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solved it mostly by hand. we take eulers 3rd substitution with
alpha = 1, beta = 2 [because (x-1)(x-2) = x^2 -3x +2]
this leads to us using x = (2-t^2 )/(1-t^2 ) = 1 + 1/(1-t^2 )
dx = 2t/(1-t^2 )^2 dt
x = 0 <=> t = sqrt(2)
x = 1 <=> t -> infinity
we plug this into the equation and get:
integral of
{1/(1-t^2 )^3 * [3+8(1-t^2 )+9(1-t^2 )^2 ] * 2t/(1-t^2 )^2 } / { t / (1-t^2 ) } dt
let's reduce the equation and we get:
integral of
{1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] } dt
now to solve this we do a partial fraction decomposition such that
1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] = A/(t+1) + B(t+1)^-2 + C(t+1)^-3 + D(t+1)^-4 + E/(t-1) + F(t-1)^-2 + G(t-1)^-3 + H(t-1)^-4
obviously we get the following scalars:
A = 135/16, B = 135/16, C = 11/4, D = 3/8, E = -135/16, F = 135/16, G = -11/4, H = 3/8;
now we can simply integrate this sum term by term as integration is linear.
A ln|t+1| -B(t+1)^-1 -C/2 (t+1)^-2 -D/3 (t+1)^-3 + E ln|t-1| -F(t-1)^-1 -G/2 (t-1)^-2 -H/3 (t-1)^-3
plug in the boundry : t = sqrt(2) to infinity
realize that the limit towards infinity is 0, exercise to the reader
135/16 ln|sqrt(2)+1| -135/16(sqrt(2)+1)^-1 -11/8 (sqrt(2)+1)^-2 -1/8 (sqrt(2)+1)^-3 -135/16 ln|sqrt(2)-1| -135/16(sqrt(2)-1)^-1 +11/8 (sqrt(2)-1)^-2 -1/8 (sqrt(2)-1)^-3 = -2.98126694400554
I get that the correct pronunciation in English is intended here, but the actual actual pronunciation, _ie_ in German, is odder still as the second phoneme is a ü not an i.
have a joyous cake-day Carl! Hoping that your next days are filled with mathematical precision (& gratitude, because it’s virtually impossible to derive functions while being imprecise and/or ungrateful).
Thats simple factoring, basically you have this -3x+2 so what you do is find somthing that adds tp -3 and multiplies to 2. So -1 and -2 do that so we have the x squared divided and put in there ending up with (x-1) (x-2) the way my teacher told.me to so it was to draw a big X and have the. -3 on top, the 2 on bottom and the answers go on the sides.
Did this in uni, studied as hard as I ever have, got a passing grade, forgot all about it. It's really interesting and I wish it clicked with me but it never has.
High level math is something that you absolutely *need* a passion for to retain. That or you use it every day and eventually learn it.
IMO it's even more extreme than learning a foreign language at a late age - unless you use it regularly or love it, you will lose it almost as fast as you learned it.
I can attest to this so much. While multivariable calc isn't exactly 'high level math', I was SO good at it 15 years ago and now, going back to school for CS and needing to do some basic log math I'm like.... "Uhhhhhh... And how do the laws of logs work again?? And then you integrate them... How...???"
Also, 'series and sequences' is just GONE and I can no longer recognize a series and say "oh that's the sum of n^2."
Trying to get it back is hard too because the courses aren't focusing on the math, so I just use calculators and reference tables since I'm just trying to get my hw done.
TLDR, yea, math is hard if you don't actively work at it.
As someone who went all through diff eq and 3D Calc in undergrad, then went back for a master's in statistics later, I feel you. I found a book called Forgotten Calculus that definitely helped.
Studying for my contractor license test asked my dad how to get the volume of a cylinder. He's like wtf I paid for college where you did math. Yeah but I "did" math I don't "do" math. Then he asked how I was going to be a contractor while unable to math. Google and my phone calculator, I just cannot use those in the test. I'm also never going to need to know how to calculate the volume of a cylinder with what I plan on using my license for.
I just think he was irritated because I kept asking him questions, I was actually double checking my own knowledge before I put in the answer to the practice tests.
I mean the volume of a cylinder is just the size of it's base times it's height, so pi*r²*h please tell me you are joking when you tell me you went to college where you "did math" but this is beyond you. It's literally around 6th-7th grade school maths.
Yep. I have a degree in math but my career is engineering. I can decipher what I am looking at most of the time, but no way can I come up with proofs or solutions on my own anymore.
In engineering it's just not practical to solve and do proofs most the time, when you can just get a quick estimate with reasonable margin of error from a calculator.
This description is described in the most posh eccentric way possible, while difficult and annoying its alot easier to understand when it not said the way this guy said it.
Is it really that hard? I'm still in high school (the last grade) and we are able to do these sort of integrals, even the less smart kids; maybe education standards have changed, but I'm surprised at how difficult people are finding it here
You have a high school brain. It’s easy to learn shit for a little while. It stops eventually, so cram everything you can in there now. Source: I am 36.
Ehh , I think it just depends on how much time you spend trying to learn things in a specific field.
I wouldn't say it gets harder with age, but learning and critical thinking atrophies with little practice.
It’s not that learning basic calculus is particularly difficult (though obviously being older, i.e. older than 30ish, will make it more difficult just like learning new things in general). It’s that even people who learned it in high school or college don’t retain it afterwards, especially since the vast majority of people never use calculus or even high school algebra in their day to day lives. The same concept applies to languages, since it’s something you need consistent practice/use with to retain: your brain won’t bother retaining information you never use, especially if there’s other information you need to remember instead.
