help-28

[split swift]

the integral is the signed area

if its positive, it just means the function bounds more positive area than negative

it doesn't mean it doesn't bound any negative area at all

[torn jolt]

ahhhhhhhhhhhhhhhhhhh i see

[split swift]

with that in mind, a counterexample should be easy to find

[torn jolt]

i hope one day i will be better at math

what about x^3 + 1/2

on interval [-1,1]

that function have some negative value but integral positive

i really good at claculating problems but not so good at concept

[split swift]

you could just take f(x)=x over [-1,2]

but your example works as well

or otherwise, what Jessica was hinting at is that you can take any function with positive integral

then move a single point to somewhere negative

because changing the value of an integrand at a single point doesn't change the value of the integral

e.g., modify the constant function f(x) = 1 to g(x) = {-1 if x=0, 1 otherwise} for another simple example

[full forum]

@torn jolt Has your question been resolved?

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[torn jolt]

Help pls I’m trying to figure out how to solve this and I’m just getting nowhere

Have to find left/right limits according to table

[ivory cairn]

seems accurate so far

[torn jolt]

This is the information given to me lol

I have to figure out the limits which is a b c

[ivory cairn]

do you know what the left limit is?

[torn jolt]

No honestly I don’t really understand or know what that is. This is from a practice worksheet for my calc 1 class and I’ve been sick this whole week so I pretty much know nothing about this

I just know the answers are in infinity form

Is it 3.999 and 4.001

Left/right

[ivory cairn]

the left limit of f(x) is $\lim_{x\to a^-}f(x)$ means the value of f(x) as x approaches a from just below a
in the case of x = 4, it means you'd want to look at values just under 4

[glossy valve]

Zybikron

[torn jolt]

And the right limit is just above 4?

[ivory cairn]

yes

[torn jolt]

How would I use that to figure out a b and c then

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

@ivory cairn sorry to ping u I was gone for a while

[torn jolt]

<@&286206848099549185>

.close

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[charred geode]

integration?

[tiny stirrup]

yeah

[full forum]

@tiny stirrup Has your question been resolved?

[charred geode]

you can use integration to find the area under each curve and compare them

[full forum]

@tiny stirrup Has your question been resolved?

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[obtuse gate]

Could i get some help how to solve this math question?

[amber palm]

!showwork

[full forum]

Show your work, and if possible, explain where you are stuck.

[obtuse gate]

I am not sure how to solve it. I am wondering what is the correct way to go about it.

2x+10y=26,98 and than do I put either x or y to 0?

[atomic steeple]

You can solve for x or y and then plug and chug for the other

[obtuse gate]

How do I solve for x and y precisely?

Is it like I said by turning one of the variables to 0?

Or do i for example:

2x+10y=26,98
-2x -2x

10y=26,98-2x

       10

[full forum]

@obtuse gate Has your question been resolved?

[atomic steeple]

Yes and y = (26,98 - 2x)/10
Insert y into the other formula

Then solve for x

Once you have x, you can then solve for y

[obtuse gate]

.close

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[nova moth]

Hello I don’t know how to do this 3d vector question

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@nova moth Has your question been resolved?

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[golden shore]

Hello

[golden shore]

type "hello"

[delicate cobalt]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stiff musk]

your grammar was incorrect there

[amber palm]

[golden shore]

that's an incomplete sentence, I suppose

[stiff musk]

i dunno, i was just following your instructions

[golden shore]

I see

I would like to discuss

Static Equilibrium

[amber palm]

mechanical?

Fluids?

[stiff musk]

cool, i'm eager to learn what that might be

[golden shore]

"center of gravity"

is a sub-topic included in Static Equilibrium

okay so there's a statement written on my book

[stiff musk]

does this have anything to do with center of mass, and if not, what's the difference

[amber palm]

static equilibrium where?

theres one in fluids

Laws of motion

both of them are really the same thing but the application changes

one is where you balance pressure and the other is with mass and tension and normal forces

[golden shore]

How do I find "center of gravity" of an object?

