How do I solve this?
How do I solve this?
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What are you trying to solve, little dude?
Solve for the area under the graph?
...
>How do I solve this?
I'm going out on a limb here, but id say probably Math.
shove it up your ass
Simple
if u put 3x-x3=y then differentiate there might be a chance
You need to provide more information. The area under the graph (without constraints) is tending towards infinite.
Take the common factor between 3x and x3 which is x
or u can try putting x=y
I'll have an exam on it soon, but I'm also too lazy to solve this shit. Try to consider the root as (3x - x^3)^(1/3) and do some substitution. If smartasses here won't be able to do this, I could bring my integral textbook so the way must be there.
This is easy.
Just substitute infinitroniom for x and subtract the entire equation up your ass.
integral-calculator.com just type the integral, It will show you the steps
the cube root of a number is the same as the exponent of 1/3
>reconstruct the function
>create the primitive function
>solve it without any integrationvalues idk how to fucking do that though
t. brainlet
download an app called photomath. it really works.
You moron, i literally teach math at university
Not op but someone show da wey
you don't need bounds
it's indefinite
either it's a shit university, or you're just a common larper
someone that teaches math would know an indefinite integral when they saw one
You can start by taking out that dot before the dx you fucking heathen
That is my point. It is indefinite... So what are you looking for here? The area under the graph is infinite.. What is the problem to solve?
I teach non-linear programming at Aarhus University in Denmark. Look it up.
lol
even wolfram won't show you how to solve this shit
There is no = sign... so there is nothing to solve.
this
simplify
substitute
Ostrogradsky's method
the usual stuff
Ostrogradsky's method
Julio Profe where are you?
Answer is a hypergeometric function. I don't think you want to do this problem
Actually it isn't helping, they don't say how to solve such things.
U-substitution... but first, you must do something with the cubed root. My guess is that you can set (3x-x^3)^1/3 equal to f(x) and take the natural log of both sides, leaving you with ln(3x-x^3)/3, from here you can factor out a 1/3, and you're left with ln(3x-x^3), this is equal to ln3x/lnx^3, from here, you can set U equal to x^3, which will cause dx to equal du/3x^2, so you have... i don't know, this x^3 is killing me here... there has to be a way to simplify the cubed root that I'm just missing so that U substitution is possible.
are you fucking retarded?
you cant
if you can find a way to take the integral of ln(3x-x^3), all you have to do at the end is multiply your answer by 1/3, add a constant to it (+ C) to make up for the lack of intervals, and then get rid of the natural log of the f(x) by taking the e of both sides...
Well if i am this retarded, then why don't you come up with an answer?
use a binomial expansion
>What is the problem to solve?
really?
trig sub??
https ://www.integral-calculator.com/
here you go faggots
the solution is a function of x, you fucking brainlet
>and you're left with ln(3x-x^3), this is equal to ln3x/lnx^3
Sorry but I doubt it, ln a + ln b = ln ab
But thanks anyway, I'll try that ln thing
lul
= dx
if that helps?
1. study maths
2. solve
3. ...
4. profit!
It's integral calculus, so you're better off just leaving it up to the professionals.
te he he 69
>1. study maths
That's actually what I'm trying to do here
Google the chain rule
0
you aint doin it right
>Try to consider the root as (3x - x^3)^(1/3) and do some substitution
That's the right step actually
He is not deriving though. And the U substitution doesn't seem to work. Integration by parts seems to not work as well because it is not undoing any product rule shit. I only took Calculus 2 so I can't find a way to do this shit. Not even with integration tables.
I've tried something the user above advised (now his post is deleted though). I've got the integral of ln (3x-x^3)^1/3 ( it seems to be correct). Can it help in calculating the original integral?
gotta factor out an x under the sqrt and then use trig substitution
Imgur com Sup Forums2BAIm
you kys
good luck tbh
Explanation and answer
I'll try it but I don't think it'll work. We got 1/3 as a degree thus the thing under the root sign can have any value, which isn't good for trigonometric substitutions. Fix me if I'm wrong.
can get to there but not farther
dont forget +k
Integration by parts always if you’re too lazy to memorize your integrals.
That's kinda... wrong