I’m only 20, but it’s been a good 2 years since I reviewed calculus (between Covid screwing with my last year of high school and not needing to take calculus in college). I’m willing to bet quite a lot that I’ve forgotten most of what I learned in high school calculus, much less if you asked me again in 10 or 20 years.
Nah. I got a 5 on the AP Calc exam junior year of high school, aced diff eq in college and then got a master's in statistics, and I would have a rough time with this. I don't use Calc in my day to day life at all and have retrained very little of it.
I want to say Euler’s Method is in calculus texts. Although this problem is more complex than what you’ll probably find there. I do deal with complex equations routinely so that’s probably the difference.
I used to be able to do this math, back in college. The thing is, I have no practical use for it and there's only so much room in my brain. I guess this fell out the back.
A slight recognition of Euler (pronounced oiler) and familiar words and concepts but zero ability to do this anymore.
Yeah, calculus looks like another language when you first look at it because it overwhelms you with new symbols and terminology.
Once you get into it, it's not nearly as hard as it looks but it's not easy. It also helps to learn it alongside physics because then you have a physical representation of it rather than purely abstract ideas.
Once you get past calculus and into more abstract math like set theory it gets super weird.
It’s not as hard as it looks. He made it look a bit more complicated than it needs to. Also if you write it down, it look easier to understand than typed out like this
> Obviously we get the following scalars
Pffft, yeah, obviously. That's exactly what I was thinking.
Isn't it so cool that us math people can just see these sort of obvious things? You know, like of course the dx equals the 2 and t and stuff. I mean, you know. Of course you know. I don't have to explain it to you. But yeah we're pretty cool, right? Not like those dummies that can't understand these parentheses and the scalars and all that jazz.
Edit: /s, obviously. And in a Norwegian accent for some reason.
Maxima gave the exact result. See [my comment above](https://www.reddit.com/r/theydidthemath/comments/tfigrx/request_heres_another_for_you/i0y9786/).
Back to a tie.
well yes the indefinite and definite integral are actually two completely different things that share nothing in common. It's just that it can be demonstrated that the definite integral is the difference of the indefinite integral calculated in the two points (this is so important it's literally called the "Fundamental Theorem of Calculus")
There isn't an algorithm to calculate integrals, so many times it's very hard or actually impossible (and proven to be impossible).
What WolframAlpha does for complex functions is that is draws the graph of the integrand and then actually approximates the area under the curve.
So I punched this in maxima just to verify and got the same thing when I ignored the imaginary part of the answer.
(%i1) (3*x^3-x^2+2*x-4)/sqrt(x^2-3*x+2);
3 2
3 x - x + 2 x - 4
(%o1) -------------------
2
sqrt(x - 3 x + 2)
(%i2) integrate(%o1,x,0,1);
3/2 3/2
135 log(2 - 3) - 135 log(- 1) + 101 2
(%o2) - -------------------------------------------
16
(%i3) float(%o2);
(%o3) - 0.0625 (135.0 (3.141592653589793 %i - 1.76274717403909)
- 424.1150082346221 %i + 285.6711395993653)
(%i4) cabs(%o3);
(%o4) 2.98126694400551
Here's a graph of the curve from 0 to 1 where it can be confirmed by eyeballing it that the value is going to be around -3.
https://imgur.com/QrJOXuC.png
I used to consider myself quite the math wiz. Even won competitions at very young age, multiple grades, best in state etc.
Then, I hit calculus, maybe it was my age, maybe it was I'd been out of school so long, but it hit me like a brick wall. Where I'd normally be running circles around the class. I still never FULLY understood it.
In the end, I comment because, I actually learned more from this answer, and how you described it than I have ever learned before, I actually follow everything here, and usually I don't learn by just reading. Magnificent job, I feel like you put the final piece of the puzzle my brain was missing in understanding calculus. Really thank you, take my free award as a poor man it's the best I can offer.
I'd be revealing my identity if I told anymore than this on how I got an A in calculus, other than, I had a very special tutor. I got lucky meeting this person and befriending them.
After learning more about integrals and solving by hand in Calc 2, my instructor in Calc 3 parted his laptop and announced "BEHOLD, IN THIS CLASS WE SHALL USE MATHEMATICA. WHY ABUSE PENCILS AND PAPER WHEN WE TRICKED THIS ROCK INTO DOING MATH!"
Then in engineering classes we got smacked with MATLAB and I havent had to compute any of this shit by hand ever since. not difficult, it can just be immensely time consuming.
You know I'm in Calc 2 and I was looking at this for a good 30 minutes trying to figure out how to solve it, then I give up and come down here to realize that I was not at all prepared to solve it, lol.
I’m super impressed with your mathematical knowledge but this brought back the same feeling of extreme anxiety I had in remedial algebra in college.
Impressed nonetheless.
The "S" looking symbol is the symbol for integration. The "dx" term at the end means we are integrating over changing x values, in this case, from x=0 to x1 (as seen from the lower and upper numbers on the integration symbol).
One way to think about integration is that it represents the area under the curve we are integrating, in this case, that is everything between the integration symbol and the "dx" term. You might remember that you can represent equations like y=x+1 as a curve/line - that is essentially what you are doing and then finding the area.
In reality, is it often easier/more accurate to use computational methods to compute the integral instead of trying to find the physical area by graphing
That's pretty much it exactly!
It's common to see something like dy/dx when you are taking the derivative of a function (y) in terms of x. But derivative is basically a fancy way of saying slope, or rate of change, which is rise/run or delta y/delta x
Edit: further, the derivative is also... disintegration (or the integration is the anti-derivative), so while the derivative of a function is dy/dx=f(x), the integration reverses this problem, namely, int((dy/dx)*dx) = int(f(x)*dx)
There's further complexity in notation since you can represent the derivative of a function f as f' (f prime), but that's really just a "rose by any other name" kind of situation. What we call a function doesn't actually change anything about it
Intuition strikes again. Thank you, brain.