[amber palm]

well

first of all

"find" is kind of the wrong word, because it dosent exist

we assume it does to make our physics easy

[twin wolf]

written in* not on

(sounds better)

[stiff musk]

unless maybe you literally wrote it on the cover

[hot herald]

they have different meanings, someone could've actually physically written something on the book

[amber palm]

and there are different ways to do it, there are centre of gravities of specific shapes like rectangle at height/2

[twin wolf]

thats possible

[rugged void]

why are we discussing whether or not the text was written or or in the book?

[twin wolf]

i think in this context its likely not that

[rugged void]

guys

!topic

[full forum]

Please read the channel description before posting, and stay on topic.

[amber palm]

let the man physics

[golden shore]

I had a long paragraph talkin about this

it is better to watch the diagram first before reading the following context:

my books wrote that if I hung an object with two different connecting points (which connects the hook to the object) then the intersection of the two vertical line which crosses the connecting points respectively.

such intersection

is the center of gravity

[stiff musk]

i'm gonna guess you haven't spent much time in a @golden shore help channel

[golden shore]

and the book provides a way to find it

[rugged void]

i have...

please don't remind me

[stiff musk]

hahaha fair

[amber palm]

like the integration formulas?

[golden shore]

how it works tho?

[rugged void]

there was this one time with complex numbers and how e^180i = e^(i times pi) bc degrees and radians

;-;

i failed my mission

[golden shore]

there's no mathematical proof for the statement

[amber palm]

i see

[golden shore]

and I wonder how to interpret it

[amber palm]

well firstly

[rugged void]

you have reactivated my PTSD

[amber palm]

.

so dont think of it as a like actual thing

its like a free body diagram

drawn for convinience (engineers are lazy)

[stiff musk]

wait, physicists don't draw free body diagrams? they did in my classes

[amber palm]

they do, but thats not how the body actually is

like you draw a dot for a human

assuming the mass is spread uniformly

[amber palm]

have you ever

spun a basketball on ur finger?

[stiff musk]

well within the context of classical mechanics surely centers of mass exist, presumably we're not dealing with quantum issues here

[rugged void]

do NOT go there

[golden shore]

I did tried it

[amber palm]

try it with a book

like rn

[rugged void]

LMAOOOOOO

fr best analogy

[amber palm]

it will help you understand

[twin wolf]

you actually have wrong grammar there in that form

[rugged void]

bro fr prepared a movie for us

[amber palm]

the point where the book actually spins, is the centre of gravity

[rugged void]

wouldn't it be incorrect grammar

[stiff musk]

bro spin attempt #1 failure

[twin wolf]

well youre not allowed to correct me because my name isnt please correct my grammar

[rugged void]

:O

[amber palm]

[rugged void]

i've seen worse, my friends are clowns

i was lying
i dont have friends

💀

ok sorry

[golden shore]

does it means I cannot spin anything with my finger if it is not set on the center of gravity.

[rugged void]

back to the topic

you can

[amber palm]

[golden shore]

can such phenomenon be explaind through the concept of torque

[rugged void]

but it'll produce a sideways acceleration

so it'll fly off

[golden shore]

are they related.

[amber palm]

you are making an assumption

you assumed you can spin anything

[golden shore]

ohhh

[rugged void]

i think torque occurs if its not on the centre of mass

[amber palm]

but yes, you cant just rotate it unless its being balanced

[amber palm]

AT the very point youre spinning it

[rugged void]

i mean center of gravity

[amber palm]

there is no torque

[rugged void]

yes

ofc

[golden shore]

why, im so curious

[amber palm]

remember how we defined torque a few hours back

[rugged void]

because torque is angular acceleration

[amber palm]

you never tried the door thing did you

[rugged void]

like sideways

[amber palm]

[rugged void]

what door thing? now im curious

[stiff musk]

where is austin's self portrait when we need it

[rugged void]

what with the 🚪

[twin wolf]

try pushing a door close to its hinges

[amber palm]

same height from the floor, you try to open the door by pushing it while slowly going towards the hinge

[rugged void]

hmmm

imma do that

i see

[amber palm]

you will see WHY torque has length in its formula

[rugged void]

my body tryna open the door

[twin wolf]

jokes on you i open all my doors like this

[rugged void]

gigachad

[amber palm]

..

man

thats a lot of word done

[rugged void]

https://tenor.com/view/gigachad-chad-gif-20773266

pov @twin wolf

[twin wolf]

real

[golden shore]

it becomes harder for me to open the door, when my hand getting closer to the hinge, with same amount of force I employed.

