1886 Cambridge University Exam Integral
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- เผยแพร่เมื่อ 9 พ.ค. 2024
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your thumbnail doesn't have ln x
Welp, someone got fired
ngl i was disappointed when i saw the bait and switch math problem when i clicked on the video. we all got played.
@@jordonm5675 nah he isn't a click bait channel like that, it's probably just an honest mistake imo
I saw the thumbnail, didn’t play the video, solved it to an answer of zero, then watched the switched integral.
Well, at least the final answer is the same.
(Oops, it might be pi/2 - (-pi/2) = pi , not 0 )
@@KilamSabbayes without the ln x it's pi: integrand is positive over the whole interval so can't be zero!
You never mentioned that technically this is an improper integral, as the denominator vanishes when x=4, 0.
And the numerator is undefined when x=0.
Man that's a use of a really satisfying method to get a really unsatisfying result.
Euler's substitution
sqrt(4x-x^2) = ux
Split into two integrals
2ln(2)\int\limits_{0}^{\infty}\frac{1}{1+u^2}du - 2\int\limits_{0}^{\infty}\frac{\ln(1+u^2)}{1+u^2}du
- 2\int\limits_{0}^{\infty}\frac{\ln(1+u^2)}{1+u^2}du
This integral can be calculated by u = tan(theta) substitution
MEGA! Thank you for presenting this integral! 😊
Just let x = 4sin^2(u) and dx = 8(sinu)(cosu)du. This will simplify much more rapidly. ☺
But such substitution is not "regular". Personally, I prefer to work out from known substitution like rewrite it as 4-(x-2)² then use x-2=2sin(t)
so you did both of his substitutions in one step?
Please explain
I don’t think anyone could possibly see that substitution- this may be shorter but it’s probably harder
@@kjl3080its fairly easy to see if the denominator is split into sqrt x and sqrt(1-x)
You missed the fraction line at the cover picture
Since when does sin(x-pi/2) ever equal sin(x)… no matter the interval
yeah have no idea where that came from
@@cheesetheorange4503 this makes me feel so weird about the whole proof
Came here to say this. What is he smoking?
Wonderful editing :)
Can’t imagine a tough time consuming exam question like this unless it’s multiple guess😂 In real life it would take hours to hammer it out and it proves little as an exam question ( other than probing your character).
Good use of the 'magic box that erases part of the blackboard' at 5:51 and 15:04. It's what got me interested in his videos in the first place!
Back in 1980 I got a 5 on AP Calc AB, and a 4 on AP Calc BC, but I would not have gotten into Cambridge back in 1886. Dang! All of these steps are things I would have known how to do (though in my aged rustiness I need to be reminded), but to put this together? Really cool stuff.
Easy freshman calc. Complete the square, substitute. Arc sine. Answer pi
Nice problem professor! I just stumbled over the integrand ln(sin x) on smaller math channel, and I remembered how you solved that problem using a phase shift.
I feel like this should be a field of study: the class of non trivial integrands whose integral vanish over some interval with nonzero measure.
Very relaxing! Thanks Michael!
Another cool way to compute ∫ln(sinx)dx is to use Riemann sums, as pi/2n∑ln(sin(k*pi/2n) tends to the value of the integral as n tends to infinity.
you then have to compute ∏sin(kpi/2n) which is classic and is equal to n/2^n. You then conclude with the uniqueness of the limit.
10:30 sin(u - pi/2) = sin(u)
Is this right? 🤔 Not -cos(x)?
It is correct. The equation says that sine is symmetric around x = pi/2.
Right; take u = 2pi/3, which is in the interval of u. sin(u-pi/2) = sin(pi/6) = 1/2. But sin(u) = sin(2pi/3) = sqrt(3)/2. So definitely not a valid substitution.
it is wrong: at u=pi/2, sin(0) is NOT sin(pi/2)!!
But, the integrals should be the same: instead of sin(u-pi/2) going from 0 to 1, it's sin(u) going from 1 to 0
@@buldysk1537this would mean sin(pi/2 + x) = sin(pi/2 - x). Just insert x=pi/2 in his formula.