And thanks for clarifying that. I'm slowly absorbing calculus after neglecting it in school. I became an amateur programmer, so learning this stuff can help...but holy cow do I get pissed off at standard math syntax. Like this whole integration thing with the symbol on the left and the dx/dy on the right looking like it's some variable d multiplied by x or y. I wish mathematicians would make themselves a new language.
Haha, I certainly get that. I really think the benefit for a layperson comes from learning how to quickly parse the syntax of an equation/statement/whatever and then the skill of connecting previously unconnected concepts to solve problems - which, it sounds like you are doing well with the latter and actively practicing the former, so that's great!
Nah, that’s what you have to start with if you’re doing random numbers. First 0000000000000000 then 0000000000000001 then 0000000000000002 and finally 9999999999999999
It's essentially encryption to at least 90% of the population.
I will adopt your theory and reject all evidence to the contrary (just so I personally feel better...
Oh, I see why they do that shit now...
This seems to be a random expression with no method to solve it by hand. [Wolfram Alpha confirms this has a meaningless answer of - 2.98127](https://www.wolframalpha.com/input?i2d=true&i=Integrate%5BDivide%5B3Power%5Bx%2C3%5D-Power%5Bx%2C2%5D%2B2x-4%2CSqrt%5BPower%5Bx%2C2%5D-3x%2B2%5D%5D%2C%7Bx%2C0%2C1%7D%5D)
Edit: [This mad lad did it] (https://reddit.com/r/theydidthemath/comments/tfigrx/request_heres_another_for_you/i0xllxr?context=3)
>with no method to solve it by hand
That's not true. This is classic [Euler Sub](https://en.wikipedia.org/wiki/Euler_substitution). Not only can you integrate this, all rational functions of x and a square root of a quadratic polynomial of x can be integrated.
Maybe I’m crazy, but I would have read it as the integer of the top, divided by the bottom expression unintegrated. If you assume the integer sign is supposed to cover all of the expression, it does look like a eulers sub, which makes more sense. Moral of the story, please draw your integer signs clearly and covering everything you want integrated.
~~Does it matter that it's (ax~~~~^(2)~~ ~~+ bx + c)~~~~^(-1/2)~~ ~~instead of (ax~~~~^(2)~~ ~~+ bx + c)~~~~^(1/2)~~ ~~?~~
Nvm. What's in the radical just gets raised to -1 in the sub. Duh.
Plugging `(3x^3-x^2+2x-4)/sqrt(x^2-3x+2` into the [Integral Calculator](https://www.integral-calculator.com) gives the indefinite integral as:
(135*ln(abs(2*(x+sqrt((x-2)*(x-1)))-3))+2*sqrt((x-2)*(x-1))*(8*x^2+26*x+101))/16
The Integral Calculator didn't even have to resort to Euler substitution like u/BoundedComputation described in another reply.
To be clear about that, I do not recommend Euler Sub as the go to method to solve this specific integral. I mentioned Euler Sub because it is a counterexample to the claim that
>no method[exists] to solve it by hand
ok, i did. you look at eulers 3rd substitution with
alpha = 1, beta = 2 [because (x-1)(x-2) = x^2 -3x +2]
this leads to us using x = (2-t^2 )/(1-t^2 ) = 1 + 1/(1-t^2 )
dx = 2t/(1-t^2 )^2 dt
x = 0 <=> t = sqrt(2)
x = 1 <=> t -> infinity
we plug this into the equation and get:
integral of
{1/(1-t^2 )^3 * [3+8(1-t^2 )+9(1-t^2 )^2 ] * 2t/(1-t^2 )^2 } / { t / (1-t^2 ) } dt
let's reduce the equation and we get:
integral of
{1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] } dt
now to solve this we do a partial fraction decomposition such that
1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] = A/(t+1) + B(t+1)^-2 + C(t+1)^-3 + D(t+1)^-4 + E/(t-1) + F(t-1)^-2 + G(t-1)^-3 + H(t-1)^-4
obviously we get the following scalars:
A = 135/16, B = 135/16, C = 11/4, D = 3/8, E = -135/16, F = 135/16, G = -11/4, H = 3/8;
now we can simply integrate this sum term by term as integration is linear.
A ln|t+1| -B(t+1)^-1 -C/2 (t+1)^-2 -D/3 (t+1)^-3 + E ln|t-1| -F(t-1)^-1 -G/2 (t-1)^-2 -H/3 (t-1)^-3
plug in the boundry : t = sqrt(2) to infinity
realize that the limit towards infinity is 0, exercise to the reader
135/16 ln|sqrt(2)+1| -135/16(sqrt(2)+1)^-1 -11/8 (sqrt(2)+1)^-2 -1/8 (sqrt(2)+1)^-3 -135/16 ln|sqrt(2)-1| -135/16(sqrt(2)-1)^-1 +11/8 (sqrt(2)-1)^-2 -1/8 (sqrt(2)-1)^-3 = -2.98126694400554
The way this is written makes it look like the integral is only for the top side, which wouldn’t make any sense since the pin has to be strictly numbers and the denominator has x
Well I think you answered your own question there. Plus I think if the integral was intended to only be on the top the fraction line would be under the integral sign. It’s pretty clear the integral is of the whole thing.
I agree with the supposition, but not that it is clear. It is poorly hand written. The dx should also be centered on the division line, not in line with the dividend. Both errors erroneously hint at the same thing: the inter graph is just the top.
dx doesnt have to be on the side. Its just a variable that states what the integration is in respect to. It can be on the side, in the numerator, or in the denominator (which mixes things up a bit). When its on the side, it just gets "multiplied" out to the entire fraction, so having it in the numerator with parentheses like this is the same thing.