[rugged void]

@twin wolf btw where was that question from before rfom

[rugged void]

yea

that's bc torque

has less effect

[stiff musk]

r x F baby

[rugged void]

closer to the joint

[amber palm]

which is why the distance from the hinged point matters

[twin wolf]

just a textbook my school makes us uses

[rugged void]

no like what course/grade

??

[twin wolf]

12th

[stiff musk]

Yoyo

[rugged void]

okay

@glossy raft if you need help this channel is occupied

lmao

[golden shore]

but the hinge case does not explain the statement

[stiff musk]

do you know where to push a door for maximum torque

[rugged void]

@stiff musk IT'S HAPPENING AGAIN

IT'S A CLASSIC PCMG CASE

[stiff musk]

it always happens

[rugged void]

AAAAAAAA

PleaseCorrectMyGrammar PTSD™

[golden shore]

why the center of the weight happens to be the intersection?

[twin wolf]

close to its hinge!!!

[rugged void]

fr

[amber palm]

thus you have length in the formulas of torque, now since you need length to calculate torque, how would there be any torque at the very point the body is spinning

[rugged void]

https://tenor.com/view/talos-talosian-star-trem-star-trek-big-head-gif-12496470

...

@glossy raft

[amber palm]

symmetry

[stiff musk]

try calculating the center of mass of that noggin

r x F

[rugged void]

well that was quick

[stiff musk]

whoa impressive

[rugged void]

ikr

[amber palm]

The reason symmetry matters is very obvious, read my book example is gone

[stiff musk]

at least we didn't lose the book spinning video

[golden shore]

I see

[rugged void]

a true treasure to mankind

[amber palm]

you need to apply your force to balance the "mg"

[amber palm]

since it has mass in it

it needs to be at the centre of gravity

because we have defined centre of gravity to be the point where the entire weight of the body is

[rugged void]

actually no nax

[amber palm]

which is why symmetry is important

[golden shore]

because my tip is on the center of gravity of the book

[rugged void]

centre of gravity is where the relative distribution of mass is equal

[golden shore]

therefore it would not falls from my hand

[rugged void]

no

it is falling

ur fist is propping it up

in the first image

[amber palm]

it needs to be precisely at the centre

[amber palm]

i see

[rugged void]

and also it doesn't need to be symmetrical

ever see a crane?

[golden shore]

they did not make any contact

[rugged void]

it has the load-bearing-thing

on the back

[golden shore]

symmetrical in terms of mass

[rugged void]

oh alright, might just be elasticity

[amber palm]

no

[rugged void]

oh alright then

[amber palm]

symmerical in uniform shapes

[rugged void]

no

[amber palm]

rectangles

circles

[amber palm]

why not?

a rectangle has its centre of gravity at h/2

[rugged void]

well

not everything is a rectangle

;-;

[amber palm]

the spheres is at the midpoint

[amber palm]

.

[rugged void]

hang on

OH

[amber palm]

uniform shapes

[rugged void]

I THOUGHT U MEANT SYMMETRICAL IN UNIFORM SHAPES

[amber palm]

uh

[rugged void]

semantics issue

[amber palm]

i am confused but yes i suppose

[rugged void]

something @golden shore's username would be worried about

which, incidentally, has multiple grammatical errors itself

[golden shore]

there's also another statement

about the same topic

[amber palm]

?

[golden shore]

i dont know how to describe the blue area

[rugged void]

what is this supposed to signify

what blue area

[golden shore]

it is the area that the eraser contacts with the table

[rugged void]

PCMG douses the channel in a swathe of blue

[golden shore]

im not sure if there's a specific term for it

[rugged void]

ok we see the blue area

what about it

?

