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/sci/ - Science & Math

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File: 3 KB, 325x156, integrate.jpg [View same] [iqdb] [saucenao] [google] [report]
8682139 No.8682139 [DELETED]  [Reply] [Original] [archived.moe]

Can /sci/ integrate this?

>> No.8682146

>Can /sci/ type this into WolframAlpha?


>> No.8682148

No. I can prove that it is integrable and approximate it if you want.

That said, finding the antiderivative is not trivial as you need to factorize the denominator into something no one would even imagine and could only be found via brute force.

And the result is a clusterfuck of fractions involving powers, roots, logarithms and even inverse trig functions. As my president would say: SAD!

>> No.8682153


>these plebs using Wolfram Alpha

so it's a no then?

>> No.8682156

I literally said 'No.'

Can you read? I guess that's a no aswell.

>> No.8682159

>Wasting your time manually solving something that a computer can do in a second.

You're probably also impressed by people who can recite Pi to a stupid number of digits.

>> No.8682163


If you're not smart enough to integrate it give me your best approximation then.

>> No.8682170

Give me an interval and I can approximate it through partial riemann sums.

>> No.8682172


You pick whatever you want.

>> No.8682177


Why wouldn't you be impressed by someone having an amazing memory?

>> No.8682178

From 0 to 1

>> No.8682185

Only intelligent people impress me.

>> No.8682190


Having an amazing memory makes understanding difficult concepts easier.

>> No.8682209

From 0 to 1, using 5 partial sums, evaluating at the midpoint of each partition and using only 4 decimal places of each sum I get 0.07616

It is pretty off but by hand I won't bother to get something better.

>> No.8682224

It is correct to 3 decimal places and from what I remember in high school physicists only use 2 decimal places so I guess you can call me the master integrator from now on.


>> No.8682233
File: 24 KB, 608x494, plottttt.jpg [View same] [iqdb] [saucenao] [google] [report]

You should be able to figure out the integral from this picture, unless of course you're not mathematically gifted.

>> No.8682236

>from what I remember in high school physicists only use 2 decimal places


>> No.8682244


You're welcome

>> No.8682247


>You're welcome

What did he mean by this?

>> No.8682249

This is really just partial fractions. The bottom is quintic but has one obvious real root. You then factor the quartic and it will be ugly most likely but you seriously can just partial fractions this bitch

>> No.8682256

Back in high school for physics class they would tell us to just round to 2 decimal places most of the time.

>> No.8682260



ugly doesn't even begin to describe this

>> No.8682268

I see, just thought you meant that actual physicists only round to two decimal values

>> No.8682286

How can you guys claim that you're so smart yet you cant integrate that?

>> No.8682297

my bad

>> No.8682299

an amazing long term memory used to be valuable, but supply and demand bruh

now we have cheap memories that easily outclass humans

what matters for intelligence is that size of your working memory, which is completely different from the memory needed for reciting digits of pi

>> No.8682302

>my bad
You aren't wrong.

You are supposed to factorize the denominator and then partial fractions.

But lets just say, some fractions are more partial than others.

>> No.8682326

If you can't integrate it,then just leave /sci/

>> No.8682358
File: 583 KB, 800x900, Marisa is confused.png [View same] [iqdb] [saucenao] [google] [report]

I'm shit at this and not much of a /sci/ person in all honestly. I would Imagine you could do one of the following:

Rewrite x^2/x^5+4 as log|x^5+4|, then, when integrating turn it into an exponential e^x^5+4.

Alternatively multiply by bottom brackets and integrate the solution. I need help on this sorta stuff desu though so help would be appreciated.

>> No.8682364


this dumb weeaboo had been shitting up the board in search for attention for days now.
just letting you guys know so that you don't reply to him and he might just leave

>> No.8682374

This is my first time on the board in months, I've been doing further maths for while and this shit has been causing some grievance, sorry for the misunderstanding.

>> No.8682375

I was about to tell him how wrong he was but I see, he is just pretending to be retarded.

Or maybe he is retarded but he also knows that he is retarded so he is roleplaying as a retarded person who doesn't know he is retarded.

Really makes you think

>> No.8682378

No, I'm probably retarded, what am I doing wrong here? I've never been that great at maths, I just want a quick solution.

>> No.8682379

You can't
>Rewrite x^2/x^5+4 as log|x^5+4|

That doesn't even make sense.

>> No.8682389
File: 185 KB, 714x337, Capture.png [View same] [iqdb] [saucenao] [google] [report]

Would it have to be in the context shown here then? (where you are integrating the exponent of an exponential).

>> No.8682390

Just residue theorem it nigga

>> No.8682395

What the fuck? That has nothing to do with this.

>> No.8682400

I figured that if you can integrate factors in that manner then you would be able to do the same in this case. Sure its something completely different, though I'd assume a link (though obviously false as proven).

>> No.8682404

Adding on to this, I'd assume the correct way is using partial fractions then integrating them individually?

>> No.8682414

In your example of differential equation that exponent of an integral is the general solution of that kind of DE.

You can't really put the log inside the integral and then put an e outside. It is not the same.

You could put an e and a log inside but then that would be completely useless.

Yes, this is what you have to do and there is no other way.

>> No.8682473

Great, thanks for the help! Sorry if I came off as retarded, I'm not very good at this.

>> No.8682600

Too lazy to compute the residues, but residue integration seems to work if you choose the contour [math]C_{R,\epsilon}[/math] that comprises a small semicircle around [math]-\sqrt[5](4)[/math] of radius e, then the segment [math][-\sqrt[5](4)+\epsilon, R][/math], then a semicircle from -R to R, then the segment [math][-R, -\sqrt[5](4)-\epsilon][/math].
As you let R go to infinity, the integral over the big semicircle goes to 0. Then you let epsilon go to 0 and then you have to compute the limit of the integral over the small semicircle (probably a dominated convergence argument). That tells you that the integral you want plus the limit of the small integral is 2pi i times the sum of the residues, which I won't compute but that wolfram can compute.
Still, it works very well.

>> No.8682627

Easy, just Taylor expand it and integrate term by term.

>> No.8683665

Take your pedophile cartoons back to >>>/a/

>> No.8683954

Yes. Any fractions of polynomials are easily solvable.

>> No.8683979

why can't i use integration by partial fractions here?

im a brainlet and i need an explanation

>> No.8684061

>all these retards who cant't do it
You literally just factor the denominator, decompose into partial fractions, separate the integral and integrate each part as u'/u or the inverse of a first or second degree polynomial which you should kill yourself if you're a math major and can't do it with your eyes closed.

>> No.8684066

holy shit im a complete brainlet and i got it >>8683979

holy shit

>> No.8684142

that's the solution btw


>> No.8684151

o well i lied

cant belive someone took the time to do that

>> No.8684154


>> No.8684158

let me believe what i want to believe

>> No.8684159

a computer took the time to do that
i don't use wolfram cause it's shit shows no steps
just type it in there just to get a scope of how much work is required to get the solution

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