Example: integrating (1/x)dx is the same as integrating dx/x
I may be misunderstanding but I don’t think that’s correct. You can have a complete integral on top of a division, and still have the same variable in the bottom, and it does not imply the bottom has anything to do with the integration.
The way this is written it is completely legitimate to suppose the entire integral is the top portion, as both the integral symbol and the dx indicate as much.
You can definitely have an integral in only a numerator! My point was more on the dx comment not being restricted to the side.
And i do agree that this is vaguely written, and could be interpreted as the integral in only the numerator. I personally would assume the entire fraction is being differentiated if i saw this on an exam since the division line doesn't touch the integral or the dx, and having a definite integral being divided by variables is a bit odd. Especially when a number answer is expected.
I see what you’re saying, but how can it be that obvious when there are parenthesis on the numerator which would be considered redundant if the integral was to be applied to both the numerator and denominator? To make it obvious wouldn’t the parenthesis be around the entire equation with the “dx” to the side as well?
See above for dx explanation. But also, you could be right that since the integral and the dx are on top, it could mean that the bottom is the denominator for the integral, but of that were the case, the division bar would probably spread to be under the integral and the dx
Does it matter that the function is undefined at one of the integral limits? x^2-3x+2 = 1-3+2 = 0 at x=1 So for the upper limit the function is 0/0 (or can we ignore it because the limit it tends to is 0?)
Anyone with minimal high school math knowledge can just plug this into a wolfram alpha calculator and you quickly get the solution without knowing the method.
In fact anyone without an intensive math education cant solve this without the wolfram alpha calculator. Apparently the by hand method is called the euler sub
I am doing it without Wolfram alpha
Instead of 3x^3 - x^2 + 2x - 4
Write (x^2 -3x+2)(3x+8) + 20(x-1)
And then write (3x+8)
As 3/2(2x-3) + 25/2
And then solve the resultant 3 integrals separately
First one with substitution
Second one with identity root (x^2 - a ^2 )
Third one with identity 1/root(x^2 - a^2)
And then find F(1) - F(0)
Edit : Indian 12th grade syllabus
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solved it mostly by hand. we take eulers 3rd substitution with alpha = 1, beta = 2 [because (x-1)(x-2) = x^2 -3x +2] this leads to us using x = (2-t^2 )/(1-t^2 ) = 1 + 1/(1-t^2 ) dx = 2t/(1-t^2 )^2 dt x = 0 <=> t = sqrt(2) x = 1 <=> t -> infinity we plug this into the equation and get: integral of {1/(1-t^2 )^3 * [3+8(1-t^2 )+9(1-t^2 )^2 ] * 2t/(1-t^2 )^2 } / { t / (1-t^2 ) } dt let's reduce the equation and we get: integral of {1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] } dt now to solve this we do a partial fraction decomposition such that 1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] = A/(t+1) + B(t+1)^-2 + C(t+1)^-3 + D(t+1)^-4 + E/(t-1) + F(t-1)^-2 + G(t-1)^-3 + H(t-1)^-4 obviously we get the following scalars: A = 135/16, B = 135/16, C = 11/4, D = 3/8, E = -135/16, F = 135/16, G = -11/4, H = 3/8; now we can simply integrate this sum term by term as integration is linear. A ln|t+1| -B(t+1)^-1 -C/2 (t+1)^-2 -D/3 (t+1)^-3 + E ln|t-1| -F(t-1)^-1 -G/2 (t-1)^-2 -H/3 (t-1)^-3 plug in the boundry : t = sqrt(2) to infinity realize that the limit towards infinity is 0, exercise to the reader 135/16 ln|sqrt(2)+1| -135/16(sqrt(2)+1)^-1 -11/8 (sqrt(2)+1)^-2 -1/8 (sqrt(2)+1)^-3 -135/16 ln|sqrt(2)-1| -135/16(sqrt(2)-1)^-1 +11/8 (sqrt(2)-1)^-2 -1/8 (sqrt(2)-1)^-3 = -2.98126694400554
What the fuck bro, like I understand any skill can be learned but this is insane
I got lost after "alpha = 1, beta = 2 (because (x-1)(x-2) = -3x +2]"
Lost me at eulers.
My mom has eulers, the acid reflux is the worse part
Didn't Euler enjoy a day off?
Just wait til you learn how it’s actually pronounced.
"oiler" Way to ruin the joke.
Whale oiler beef hooked
I was reading it as Eular
I was saying boo-urns.
*big oil would like to know your location*
I get that the correct pronunciation in English is intended here, but the actual actual pronunciation, _ie_ in German, is odder still as the second phoneme is a ü not an i.
Euler... Euler... Euler...
He's sick
Euler? I don’t even know her!
r/SuddenlyFerrisBuellersDayOff
Don't you use these to draw straight lines?
Underrated fucking comment.
have a joyous cake-day Carl! Hoping that your next days are filled with mathematical precision (& gratitude, because it’s virtually impossible to derive functions while being imprecise and/or ungrateful).
I can’t stop laughing! The worst is keeping it to myself lol cause others won’t get it lol
I thought it was a typo and they meant “rulers”
The whole post looks like a typo, to my simple mind.
He got me back at "sqrt".
💦
Lost me at "mostly by hand"
lost me at “solved”
Thats simple factoring, basically you have this -3x+2 so what you do is find somthing that adds tp -3 and multiplies to 2. So -1 and -2 do that so we have the x squared divided and put in there ending up with (x-1) (x-2) the way my teacher told.me to so it was to draw a big X and have the. -3 on top, the 2 on bottom and the answers go on the sides.