[golden shore]

if the vertical line which crosses the center didnt contact with the blue area

[rugged void]

yes

[golden shore]

then the object will falls

[rugged void]

ni

*no

[golden shore]

that's the statement written on my book

[rugged void]

well

that's because

the blue area is defined as the area of contact

im assuming

so if it's unsupported

[golden shore]

yes

[rugged void]

ofc it will fall

so that's why

if the eraser isn't touching anything but the COG is, it won't fall

[golden shore]

can it be explained by torques and Static Equilibrium

[rugged void]

no idea what those r

[amber palm]

torques

uh

[rugged void]

and no idts

[amber palm]

i dont know

[rugged void]

you can't just link everything to torque PCMG

[amber palm]

but static equilibrium

yes

[golden shore]

really

[rugged void]

yes

[amber palm]

its pretty obvious

[rugged void]

really

[golden shore]

but the eraser are going to spin when it falls

there's some torque involve in the scenarios

[amber palm]

thats because there isnt an equilirium

[rugged void]

guys

now I have a question

[golden shore]

why it is true tho?

[rugged void]

helpers, i have a question

WHY ISN'T THIS IN THE PHYSICS DISCORD

[amber palm]

[rugged void]

??

[amber palm]

you should ask this in the physics discord

[rugged void]

this has devolved from a simple PCMG question (oxymoron intended) to... whatever this is

@golden shore has your question been solved

[golden shore]

no, I wonder why it is true

[rugged void]

because a force is being applied smwh other than the COG, so angular acceleration occurs

i just answered

[stiff musk]

physics stuff is generally allowed here, but it helps a lot if there's at least some math content to the question haha

[rugged void]

@stiff musk im losing my mind here, classical is not my strong suit

im more of the particle or astronomy guy

[stiff musk]

i don't know what the question actually is here

[rugged void]

neither do i

then again, it's PCMG. usually never is

(jk)

[stiff musk]

ok i guess it's this?

[rugged void]

OH GOD

[golden shore]

this is also one of my questions

[rugged void]

I FORGOT ABOUT THAT

...

"one of my questions"?

[stiff musk]

what are the other ones

[rugged void]

@stiff musk
IT'S ALL COMING BACK TO ME
THE PCMG CHANNELS

[golden shore]

the other ones are solved

[stiff musk]

well the center of gravity iirc is just
$$\frac{\int x \rho(x),dx}{\int \rho(x),dx}$$

[glossy valve]

Bungo

[golden shore]

I suppose the hooking method I just delivered earlier will alway works despite the shape of the object

[stiff musk]

where x = position vector of a point and rho(x) is the mass density at that point

[rugged void]

@stiff musk u sure?

i have no idea

[stiff musk]

i think so yea

[rugged void]

hm lemme google

hang on

[stiff musk]

note, the integral in the denominator is just the total mass

[rugged void]

the center of gravity is a POINT

[stiff musk]

yep

[rugged void]

so how does that define a point

[stiff musk]

the denominator is a scalar, the numerator is a vector

(point)

[rugged void]

i havent done vector calc

[stiff musk]

x is the position vector

[rugged void]

oh i see

[stiff musk]

i'm too lazy to put a boldface or whatever

[rugged void]

alr tehn

nah is fine

@golden shore

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[golden shore]

I have one more tho

which also confused me a lot

[rugged void]

sighs
@stiff musk

[stiff musk]

do tell

[golden shore]

So all of these circles

[rugged void]

more physics

[stiff musk]

look at those equations, looks like what i wrote if dx is counting measure

[rugged void]

discord.gg/physics

[rugged void]

alpha particles?

[golden shore]

(I suppose) is some small particle of a object

[stiff musk]

those are just the discrete version of my equation, and they split it into x and y components

the W's correspond to my rho

(it's the mass density)

[golden shore]

Why are they equivalent?

[rugged void]

imma brb

[golden shore]

the two scenarios

[stiff musk]

my x is a vector, think of it as the vector (x, y) if you're in 2 dimensions, or (x, y, z) if you're in 3

[golden shore]

G is the center of mass

[rugged void]

we can't read chinese/japanese/other asian language

[golden shore]

What is p(x)?

