calculus question | Mathematics | Page 1

visual dagger Feb 10, 2024, 5:26 AM

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I have a general idea what to do but I’m unable to integrate the function(someone said you do it by parts but i still can’t do it)

spare aspenBOT Feb 10, 2024, 5:26 AM

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visual dagger I have a general idea what to do but I’m unable to integrate the function(someon...

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visual dagger Feb 10, 2024, 5:26 AM

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Any help is appreciated, just ping me

visual dagger Feb 10, 2024, 9:16 AM

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Please

wet parrot Feb 10, 2024, 9:17 AM

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hi

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let me see

visual dagger Feb 10, 2024, 9:17 AM

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Hiii

mellow goblet Feb 10, 2024, 9:36 AM

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what @wet parrot

wet parrot Feb 10, 2024, 9:36 AM

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solve it

mellow goblet Feb 10, 2024, 9:37 AM

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what's f^2

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is it f(x)^2 or f(f(x))

wet parrot Feb 10, 2024, 9:37 AM

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f(x)^2

mellow goblet Feb 10, 2024, 9:38 AM

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ok

visual dagger Feb 10, 2024, 9:42 AM

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mellow goblet is it f(x)^2 or f(f(x))

(f(x))^2

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So basically for the limits I know that f(1)=1 and f(0)=3

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But either I need to find f(x) or directly integrate it somehow

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By parts wasn’t really working

mellow goblet Feb 10, 2024, 9:48 AM

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same here

wet parrot Feb 10, 2024, 9:56 AM

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bro

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since $f$ is a root $f^3(x)+26f(x)x=27$ holds true for all $x$ inputs. this implies $f(0)=\sqrt[3]{27}=3$

steady escarp Feb 10, 2024, 9:58 AM

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yes so f(0) = 3

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gayre already said that

wet parrot Feb 10, 2024, 9:59 AM

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how

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fifth root of 27 isnt 3???

steady escarp Feb 10, 2024, 9:59 AM

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you put a 5 where a 3 should be

wet parrot Feb 10, 2024, 9:59 AM

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OHH THE EQUATION IS A CUBIC

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BRUH

eternal patioBOT Feb 10, 2024, 9:59 AM

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Romans 12:19

wet parrot Feb 10, 2024, 10:00 AM

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from the original equation we can find as many values for f(x) as we like

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but

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ok

mellow goblet Feb 10, 2024, 10:02 AM

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eternal patio **Romans 12:19**

get f^2 from this

steady escarp Feb 10, 2024, 10:03 AM

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how about first subbing u=f(x)

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then f'(x) also comes from the cubic by implicit differentiation

wet parrot Feb 10, 2024, 10:10 AM

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it sucks

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trust

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$f'(x)=-\qty(\frac{f(x)}{x}+\frac{26}{3f(x)})$

steady escarp Feb 10, 2024, 10:13 AM

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shouldn't that 27 be gone

wet parrot Feb 10, 2024, 10:14 AM

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QUACK

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mb i dont differentiate constants

eternal patioBOT Feb 10, 2024, 10:15 AM

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Romans 12:19

visual dagger Feb 10, 2024, 10:19 AM

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Someone who solved it and told the right answer said that we integrate by parts

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But they have little ego issue and when I asked I’m not able to do it by parts they told me to get good

wet parrot Feb 10, 2024, 10:22 AM

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LMAO

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ok wait let's do it by parts 😁

steady escarp Feb 10, 2024, 10:23 AM

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then i get this

wet parrot Feb 10, 2024, 10:23 AM

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eternal patio **Romans 12:19**

oh this helps.