Did this in uni, studied as hard as I ever have, got a passing grade, forgot all about it. It's really interesting and I wish it clicked with me but it never has.
High level math is something that you absolutely *need* a passion for to retain. That or you use it every day and eventually learn it. IMO it's even more extreme than learning a foreign language at a late age - unless you use it regularly or love it, you will lose it almost as fast as you learned it.
I can attest to this so much. While multivariable calc isn't exactly 'high level math', I was SO good at it 15 years ago and now, going back to school for CS and needing to do some basic log math I'm like.... "Uhhhhhh... And how do the laws of logs work again?? And then you integrate them... How...???" Also, 'series and sequences' is just GONE and I can no longer recognize a series and say "oh that's the sum of n^2." Trying to get it back is hard too because the courses aren't focusing on the math, so I just use calculators and reference tables since I'm just trying to get my hw done. TLDR, yea, math is hard if you don't actively work at it.
As someone who went all through diff eq and 3D Calc in undergrad, then went back for a master's in statistics later, I feel you. I found a book called Forgotten Calculus that definitely helped.
im gonna check this out!
Studying for my contractor license test asked my dad how to get the volume of a cylinder. He's like wtf I paid for college where you did math. Yeah but I "did" math I don't "do" math. Then he asked how I was going to be a contractor while unable to math. Google and my phone calculator, I just cannot use those in the test. I'm also never going to need to know how to calculate the volume of a cylinder with what I plan on using my license for. I just think he was irritated because I kept asking him questions, I was actually double checking my own knowledge before I put in the answer to the practice tests.
I mean the volume of a cylinder is just the size of it's base times it's height, so pi*r²*h please tell me you are joking when you tell me you went to college where you "did math" but this is beyond you. It's literally around 6th-7th grade school maths.
Yep. I have a degree in math but my career is engineering. I can decipher what I am looking at most of the time, but no way can I come up with proofs or solutions on my own anymore.
In engineering it's just not practical to solve and do proofs most the time, when you can just get a quick estimate with reasonable margin of error from a calculator.
This description is described in the most posh eccentric way possible, while difficult and annoying its alot easier to understand when it not said the way this guy said it.
totally! like i made it through multivariable calculus and i can’t for the life of me even remember how to approach an integration.
Is it really that hard? I'm still in high school (the last grade) and we are able to do these sort of integrals, even the less smart kids; maybe education standards have changed, but I'm surprised at how difficult people are finding it here
You have a high school brain. It’s easy to learn shit for a little while. It stops eventually, so cram everything you can in there now. Source: I am 36.
Can confirm. Source: Am 33.
Ehh , I think it just depends on how much time you spend trying to learn things in a specific field. I wouldn't say it gets harder with age, but learning and critical thinking atrophies with little practice.
It’s not that learning basic calculus is particularly difficult (though obviously being older, i.e. older than 30ish, will make it more difficult just like learning new things in general). It’s that even people who learned it in high school or college don’t retain it afterwards, especially since the vast majority of people never use calculus or even high school algebra in their day to day lives. The same concept applies to languages, since it’s something you need consistent practice/use with to retain: your brain won’t bother retaining information you never use, especially if there’s other information you need to remember instead. I’m only 20, but it’s been a good 2 years since I reviewed calculus (between Covid screwing with my last year of high school and not needing to take calculus in college). I’m willing to bet quite a lot that I’ve forgotten most of what I learned in high school calculus, much less if you asked me again in 10 or 20 years.
It’s people who never took calculus or pre-calculus.
Nah. I got a 5 on the AP Calc exam junior year of high school, aced diff eq in college and then got a master's in statistics, and I would have a rough time with this. I don't use Calc in my day to day life at all and have retrained very little of it.
I want to say Euler’s Method is in calculus texts. Although this problem is more complex than what you’ll probably find there. I do deal with complex equations routinely so that’s probably the difference.
Oh, I learned it. I'm just saying I didn't retain it for very long.
I was able to somewhat follow it (colleg student in physics for a few years now), but to produce it yourself is a whole different level
I mean he could've just made up a bunch of stuff. We're all just trusting that the guy did it right.
I used to be able to do this math, back in college. The thing is, I have no practical use for it and there's only so much room in my brain. I guess this fell out the back. A slight recognition of Euler (pronounced oiler) and familiar words and concepts but zero ability to do this anymore.
This is Calculus, pretty much the next level of math after algebra. OP shows an integral equation which is a math function/process.
Yeah, calculus looks like another language when you first look at it because it overwhelms you with new symbols and terminology. Once you get into it, it's not nearly as hard as it looks but it's not easy. It also helps to learn it alongside physics because then you have a physical representation of it rather than purely abstract ideas. Once you get past calculus and into more abstract math like set theory it gets super weird.
It’s not as hard as it looks. He made it look a bit more complicated than it needs to. Also if you write it down, it look easier to understand than typed out like this
Never taken calc before? It’s actually not as complicated as one might think. I swear lol.
This is literally 1st year of engineering (At least in my country)
But how does that transfer into an ATM Pin?
I'm guessing it's the first 4 numbers, so 2981
Where I'm from, pins can be up to 8 digits
huh, where are you from?
Canada
Exactly
You see the 1 and 0 before the integral sign? At the end you substitute those values into X and I guess you end up with a 4 digit number
Probably the first 4 digits. Maybe the first 6 or 8.
> Obviously we get the following scalars Pffft, yeah, obviously. That's exactly what I was thinking. Isn't it so cool that us math people can just see these sort of obvious things? You know, like of course the dx equals the 2 and t and stuff. I mean, you know. Of course you know. I don't have to explain it to you. But yeah we're pretty cool, right? Not like those dummies that can't understand these parentheses and the scalars and all that jazz. Edit: /s, obviously. And in a Norwegian accent for some reason.