[rugged void]

translate

its rho x

mass density at that point

[stiff musk]

my p(x) corresponds to your W_i

mine is just the "continuous" version

[golden shore]

it says the situation that these w1,w2,... forces applied on these particles in the case 1

[stiff musk]

in the discrete version, p(x) becomes W_i, the integrals become sums

[rugged void]

@stiff musk the feeling of it being a PCMG channel is back

[stiff musk]

did it ever leave?

[golden shore]

can be simpliied to the case 2

[rugged void]

once or twice briefly

[stiff musk]

oh those are probably mass times gravity accel then?

sure, you can multiply my whole equation by g if you want force

[golden shore]

where the sum of these forces is applied merely on the center of the mass G

[stiff musk]

center of mass is a position though, not a force

[golden shore]

How to interpret it

[rugged void]

sighs

[golden shore]

how it works

[golden shore]

why can the situation be simplified

[stiff musk]

what that is saying (i guess, i don't read the language) is that if you have a multiparticle system with gravity acting on all the particles, then for the purpose of translational motion, you can treat the system as if all the mass is concentrated at the center of mass (what you're calling center of gravity)

the center of mass will then obey the usual F = ma equations as if it were a single point mass

the individual particles may fly around in crazy ways but the CM will have your usual nice clean parabolic motion

[golden shore]

I see

[rugged void]

wait stop stop stop

PCMG i think one thing is important

in the case of random particulate matter

gravity is negligible

at the subatomic level, it's basically nothing

keep that in mind

[stiff musk]

see this example from kleppner and kolenkow

[rugged void]

yes?

@golden shore done?

no?

[golden shore]

I think so, even though i cannot interpret the context on the picture

[rugged void]

neither can i tbh

[stiff musk]

it's some kind of rope with three strings at the end and a ball attached to each string

the man throws it at an animal to capture it

the three balls can move in hard to predict ways due to the strings
but the center of mass of the system moves in the predictable parabola
the man can throw it as if it were a single ball and overall it behaves as expected

[rugged void]

devolves into ball demon

[stiff musk]

or see this picture

here there is a stick with a ball at each end

[twin wolf]

I’ve never heard of a bola

[golden shore]

i cannot visualize the center of mass in the system which consists of three balls and a bola.

[stiff musk]

hurling through the air

the rotation is hard to predict but the center follows the parabola

[rugged void]

ayyyy he's back

[stiff musk]

me either, except in that book haha

[rugged void]

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[golden shore]

I think i will leave it now.

yes it is done

[rugged void]

hurrah

[golden shore]

thank you guys for all the explanation.

[rugged void]

np

this was actually fun

[golden shore]

thank you

[twin wolf]

No problem I did so much!

[rugged void]

yes

opened a door

with the hinges

agreed

[twin wolf]

Yes

[rugged void]

@golden shore he's the real gigachad here fr

[stiff musk]

see more examples in this video

at around 1:45

https://www.youtube.com/watch?v=nyJeaUe7wXM&list=PLPyapQSxH6mbsY2g78ZaMfZx-3EEAxLia

[rugged void]

BRO THAT THUMBNAIL

[golden shore]

real life related example

that's my fav

[rugged void]

https://tenor.com/view/spit-take-laugh-lmao-gif-27296759

[golden shore]

thats terrific

as an entertainment while eating my lunch

i will close the channel

now

since my question is solved

.close

[full forum]

Channel closed

Closed by @golden shore

Use .reopen if this was a mistake.

[twin wolf]

The “an” is unnecessary by the way

[full forum]

Available help channel!

Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[obsidian swift]

Am I right until the arrow

[gilded quarry]

check your limits after you change 2theta to u

[obsidian swift]

what do you mean

like plug in my limits

.close

[full forum]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
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[sly prairie]

Can someone explain me why this function is defined on ]0,e^1/3[ please ?

[sly prairie]

I mean my problem is the e^1/3

[slate violet]

Seems to be when y(x) is positive

Look back at the problem to see if there's a condition where y(x) needs to be positive

[sly prairie]

ok I see

[slate violet]

Yeah makes sense now

[sly prairie]

"x,y,λ /= 0

"

Thank you ^^

[slate violet]

Npnp!