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just use by parts on it

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yeah

steady escarp Feb 10, 2024, 10:23 AM

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$-3^2\left(\int{f}\right)\left(0\right)\ln\left(3\right)-\int_{0}^{1}\frac{f\left(x\right)^{2}}{3}\left(\int {f}\right)f'\left(x\right)dx$

wet parrot Feb 10, 2024, 10:23 AM

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cuz we already know f(0) and f(1) it makes it a LOT easier

mellow goblet Feb 10, 2024, 10:23 AM

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https://tenor.com/view/kitten-small-swimming-floor-crawl-gif-4536438629386526549

Tenor

visual dagger Feb 10, 2024, 10:23 AM

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Parts is getting hard for me

wet parrot Feb 10, 2024, 10:24 AM

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let me try

visual dagger Feb 10, 2024, 10:24 AM

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Not able to integrate

mellow goblet Feb 10, 2024, 10:24 AM

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I'm here for emotional support

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you can do it gayre

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you can do it gayre

eternal patioBOT Feb 10, 2024, 10:24 AM

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Djake3tooth

mellow goblet Feb 10, 2024, 10:24 AM

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what's $\qty(\int f)$

eternal patioBOT Feb 10, 2024, 10:24 AM

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Coffey

mellow goblet Feb 10, 2024, 10:24 AM

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is it the primitive

steady escarp Feb 10, 2024, 10:24 AM

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yes

mellow goblet Feb 10, 2024, 10:25 AM

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indefinite integrals are propaganda

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don't do it

wet parrot Feb 10, 2024, 10:25 AM

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eternal patio **Djake3tooth**

0?

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0?!?

steady escarp Feb 10, 2024, 10:25 AM

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is gone because f(1)=1

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and ln(1)=0

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or wdym

visual dagger Feb 10, 2024, 10:26 AM

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It's 4
Also there seems to be a problem in the question
The numbers seem inverted
243 - 128log3 is better
f(1) is 1
f(0) is 3
just integrate normally
by parts
Apply limits
Should leave you with a 1/12
But the number still looks wrong because I got 243-128log3 instead so
Could've made a silly error
Or they printed wrong

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The person told me all this

steady escarp Feb 10, 2024, 10:26 AM

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and what is u and v for the by parts integration?

wet parrot Feb 10, 2024, 10:26 AM

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steady escarp and what is u and v for the by parts integration?

yeah exactly

visual dagger Feb 10, 2024, 10:26 AM

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I tried various combinations but didn’t get it

wet parrot Feb 10, 2024, 10:26 AM

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both integrals suck

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mb mb

visual dagger Feb 10, 2024, 10:28 AM

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Yes

steady escarp Feb 10, 2024, 10:28 AM

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maybe first applying the cubic

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$\int_{0}^{1}\frac{\left(27-26xf\left(x\right)\right)\ln\left(f\left(x\right)\right)}{f\left(x\right)}dx$

eternal patioBOT Feb 10, 2024, 10:29 AM

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Djake3tooth

wet parrot Feb 10, 2024, 10:29 AM

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eternal patio **Djake3tooth**

27-26xf(x)

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lmao that's just f^3(x)

steady escarp Feb 10, 2024, 10:30 AM

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exactly

visual dagger Feb 10, 2024, 10:30 AM

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Yes they just divided by f(x)

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Making it ^2

wet parrot Feb 10, 2024, 10:30 AM

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oh you made it into that

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how'd that help tho

steady escarp Feb 10, 2024, 10:30 AM

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yes

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idk

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i'm trying

visual dagger Feb 10, 2024, 10:30 AM

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I’ll try that

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Too

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It’s messed up

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,w integrate ln(f(x))/(f(x))

eternal patioBOT Feb 10, 2024, 10:35 AM

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visual dagger Feb 10, 2024, 10:37 AM

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steady escarp Feb 10, 2024, 10:37 AM

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$\frac{1}{y}=\frac{y^{2}}{27}+\frac{26}{27}x$

eternal patioBOT Feb 10, 2024, 10:37 AM

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Djake3tooth

visual dagger Feb 10, 2024, 10:59 AM

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😭😭😭

visual dagger Feb 10, 2024, 4:46 PM

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Is anyone willing to

steady escarp Feb 10, 2024, 4:57 PM

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Can you ask that kid with ego problem again?