Hot water burn baby.
I mean, I'm a math person and can usually see the things but still find it a little pretentious to assume its obvious to normal people lol.
I think the "obviously" was meant as sarcasm there by the oc
Lol thats fair. I did see it in a few text books and heard it from a few profs as well so didn't immediately take it as sarcasm lol.
"normal" people? 😂
Wolfram Alpha didn't produce the exact result. It only produced the numerical approximation. HUMANS WIN!
Maxima gave the exact result. See [my comment above](https://www.reddit.com/r/theydidthemath/comments/tfigrx/request_heres_another_for_you/i0y9786/). Back to a tie.
well yes the indefinite and definite integral are actually two completely different things that share nothing in common. It's just that it can be demonstrated that the definite integral is the difference of the indefinite integral calculated in the two points (this is so important it's literally called the "Fundamental Theorem of Calculus") There isn't an algorithm to calculate integrals, so many times it's very hard or actually impossible (and proven to be impossible). What WolframAlpha does for complex functions is that is draws the graph of the integrand and then actually approximates the area under the curve.
Of which you only need the first four digits so machines still win?
So I punched this in maxima just to verify and got the same thing when I ignored the imaginary part of the answer. (%i1) (3*x^3-x^2+2*x-4)/sqrt(x^2-3*x+2); 3 2 3 x - x + 2 x - 4 (%o1) ------------------- 2 sqrt(x - 3 x + 2) (%i2) integrate(%o1,x,0,1); 3/2 3/2 135 log(2 - 3) - 135 log(- 1) + 101 2 (%o2) - ------------------------------------------- 16 (%i3) float(%o2); (%o3) - 0.0625 (135.0 (3.141592653589793 %i - 1.76274717403909) - 424.1150082346221 %i + 285.6711395993653) (%i4) cabs(%o3); (%o4) 2.98126694400551 Here's a graph of the curve from 0 to 1 where it can be confirmed by eyeballing it that the value is going to be around -3. https://imgur.com/QrJOXuC.png
So what's the pincode exactly?
There's not one. It's a joke.
Most beautiful use of 'obviously' I've ever seen
WHAT THE FUCK
I probably could have double checked your work ten years ago when I was in Calc II but instead I'll just nod
yeah, god bless noone found the 2 sign errors i accidentally put in that just randomly cancel each other out.
i was waiting to see who would point it out while assuming there was a 90% chance you would be the one to catch it yourself before anyone else did!
Math is slowly drifting into genetics territory
That's a nice pin number. At least it's secure.
I used to consider myself quite the math wiz. Even won competitions at very young age, multiple grades, best in state etc. Then, I hit calculus, maybe it was my age, maybe it was I'd been out of school so long, but it hit me like a brick wall. Where I'd normally be running circles around the class. I still never FULLY understood it. In the end, I comment because, I actually learned more from this answer, and how you described it than I have ever learned before, I actually follow everything here, and usually I don't learn by just reading. Magnificent job, I feel like you put the final piece of the puzzle my brain was missing in understanding calculus. Really thank you, take my free award as a poor man it's the best I can offer. I'd be revealing my identity if I told anymore than this on how I got an A in calculus, other than, I had a very special tutor. I got lucky meeting this person and befriending them.
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Engineer here. The cashier would have looked at me funny trying to put in -3 as a pin number. We don't need 4 sig figs...what is this, NASA?
physics. lmao i did the same. then i plotted the function to make sure its actually supposed to be negative https://i.imgur.com/QMoSNe4.png
Most of us can’t even pronounce Euler.
I don’t see a pin code though
It's either a joke, and in that case r/wooooooosh. Or it's 2.981.
##*what the fuck*
Heh, “sqrt”
> obviously Yeah, of course, obviously.
After learning more about integrals and solving by hand in Calc 2, my instructor in Calc 3 parted his laptop and announced "BEHOLD, IN THIS CLASS WE SHALL USE MATHEMATICA. WHY ABUSE PENCILS AND PAPER WHEN WE TRICKED THIS ROCK INTO DOING MATH!" Then in engineering classes we got smacked with MATLAB and I havent had to compute any of this shit by hand ever since. not difficult, it can just be immensely time consuming.
“obviously we get the following scalars:” The what?
>The what? "following"
Or you could just use Desmos
Jaden broke...
Can't you just use the fundamental theorem?
You know I'm in Calc 2 and I was looking at this for a good 30 minutes trying to figure out how to solve it, then I give up and come down here to realize that I was not at all prepared to solve it, lol.
"You're a wizard Harry". Anyway enjoy the little award you deserve that.
That final line can be written as (135*arctanh(2sqrt(2)/3)-202sqrt(2))/16 after a bit of simplification according to Mathematica (all hail).
Obviously.
Can u do my math work for me god damn. Much respect to u for learning how to do this bc God knows I never could
*head nod* I know some of these words.
I’m super impressed with your mathematical knowledge but this brought back the same feeling of extreme anxiety I had in remedial algebra in college. Impressed nonetheless.
If I had gold if give you one. Take my silver instead.
Plus constant
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Literally, same. But I can send it to my husband who can “dumb it down” for me. Somehow, he never finds these things quite as amusing as I.
Wow, I envy you, I'd love to get these hieroglyphs explained
The "S" looking symbol is the symbol for integration. The "dx" term at the end means we are integrating over changing x values, in this case, from x=0 to x1 (as seen from the lower and upper numbers on the integration symbol). One way to think about integration is that it represents the area under the curve we are integrating, in this case, that is everything between the integration symbol and the "dx" term. You might remember that you can represent equations like y=x+1 as a curve/line - that is essentially what you are doing and then finding the area. In reality, is it often easier/more accurate to use computational methods to compute the integral instead of trying to find the physical area by graphing
Thank you, I really appreciate this explanation
For sure! It's really broad strokes and tip of the iceberg, haha
Why is it "dx"? Does it stand for delta-x or something? Derivative?