[sly prairie]

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[marsh tusk]

um

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[marsh tusk]

uhh

[hybrid fjord]

The timing 😭

[marsh tusk]

lol

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[hollow loom]

How do u solve these

[hollow loom]

Do u just find derivative and sub x in to fx

[thick hedge]

,rotate

[thick hedge]

first princples

so use the definition fo the derivative in terms of limits

[hollow loom]

Umm can u help me a bit

[thick hedge]

use this

[hollow loom]

Ohk thx also what does the denominator for the second one mean

[thick hedge]

use the first one

[hollow loom]

What do i do after that

Oh i get it

Nvm thx

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[hardy folio]

how to find the vertices of the circle 4x^2+4y^2=1 ?

[hardy folio]

ik that for standard form -> x^2/a^2+y^2/b^2=1 a & b are the radius but i'm confused on finding the radius when x & y have coefficients

[stiff summit]

A circle has no corners

[hardy folio]

Sorry, how to find the radius

[stiff summit]

You can divide both sides by 4

[hardy folio]

ok i got it thank you @stiff summit

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[bitter tide]

Hi, Idk how to solve it properly, my first try was something like x*35/1.7, but its is incorrect. Any help?

[glacial veldt]

Have you been told if the amount of light hold increases lineraly or exponentially?

[bitter tide]

no (

[bitter tide]

I guess my solution I tried is related to linear increase, so I need to try exponential?

[glacial veldt]

Hm, well, as far as I remember this from optics lecture it is indeed linear

[bitter tide]

yeah, but it's math xD

i guess it shouldn't be solved using physics formulas

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[next basalt]

Can someone help me understand what i'm doing wrong?

$\int_0^1 \frac{x+1}{x^2+5x+6} dx$

[glossy valve]

Merineth

[next basalt]

according to wolf the integral should be two logs?

[thick hedge]

I suggest you multiply and divide by 2 first and then add and subtract 3 to the numerator

Instead of using partial fractions

[next basalt]

Why can't i use partial fraction?

I would've never come up with multiplying and divide by 2 and add and subtract 3

shouldn't partial fraction work just as fine?

[thick hedge]

You could use partial fractions too

[next basalt]

Did i atleast figure out A and B correctly?

[proven mango]

$\int\frac{1}{x+2} dx$ isn’t equal to something in arctan

it should be ln

[next basalt]

Why?

[proven mango]

you can u sub for x+2

[next basalt]

oh right

but why can't i make both of them sqrt and then ^2 ?

[glossy valve]

y0shi

[next basalt]

Isn't that technically ok?

[proven mango]

then you have to take the derivative of $sqrt(x)$

[glossy valve]

y0shi

[proven mango]

since you’re technically u subbing

u for sqrt x

[next basalt]

No i'm not?

I'm not subbing anything

i'm rewriting

[proven mango]

when doing arc tan

you are changing the bottom x term

to something in terms of u^2

and since you are subbing x for u

you are applying u sub

[next basalt]

$\frac{1}{2+3} = \frac{1}{\sqrt{2}^2 + \sqrt{3}^3}$

[glossy valve]

Merineth

[next basalt]

They are equivalent

[proven mango]

yeah but that’s for constants

[next basalt]

So?

x is a value the same as constant

$x = \sqrt{x}^2$

[glossy valve]

Merineth

[proven mango]

$\int\frac{1}{\sqrt(x)^2+\sqrt(2)^2}dx$ let’s take this for example

[next basalt]

yes

[glossy valve]

y0shi

[next basalt]

apply rule for arctan on it

that is perfectly viable

[proven mango]

you are trying to get this in the form of $\frac{1}{u^2+a^2}$ yes?