You are correct and Michael is wrong. Since the integral of ln(sin x) from 0 to pi/2 is the same as the integral over ln(cos x), the third term should be a copy of the second term (with the cosine), and then the substitution u = x + pi/2 works to give the result Michael wants.
20:09 This is not the thumbnail 🤨
From what it said in the thumbnail the answer is π
First √(4x-x²)=√(4-x)*√(x)
u=√(4-x)
du=-[1]/[2√(4-x)]dx
-2du=[1/[√(4-x)]dx
u²=4-x
u²-4=-x
4-u²=x
√(4-u²)=√(x)
Change of limits
u(0)=2
u(4)=0
So {0}S{4}([dx]/[√(4x-x²)])=
{2}S{0}([-2du]/[√(4-u²)])
We can take the 2 out from the integral, and change the limits of integrations with the negative sign
2*{0}S{2}([du]/[(4-u²)])
Now trig substitution
u=2sin(θ)
du=2cos(θ)dθ
θ=arcsin([u]/[2])
θ(0)=arcsin(0)=0
θ(2)=arcsin(1)=[π]/[2]
u²=4sin²(θ)
4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ)
So
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])=
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|])
Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ)
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])=
2*{0}S{[π]/[2]}(dθ)=
2*θ{[π]/[2]\0}=
2*([π]/[2]-0)=π
Obi Wan Kenobi: "These are not the droids you are seeking."
If you do the substitution x=2(1-cos t) then this becomes integral of the even function ln(2-2 cos t) from 0 to pi, so 1/2 of the same integral from -pi to pi. Now this is a special case of Jensen's Lemma, which says that integral of ln | a - exp(it) | is equal to ln max (1, |a| ), for an complex number a. Just set a=1 to recover the case at hand.
(This result is used to define what is called the Mahler measure of a polynomial)
10:55 I got lost at this step. How can sin(x) subbed to make cos(t) result in cos(x) with the same bounds?
Wow, I haven't seen this coming, that the area on the interval from 1 to 4 (no pi around) exactly compensates the negative area from 0 to 1!
Nice to see my old room in the thumbnail :)
There is a way simpler way to do the first integral: Just substitute t^2 for x. You get 2 times integral of 1/sqrt(1-t^2) from zero to one 2 times a half circle area of radius one. pi.
Magnificent!
He always know the good place to stop 😮
this is simpler to solve by substituting x=t**2 and the t=cos(y) and the integral becomes trivial
I did it in my head. The answer’s 7.
I don't understand why you could say u and x are both just dummy variables when u is in terms of x.
What are the limitations as to which types of integrals that are computable in practice? - when do we just have to give up straight away?
does the integral from 0 to 1 of the same function have a nice value?
This got me thinking, I always pause when I do a trig substitution, because the range of (cos and sin) trig functions is -1 to 1, so you can't always use it, but with the explicit example of scaling the domain by a change of variables to x = 4y and thus enabling the use of trig substitutions, but this only works for definite integrals, I wonder if there's ever a case where you can do something similar with indefinite integrals and you take a limit or something. To give a crude example x = y * 1/h where h -> infinity. This won't work in most cases, but you might be able to get a cancelation in the integral, and therefore you technically convert an indefinite integral to a definite one where the bounds are between -1 and 1 so you can use cos or sin substitution. I've not fully thought this through, I've only had the idea seeded by the video.
integral of 1/sqrt(x-x^2)dx,x=0,1 is a lame integral. You perform the change of variable u=sqrt(x) then you perform the change of variable u=sin(t).
So the first step is to change the problem? is that how they do things at Cambridge?
I solved the thumbnail integral before watching the video 😭
What level of the Tripos did this question appear in?
Seems like a ton of these types of integrals end up with a ln2 in the solution. Just an observation
W Respect for Cambridge math!
Integ 0 to 2 is (-2 Catalan). The question is why? Something is going on. Maybe the world is flat we are just looking at it wrong.
I doubt anyone that has not seen the solution could solve this on a test. So what were they testing?
THANKS PROFESOR. !!!!, VERY INTERESTING !!!!