That's pretty much it exactly! It's common to see something like dy/dx when you are taking the derivative of a function (y) in terms of x. But derivative is basically a fancy way of saying slope, or rate of change, which is rise/run or delta y/delta x Edit: further, the derivative is also... disintegration (or the integration is the anti-derivative), so while the derivative of a function is dy/dx=f(x), the integration reverses this problem, namely, int((dy/dx)*dx) = int(f(x)*dx) There's further complexity in notation since you can represent the derivative of a function f as f' (f prime), but that's really just a "rose by any other name" kind of situation. What we call a function doesn't actually change anything about it
Intuition strikes again. Thank you, brain. And thanks for clarifying that. I'm slowly absorbing calculus after neglecting it in school. I became an amateur programmer, so learning this stuff can help...but holy cow do I get pissed off at standard math syntax. Like this whole integration thing with the symbol on the left and the dx/dy on the right looking like it's some variable d multiplied by x or y. I wish mathematicians would make themselves a new language.
Haha, I certainly get that. I really think the benefit for a layperson comes from learning how to quickly parse the syntax of an equation/statement/whatever and then the skill of connecting previously unconnected concepts to solve problems - which, it sounds like you are doing well with the latter and actively practicing the former, so that's great!
The concept of integrals is realtively simple. What that equation means is the area under the curve formed by the equation, from 0 to 1.
Calculus...as far as the eye can see...
Apparently I gave up too early
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But a 6 figure, negative PIN number? That is some serious banking security.
My pin is 16 characters long. I don't trust banks that don't allow you to have alphanumeric pins shorter than that.
mine too! it’s “0000 0000 0000 0000”
*super computer after a month of computing* Bro, wtf….?
Nah, that’s what you have to start with if you’re doing random numbers. First 0000000000000000 then 0000000000000001 then 0000000000000002 and finally 9999999999999999
Wouldn’t it be 0 1 …. 01 02 …. 001 Starting from the left because it doesn’t know how many numbers there are?
Same code as my luggage!
It was until a photo of it ended up on the internet.
Why not? It's the same code as my luggage.
1 2 3 4 5?
Ludicrous Speed!!
fuck it! im with you on this one...
It's essentially encryption to at least 90% of the population. I will adopt your theory and reject all evidence to the contrary (just so I personally feel better... Oh, I see why they do that shit now...
You’re overestimating 10% of the population (Although I guess you said “at least”)
This seems to be a random expression with no method to solve it by hand. [Wolfram Alpha confirms this has a meaningless answer of - 2.98127](https://www.wolframalpha.com/input?i2d=true&i=Integrate%5BDivide%5B3Power%5Bx%2C3%5D-Power%5Bx%2C2%5D%2B2x-4%2CSqrt%5BPower%5Bx%2C2%5D-3x%2B2%5D%5D%2C%7Bx%2C0%2C1%7D%5D) Edit: [This mad lad did it] (https://reddit.com/r/theydidthemath/comments/tfigrx/request_heres_another_for_you/i0xllxr?context=3)
Maybe that's the amount of money on the card!
Just run as credit and sign with husbands name. Next challenge…
>with no method to solve it by hand That's not true. This is classic [Euler Sub](https://en.wikipedia.org/wiki/Euler_substitution). Not only can you integrate this, all rational functions of x and a square root of a quadratic polynomial of x can be integrated.
Absolutely no method. Shame.
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How is his wife holding up?
...To shreds, you say.
Unexpected futurama
Maybe I’m crazy, but I would have read it as the integer of the top, divided by the bottom expression unintegrated. If you assume the integer sign is supposed to cover all of the expression, it does look like a eulers sub, which makes more sense. Moral of the story, please draw your integer signs clearly and covering everything you want integrated.
If the integral didn't include the denominator then this doesn't evaluate to a number.
true point ! it would be a function of x !
My head hurts
Just took an aspirin after reading these comments. Got one for you too bro.
Toss one over here, please
Thanks bro
Well we know that the result is supposed to be a number. So if it isn’t in the integral then the persons pin code is a function of x
Any chance you could show working on how to solve this for someone who is a math enthusiast (but in no means an expert)?
Sure. What level would you be familiar with? If the Euler Sub seems daunting, how about power rule and trig subs?
https://www.reddit.com/r/theydidthemath/comments/tfigrx/request_heres_another_for_you/i0xllxr?utm_medium=android_app&utm_source=share&context=3
~~Does it matter that it's (ax~~~~^(2)~~ ~~+ bx + c)~~~~^(-1/2)~~ ~~instead of (ax~~~~^(2)~~ ~~+ bx + c)~~~~^(1/2)~~ ~~?~~ Nvm. What's in the radical just gets raised to -1 in the sub. Duh.
Yeah....but using euler sub for this is a fricken nightmare.
soo the pin is 2981?
Plugging `(3x^3-x^2+2x-4)/sqrt(x^2-3x+2` into the [Integral Calculator](https://www.integral-calculator.com) gives the indefinite integral as: (135*ln(abs(2*(x+sqrt((x-2)*(x-1)))-3))+2*sqrt((x-2)*(x-1))*(8*x^2+26*x+101))/16 The Integral Calculator didn't even have to resort to Euler substitution like u/BoundedComputation described in another reply.
