscribers in the space of a few days having only just got to 1k is unbelievable - thanks to each and every one of you.
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By drawing out a function on a graph and calling the are under the curve between 0 and 1 "a", and from 1 to 2 "b". Then you have:
3a + 2b = 7
(a + b) + b = 1
a = 3, b = -1
Intergral from 0 to 2 is just a + b again which is 3 -1 = 2.
@@maxhagenauer24 Nice solution!
That's what I did too :)
I did similar method and got 2 but with Venn diagram and sets, hehehe
Love this! 👍 I always find it interesting when the solution to a problem involving a (nearly) arbitrary function doesn't require you to figure out what that function is!
So glad you enjoyed! I totally agree.
another incredible video man your explanations are phenomenal
Thank you! I really appreciate it.
I simply added both equations.
i love these videos, keep going ❤️
@@Microplastics2 Thank you!
Let the integral from 0 to 1 be called A and the integral from 1 to 2 be called B . Then we are looking for A+B. We have a system of equations
3A+2B=7
A+2B=1
Which quickly gives
A+B=2
Nicely done!
@@dalitlegreenfuzzyman Yes, the integrals really just obfuscate the underlying simple math. But I believe it being quite instructive to show students that definite integrals are really just plain numbers, too (well, unless they diverge).
I've recently started learning linear algebra and it was fun to apply the techniques in a problem involving integrals. Great video!
@@MrKoteha Great! Glad to hear it - thanks for commenting.
Really good job with this Oscar!
Watching your videos makes me really wish that I chose Further Maths 😅
They’re really well made!
Harry :)
@@somenorthlondoner Thank you!
i := x+y
[1] 3x+2y=7
[2] x+2y=1
[1]+[2] => 4(x+y)=4i=8 => i=2
I did this question yesterday in a past paper, it’s a great question, and a great video!
Thank you!
Let I:=∫₀¹ f(x) dx, J:=∫₁² f(x) dx, rewrite the question as a system:
3I + 2J = 7……….(1)
(I + J) + J = 1…..(2)
I + 2J = 1
I + (7 - 3I) = 1
-2I = -6
I = 3
3(3) + 2J = 7
2J = -2
J = -1
Ans = I + J = 3 - 1 = 2
This is the way I did it too!
This is equivalent to asking:
3x + 2y = 7
(x+y) + y = 1
x + y = ?
Which I’m sure we all see is quite easy.
2x=6->3,y=-1,sum=2
Before watching: break up the int in the second eqn into integrals over domain [0,1] and (1,2] and then solve simultaneously for the ints then sum the values found: answer seems to be 2
Okay so it seems I was correct. Thanks for the problem!
@@FPSIreland2 Glad you enjoyed! Thanks for the comment.
The interger numbers A [0 1] and interger numbers B [1 2]
The method to solve a systems of equations 3A+2B=7
2A+2B=4
That curveed interger
A+B=2
😊
Interesting question, I think I remember doing this as revision 😂
Superb! Thanks!
@@booshkoosh7994 Thanks so much! Glad you enjoyed.
I just solved it as a system of equations. I know the integral of f(x) is simply some area function F(x)
so 3F(1) - 3F(0) + 2F(2) - 2F(1) = 7
and F(2) - F(0) + F(2) - F(1) = 1
I want F(2) - F(0). Notice that if I add those two expressions together the F(1)s cancel and I get
4F(2) - 4F(0) = 8
so F(2) - F(0) = 2
So the integral from 0 to 2 of f(x) is just 2.
Pretty easy honestly.
Strangely, I am enjoying your videos so much! It reminds me of my time at uni doing math exercises like that and being proud having solved a little riddle. I wonder whether I should have pursued math instead of chemistry. Well, whatever.
Glad you enjoyed! Thanks for the comment.
This exercise seems easy af for Oxford. 😂
Otherwise, dope content fam!
Thank you! Yes I was surprised to find it on the exam too.
Interesting problem!
Thank you for sharing such nice video, I highly appreciate that.
I will not give up on your challenge, because I really like such problems.
I have shared one of 'almost' similar kind, did you have a chance to see it?
I took a quick look at it and it seems quite tricky (in a good way)! I'll sit down and properly attempt it soon hopefully. Thanks for the comment!
@OscgrMaths You are most welcome. But sorry actually I forgot that I did not send the one that is similar to this. What I have shared is a nice one, but not similar to this.
I will type it and will share it in a few minutes.
I wanted to discuss this with someone
I have observed the equation
14(n)-3 =prime no
For n=1,10,15,20,25,35,40,65,70,80,85,95,115,125,145,160,170,175,190,200,230,275,280,295,310
And so on.. thess are just some few
I searched till 1000 , this continues
I am wondering if this follows a pattern or just a coincidence
Waiting for your feedback..
my method was to suppose the integrand of f(x)=F(x) then sub in the limits for all then got the anwnser
Nice!
Alright having a go before watching, hang on this is just simultaneous equations from secondary school! So taking the factors to letters: 3a+2b=7, a+2b=1, therefore 2a=6, a = 3, b=-1, a+b = 2, final answer of combined int = 2. Yeah seems to reflect other comments. Funny way to put what looks like on the outset to be a weird functional equation.
Yeah that's what I love about these kinds of questions! Always a slightly different approach than is expected at first glance. Thanks for the comment!
@@OscgrMaths i love your channel, it's very original and you come up with things to solve I've never seen elsewhere plus you're super understandable as well. Keep it up!
@@DanDart Thank you!! That means a lot.
As a long time (ever since ur first video) follower of yours, I think this one was a bit too easy for this channel. I think you should solve harder questions next time around. Still good work tho!
Check out the other one I uploaded at the same time! I thought it was quite easy too - I'll avoid ones this easy in the future. Thanks for the comment!
@@OscgrMaths I surely will!
nice solution
by the way you look like a math major, i have something to discuss
I got 2!! ❤
Very hard to read anything from "the side".
Please put your camera right "in front" of the board.
Really easy question
Too easy!
Are you a math major or a math teacher?
Did you time the uploads such that this one would release at the same time as someone would finish the other video? Or was this just a coincidence?
Just a happy coincidence! Thanks for the comment.
Was this really a University of Oxford interview question? It's trivial.
Yes it was from an entrance exam. The MAT is surprisingly easy sometimes.
Quite an easy one today but good video nevertheless👍👍
@@slip3102 Thank you! Glad you enjoyed.
Easiest integral❤❤
can you help me in l hospital rule
asnwer=-1 .3 isit