[glossy valve]

y0shi

[next basalt]

It is already in that form

[proven mango]

yes so u is sqrt x

and a is sqrt 2

[next basalt]

yep

[proven mango]

we have to put the entire integral in terms of u

instead of in terms of x

[next basalt]

No

[proven mango]

which requires changing dx to du

[next basalt]

No

[proven mango]

yes you do

[next basalt]

I'm not subbing

no i'm not

I can rewrite 100 as 10^2 if i wanted to

[proven mango]

okay let’s take this example $\int\frac{1}{4x^2+4} dx$

[glossy valve]

y0shi

[next basalt]

the same with 2

[proven mango]

what would be the answer

[next basalt]

2 can be rewritte as sqrt 2 ^2

That is not the same question

[proven mango]

ik

take it as an example

[next basalt]

you can't apply it

since we have a constant infront of x

Mine does not have a constant

[proven mango]

what do you mean

you can say u = 2x

[next basalt]

NO

[proven mango]

it’ll be in the form of u^2

[next basalt]

i'm not subbing!!!!!!!!!!!

[proven mango]

this is subbing

[next basalt]

**no u substituion **

[proven mango]

look

[proven mango]

yep

[next basalt]

I am not subbing

[proven mango]

x here is sqrt x for your question

[proven mango]

it’ll be better if you put that x to u so there’ll be less confusion

[next basalt]

I dont

want to

do

u

sub

sti

tution

[proven mango]

you have to

[next basalt]

No?!

I do not have to

[proven mango]

since that equation only works

if you have it all in terms of u

okay how about this

take the derivative of what you have

see if you get it back

cuz you won’t

[limpid junco]

This is basically like saying int of 8x³ is (2x)⁴/4

[proven mango]

why’re we even using inverse tan tho

cuz it doesn’t even work for something like this

[next basalt]

And i'm asking why

[proven mango]

well actually it does kinda work but still give an ln at the end

[next basalt]

Is it because the formula requires it to be $x^2$ and not $\sqrt{x}^2$ ? Even tho they are essentially the same thing?

[glossy valve]

Merineth

[proven mango]

yes

[next basalt]

Well okay then

[proven mango]

you can do sqrt x

but you also have to take the derivative of it

let’s not make it more complicated and just leave it as what you wrote there

[next basalt]

Fine, it makes no sense but i'll just live with it

What is the reason behind being able to rewrite $4 = 2^2$ but not being able to rewrite x as $\sqrt{x}^2$?

[glossy valve]

Merineth

[proven mango]

here how about this $\int\frac{1}{u^2+a^2} du$ = $\frac{1}{a} arctan(\frac{u}{a})+C$

[glossy valve]

y0shi

[proven mango]

this is true right

our inverse tan formula

but i just switched u to x so we can distinguish between them easily

this formula only holds true if an integral can end up in this form

u would be sqrt x

and a would be 2

yet here we have to change dx to du

which is why a substitution must be used

if we keep it as what you have up there

the final antiderivative doesn’t account for the chain rule

coming from the sqrt x

[next basalt]

Ok so basically constants can be written in exponential form but variables cannot

[proven mango]

not necessarily

but like doing what you did up there is equivalent of saying the integral of sin^2(x) can be solved using the power rule

[queen crater]

That's not the way to look at it

First of all, if you have any positive quantity a, you can use a = sqrt(a^2)

[next basalt]

Aah the sqrt can't handle negatives on the variable x?

[queen crater]

However, in this expression, there is only one a but two u's:

$\int\frac{1}{u^2+a^2} du$

[glossy valve]

Nel

[queen crater]

No, sqrt is always positive

Though if you know that the quantity a is negative, you can use a = -sqrt(a^2)

[next basalt]

$\frac{1}{x + 4} = \frac{1}{\sqrt{x}^2 + 2^2}$ Then how come this isn't perfectly viable?

[glossy valve]

Merineth

[queen crater]

This is viable (assuming x is positive)

[queen crater]

Read that again

[proven mango]

look how u is sqrt x in this case

[proven mango]

and the integral here has a du

whereas we have a dx in our original integral

[devout valley]

• how are you thinking of dealing with that?

[queen crater]

Do you understand what I mean yet?

[next basalt]

No i don't

But i think i'll just accept that unless i have x^2 to begin with i wont use the formula for arctan and use u sub instead

[queen crater]

How many times do you see u here?

[next basalt]

2?

[queen crater]

Right, now if you have sqrt(x)^2 and you want this to be u^2, what do you want to replace u with?