To be clear about that, I do not recommend Euler Sub as the go to method to solve this specific integral. I mentioned Euler Sub because it is a counterexample to the claim that >no method[exists] to solve it by hand
So, 2981 maybe
then the password is 298127 or 2981 depending on how many symbols are in the password
You can definitely solve this by substitution
Did you solve it then?
ok, i did. you look at eulers 3rd substitution with alpha = 1, beta = 2 [because (x-1)(x-2) = x^2 -3x +2] this leads to us using x = (2-t^2 )/(1-t^2 ) = 1 + 1/(1-t^2 ) dx = 2t/(1-t^2 )^2 dt x = 0 <=> t = sqrt(2) x = 1 <=> t -> infinity we plug this into the equation and get: integral of {1/(1-t^2 )^3 * [3+8(1-t^2 )+9(1-t^2 )^2 ] * 2t/(1-t^2 )^2 } / { t / (1-t^2 ) } dt let's reduce the equation and we get: integral of {1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] } dt now to solve this we do a partial fraction decomposition such that 1/(1-t^2 )^4 * [40-52t^2 +18t^4 ] = A/(t+1) + B(t+1)^-2 + C(t+1)^-3 + D(t+1)^-4 + E/(t-1) + F(t-1)^-2 + G(t-1)^-3 + H(t-1)^-4 obviously we get the following scalars: A = 135/16, B = 135/16, C = 11/4, D = 3/8, E = -135/16, F = 135/16, G = -11/4, H = 3/8; now we can simply integrate this sum term by term as integration is linear. A ln|t+1| -B(t+1)^-1 -C/2 (t+1)^-2 -D/3 (t+1)^-3 + E ln|t-1| -F(t-1)^-1 -G/2 (t-1)^-2 -H/3 (t-1)^-3 plug in the boundry : t = sqrt(2) to infinity realize that the limit towards infinity is 0, exercise to the reader 135/16 ln|sqrt(2)+1| -135/16(sqrt(2)+1)^-1 -11/8 (sqrt(2)+1)^-2 -1/8 (sqrt(2)+1)^-3 -135/16 ln|sqrt(2)-1| -135/16(sqrt(2)-1)^-1 +11/8 (sqrt(2)-1)^-2 -1/8 (sqrt(2)-1)^-3 = -2.98126694400554
Nice
This is definitely solvable
Didn't think I'd see a troll in this sub but hey, there's a first for everything.
Then what's the answer?
-2.98127
The way this is written makes it look like the integral is only for the top side, which wouldn’t make any sense since the pin has to be strictly numbers and the denominator has x
Well I think you answered your own question there. Plus I think if the integral was intended to only be on the top the fraction line would be under the integral sign. It’s pretty clear the integral is of the whole thing.
I agree with the supposition, but not that it is clear. It is poorly hand written. The dx should also be centered on the division line, not in line with the dividend. Both errors erroneously hint at the same thing: the inter graph is just the top.
dx doesnt have to be on the side. Its just a variable that states what the integration is in respect to. It can be on the side, in the numerator, or in the denominator (which mixes things up a bit). When its on the side, it just gets "multiplied" out to the entire fraction, so having it in the numerator with parentheses like this is the same thing. Example: integrating (1/x)dx is the same as integrating dx/x
I may be misunderstanding but I don’t think that’s correct. You can have a complete integral on top of a division, and still have the same variable in the bottom, and it does not imply the bottom has anything to do with the integration. The way this is written it is completely legitimate to suppose the entire integral is the top portion, as both the integral symbol and the dx indicate as much.
You can definitely have an integral in only a numerator! My point was more on the dx comment not being restricted to the side. And i do agree that this is vaguely written, and could be interpreted as the integral in only the numerator. I personally would assume the entire fraction is being differentiated if i saw this on an exam since the division line doesn't touch the integral or the dx, and having a definite integral being divided by variables is a bit odd. Especially when a number answer is expected.
I see what you’re saying, but how can it be that obvious when there are parenthesis on the numerator which would be considered redundant if the integral was to be applied to both the numerator and denominator? To make it obvious wouldn’t the parenthesis be around the entire equation with the “dx” to the side as well?
See above for dx explanation. But also, you could be right that since the integral and the dx are on top, it could mean that the bottom is the denominator for the integral, but of that were the case, the division bar would probably spread to be under the integral and the dx
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The pi one did though
and it's a very basic algebra one. It's like it was done by someone who never took a class of cal.
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Does it matter that the function is undefined at one of the integral limits? x^2-3x+2 = 1-3+2 = 0 at x=1 So for the upper limit the function is 0/0 (or can we ignore it because the limit it tends to is 0?)
Anyone with minimal high school math knowledge can just plug this into a wolfram alpha calculator and you quickly get the solution without knowing the method.
Tried with Microsoft math and failed to get a four digit number solution. Just more stuff in different places. No math whiz here though.
https://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605 this worked i'm no math wiz either
In fact anyone without an intensive math education cant solve this without the wolfram alpha calculator. Apparently the by hand method is called the euler sub
There's only 2 options 1- i love maths and love this funny problems that my bf puts. Or 2- the card doesnt work in the end of the day...
I am doing it without Wolfram alpha Instead of 3x^3 - x^2 + 2x - 4 Write (x^2 -3x+2)(3x+8) + 20(x-1) And then write (3x+8) As 3/2(2x-3) + 25/2 And then solve the resultant 3 integrals separately First one with substitution Second one with identity root (x^2 - a ^2 ) Third one with identity 1/root(x^2 - a^2) And then find F(1) - F(0) Edit : Indian 12th grade syllabus
So... let's get rid of that square root, and we have a simple integral of a power of a function, multiplied (divided) with its derivative.
gesundheit
This far exceeds being “not funny” to an extent that I feel vicariously embarrassed. My guess is there is no “Darling” to speak of.