visual dagger Feb 10, 2024, 4:57 PM

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He’s not online and he won’t reply to me even if I ask

steady escarp Feb 10, 2024, 5:39 PM

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i found something

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so first important thing

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we need to do by parts with $u = f(x)^2 \ln(f(x))$ and $v = x$

eternal patioBOT Feb 10, 2024, 5:40 PM

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Djake3tooth

steady escarp Feb 10, 2024, 5:40 PM

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and we have $f^{-1}(x) = \frac{27 - x^3}{26x}$

eternal patioBOT Feb 10, 2024, 5:40 PM

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Djake3tooth

steady escarp Feb 10, 2024, 5:55 PM

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now i have calculated the integral exactly

visual dagger Feb 10, 2024, 6:57 PM

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steady escarp we need to do by parts with $u = f(x)^2 \ln(f(x))$ and $v = x$

but x isn’t even there in the integral

steady escarp Feb 10, 2024, 7:46 PM

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it is

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dx

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i'm talking about this version:

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$\int{u\mathrm{d}v} = uv - \int{v\mathrm{d}u}$

eternal patioBOT Feb 10, 2024, 7:47 PM

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Djake3tooth

mellow goblet Feb 10, 2024, 9:12 PM

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"int{fg', a, b} + int{f'g, a b} = int{(fg)', a, b} = (fg)(b) - (fg)(a)"

visual dagger Feb 11, 2024, 5:22 AM

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Oh ok ok

visual dagger Feb 11, 2024, 5:23 AM

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steady escarp $\int{u\mathrm{d}v} = uv - \int{v\mathrm{d}u}$

So if I use this, will I be able to integrate?

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I’ll try

wet parrot Feb 11, 2024, 5:28 AM

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$\int_0^1 1\cdot f^2(x)\ln f(x)\mathrm{d}x\=xf^2(x)\ln f(x)\bigg|_0^1-\int_0^1x\dv{x}\qty(f^2(x)\ln f(x))\mathrm{d}x\=f^2(1)\ln f(1)-\int_0^1xf(x)f'(x)\qty(1+2\ln f(x))$

visual dagger Feb 11, 2024, 5:29 AM

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Yeah I’m taking the whole function as u

visual dagger Feb 11, 2024, 5:33 AM

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wet parrot $\int_0^1 1\cdot f^2(x)\ln f(x)\mathrm{d}x\\=xf^2(x)\ln f(x)\bigg|_0^1-\int_0^1x...

Isn’t the x bothering here?

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Yes I got till there

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Yes

eternal patioBOT Feb 11, 2024, 5:34 AM

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Romans 12:19

wet parrot Feb 11, 2024, 5:35 AM

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how is this integral any easier

visual dagger Feb 11, 2024, 5:35 AM

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Yep

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I can’t do it further

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I mean I might be able to do it by parts again

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But idk

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Split and do it by parts in both of them

wet parrot Feb 11, 2024, 5:36 AM

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visual dagger I mean I might be able to do it by parts again

maybe taking (1+2ln(f(x))) as u, and the rest as v

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but no

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that doesnt help.

visual dagger Feb 11, 2024, 5:36 AM

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But how will I integrate xf(x)f’(x)

wet parrot Feb 11, 2024, 5:37 AM

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exacrlty

visual dagger Feb 11, 2024, 5:39 AM

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,w integrate xf(x)f’(x)(1+2ln(f(x))

eternal patioBOT Feb 11, 2024, 5:39 AM

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visual dagger Feb 11, 2024, 5:39 AM

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I think of giving up on this question

wet parrot Feb 11, 2024, 5:40 AM

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i'm pretty sure @steady escarp solved it

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wait for till he replies

visual dagger Feb 11, 2024, 5:40 AM

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Sure

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I’m not sure where he used f inverse

visual dagger Feb 11, 2024, 5:40 AM

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eternal patio **Djake3tooth**

He said it here

wet parrot Feb 11, 2024, 5:41 AM

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oh wait

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@visual dagger

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something escaped us

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so far.