Stop yelling your post in all caps. Also, watch your spelling and grammar.
so, it seems i would fail it
I was doing this in my head and read it wrong, so I completed the square with the substitution u = (x-2), and ended up with arcsin of -2 and 2, which are imaginary! But the imaginary parts cancel and you get a real number. So that was a fun mistake.
Help me out, you call it a 'box' but isn't it a square?
20 minutes of viewing to end with a zero . 😀
The integral in the thumbnail is equal to pi. I have not tried to solve the one in the video yet.
Awesome job wit facebook
Rewrite the part under the root as 4(1-(x-2)^2/4) and substitute (x-2)/2=tanh t
I might have known ;-)
I just did a u sub to get it in a different form (for the thumbnail one)
From what it said in the thumbnail the answer is π
First √(4x-x²)=√(4-x)*√(x)
u=√(4-x)
du=-[1]/[2√(4-x)]dx
-2du=[1/[√(4-x)]dx
u²=4-x
u²-4=-x
4-u²=x
√(4-u²)=√(x)
Change of limits
u(0)=2
u(4)=0
So {0}S{4}([dx]/[√(4x-x²)])=
{2}S{0}([-2du]/[√(4-u²)])
We can take the 2 out from the integral, and change the limits of integrations with the negative sign
2*{0}S{2}([du]/[(4-u²)])
Now trig substitution
u=2sin(θ)
du=2cos(θ)dθ
θ=arcsin([u]/[2])
θ(0)=arcsin(0)=0
θ(2)=arcsin(1)=[π]/[2]
u²=4sin²(θ)
4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ)
So
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])=
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|])
Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ)
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])=
2*{0}S{[π]/[2]}(dθ)=
2*θ{[π]/[2]\0}=
2*([π]/[2]-0)=π
@@Ricardo_S my initial u was different I think.
If I remember correctly I got it into difference of squares by saying:
u=2+x
Then du=dx
And sqrt(x(4-x))= sqrt((2+u)(2-u))
=sqrt(4-u^2)
And the rest was trig subs I think.
You can do both subs at once but I didn’t want to type the whole trig function every time.
thanks a lot from UZBEKISTAN
In my opinion, you have to pay attention to the fact that 0 and 4 are not in the domain of the integrate.
The bounds of integration are indeed 'improper' for this case, but are simply handled by taking them to be limits approached by x.
@@Hiltok Ok, that is true, but I am afraid that the chain of reasoning, as presented by Michel Penn, is no longer valid.
If you graph 1/sqrt(4x-x^2), it is entirely above the x axis, meaning that it cannot have an integral of 0
Well, if you consider that the thumbnail is wrong and look at the actual function ...
Did you even watch the video? 😅
Mistake in the thumbnail.
From what it said in the thumbnail the answer is π
First √(4x-x²)=√(4-x)*√(x)
u=√(4-x)
du=-[1]/[2√(4-x)]dx
-2du=[1/[√(4-x)]dx
u²=4-x
u²-4=-x
4-u²=x
√(4-u²)=√(x)
Change of limits
u(0)=2
u(4)=0
So {0}S{4}([dx]/[√(4x-x²)])=
{2}S{0}([-2du]/[√(4-u²)])
We can take the 2 out from the integral, and change the limits of integrations with the negative sign
2*{0}S{2}([du]/[(4-u²)])
Now trig substitution
u=2sin(θ)
du=2cos(θ)dθ
θ=arcsin([u]/[2])
θ(0)=arcsin(0)=0
θ(2)=arcsin(1)=[π]/[2]
u²=4sin²(θ)
4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ)
So
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])=
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|])
Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ)
2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])=
2*{0}S{[π]/[2]}(dθ)=
2*θ{[π]/[2]\0}=
2*([π]/[2]-0)=π
English is impressed with tamil ! Grandfathers and babies learn maths ?
20 sec
2th problems is the key that is KING's Property
Aren't these integrals improper? Shouldn't that be part of the conversation? It drives me up a wall that being improper is just ignored. I have students that gloss over this fact so many times. In Michael's integrals here, it ultimately doesn't matter. But in general it does.
Rationalise denominators if necessary or convenient.
More than half the video was a waste! Just set x-2=2cos(theta) and go from there.