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i'm pretty sure we can find the integral

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without by parts (immediately)

visual dagger Feb 11, 2024, 5:46 AM

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Wow howww

wet parrot Feb 11, 2024, 5:48 AM

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Since $f(x)$ is a real root of $y^3+26xy=27$, we say that $f^3(x)+26xy=27$ and find $f'(x)$ by implicit differentation.

eternal patioBOT Feb 11, 2024, 5:48 AM

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Romans 12:19

wet parrot Feb 11, 2024, 5:49 AM

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can you find it and tell me what u get

visual dagger Feb 11, 2024, 5:49 AM

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one sec I’ll do it

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3f^2(x)f’(x)dx+26(ydx+xdy)=0

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Sorry idk latex and could you just check if it’s correct

wet parrot Feb 11, 2024, 5:52 AM

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ayo what

visual dagger Feb 11, 2024, 5:52 AM

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$3f^2(x)f’(x)dx+26(ydx+xdy)=0$

eternal patioBOT Feb 11, 2024, 5:52 AM

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gayre

wet parrot Feb 11, 2024, 5:52 AM

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can u isolate f'(x)

visual dagger Feb 11, 2024, 5:52 AM

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oh ok

wet parrot Feb 11, 2024, 5:53 AM

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visual dagger Sorry idk latex and could you just check if it’s correct

no prohlem

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im pretty sure it should be $f'(x)=-\frac{26f(x)}{26x+3f^2(x)}$

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what are these dx and dys in fetween tho

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wait

eternal patioBOT Feb 11, 2024, 5:56 AM

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Romans 12:19

wet parrot Feb 11, 2024, 5:57 AM

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yes it's this

visual dagger Feb 11, 2024, 5:58 AM

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mm ok sorry

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I’m not sure what we’re doing with f’(x) but I’ll wait for you

wet parrot Feb 11, 2024, 6:01 AM

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$\int_0^1f^2(x)\ln f(x)\mathrm{d}x\\overset{u=f(x)}{=}\int_{f(0)}^{f(1)}u^2\ln u\frac{\mathrm{d}u}{f'(x)}\=\int_{f(0)}^{f(1)}\frac{u^2\ln u(3u^2+26x)}{-26u}\mathrm{d}u\=-\frac{1}{26}\int_{f(0)}^{f(1)}u\ln u(3u^2+26f^{-1}(u))\mathrm{d}u\=\frac{1}{26}\int_1^3u\ln u(3u^2+26f^{-1}(u))\mathrm{d}u$

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this is where f^-1 comes into play

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cuz if u=f(x) then x=f^-1(u)

visual dagger Feb 11, 2024, 6:03 AM

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How did you go from 2nd step to 3rd

wet parrot Feb 11, 2024, 6:03 AM

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mb for not adding the du sorry

wet parrot Feb 11, 2024, 6:04 AM

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visual dagger How did you go from 2nd step to 3rd

we found f'(x) by implicit differentiation

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just plugged it in

visual dagger Feb 11, 2024, 6:04 AM

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I got it

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Yes I understood till here

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So here we integrate?

wet parrot Feb 11, 2024, 6:08 AM

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visual dagger So here we integrate?

yes we do.

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we know f(1)=1 and f(0)=3

eternal patioBOT Feb 11, 2024, 6:09 AM

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Romans 12:19

wet parrot Feb 11, 2024, 6:13 AM

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the integral is really long

wet parrot Feb 11, 2024, 6:15 AM

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mellow goblet "int{fg', a, b} + int{f'g, a b} = int{(fg)', a, b} = (fg)(b) - (fg)(a)"

yo WHAT

visual dagger Feb 11, 2024, 6:20 AM

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We didn’t learn how to integrate inverse

hollow onyx Feb 11, 2024, 6:20 AM

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wet parrot the integral is really long

Nah it's not , ur actually doing correct

visual dagger Feb 11, 2024, 6:21 AM

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Oh

hollow onyx Feb 11, 2024, 6:21 AM

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visual dagger We didn’t learn how to integrate inverse

F-1(u) is simply (27-u³)/26u

visual dagger Feb 11, 2024, 6:22 AM

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Oh shit yesss

hollow onyx Feb 11, 2024, 6:25 AM

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Yea got it ur R is 52

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must be multiple correct option then

visual dagger Feb 11, 2024, 6:25 AM

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,w integrate u^3 ln(u)

eternal patioBOT Feb 11, 2024, 6:25 AM

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hollow onyx Feb 11, 2024, 6:25 AM

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yup

visual dagger Feb 11, 2024, 6:26 AM

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Just have to apply limits one sec

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Oh yes

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Thank you all!!

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Thanks for all of your time and help

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I really appreciate it

#calculus question