[next basalt]

sqrt(x)

[proven mango]

now what about du

[queen crater]

Yes, and what do you get as an expression if you replace all the u's with sqrt(x)?

[next basalt]

x

[queen crater]

From this, what do you get

The full expression

[next basalt]

i'd get \sqrt(x) ^2

[queen crater]

The full expression

With the integral and everything

[next basalt]

$\int \frac{1}{\sqrt{x}^2 + a^2} d\sqrt{x}$

[queen crater]

Where did the du go?

[queen crater]

.

I said all

[glossy valve]

Merineth

[queen crater]

There you go

See, that's different from the arctan formula

It's d sqrt(x), not dx

[next basalt]

I see

[queen crater]

Are we going back to this now?

[next basalt]

No i already solved it

i'm on another one rn

but i think i need a hint

or a confirmation that i'm doing right or wrong

$\int_0^1 \ln(1+x^2) dx$

[glossy valve]

Merineth

[next basalt]

Should i use u sub on entire thing ?

Because i can't seem to find any formula which allow me to integrate ln anywhere

[queen crater]

That's a composition, so probably integration by parts

[next basalt]

omg..

I can't believe i missed that

FML

i'm gonna fail lmao

..

$\ln(1+x^2) \times (x+\frac{x^3}{3}) - \int \frac{2x}{1+x^2}(x+\frac{x^3}{3})$

[glossy valve]

Merineth

[next basalt]

This seem right so far?

Deriviate the ln(1+x^2) once and integrate 1+x^2

[thick hedge]

no, you integrate dx in the first part

[thick hedge]

not $1+x^2$

[glossy valve]

Why am. I here

[next basalt]

So i should deriviate ln(x) and not ln(1+x^2) ?

$f(x) = ln(x) \
g(x) = 1+x^2$

?

[thick hedge]

f(x)=ln(1+x^2) g(x)=1

[glossy valve]

Merineth

[next basalt]

huh

Where?

f(g(x)) ?

[thick hedge]

wdym?

[next basalt]

no wdym

how did you figure that out

What am i supposed to integrate vs deriviate

[devout valley]

You hopefully should have seen how you integrate ln(x) as an example somewhere...

[queen crater]

I might have confused you by saying it's a composition, my bad

Tbh you can use integration by substitution too (it's a little more delicate)

[devout valley]

...it's a similar type of problem here, just instead of pure ln(x), it's ln(1 + x^2)

[queen crater]

Point is, when you do integration by parts, you do the derivative of one part and the antiderivative of the other

The derivative of ln(u) is much easier to deal with than ln(u) itself

And you can always do the antiderivative of a constant a, that's just (ax)

[next basalt]

,rotate

[next basalt]

Is this not how i do it?

HUH

But i did it like

yesterday

[queen crater]

Nope, that's not how IBP works

You got the derivative of ln(1+x^2) correct, but you need a product

[next basalt]

I don't know anymore

slowly losing hope for my upcoming exam

I did this like yesterday or the day before

and i've already

forgotten or gotten it wrong

oh...

I should imagine there is an invisible 1

and use that as my primitve

[thick hedge]

yes

[queen crater]

for your primitive?

[next basalt]

$[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{2x}{1+x^2}x dx$

[queen crater]

What did dx ever do to you, you always forget it

Ok, but keep the bounds of integration

The first part also has bounds

[glossy valve]

Merineth

[queen crater]

[next basalt]

$[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{2x}{1+x^2}x dx \
[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \
[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{x}{u}du$

[glossy valve]

Merineth

$[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{2x}{1+x^2}x dx \\
[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \\
[\ln(1+x^2)x]_0^1 - \int_0^1 \frac{x}{u}du$
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   \xparse function is not expandable 
l.50 [\ln(1+x^2)x]
                  _0^1 - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \\
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

[next basalt]

[queen crater]

If you use a \function, you follow with {...}, not (...)

[devout valley]

$\ln(x^2 + 1)$

[glossy valve]

@devout valley

[devout valley]

it's not that

[queen crater]

[devout valley]

$[ \ln(1+x^2)x ] - \int_0^1 \frac{2x}{1+x^2}x dx \
[ \ln(1+x^2)x ] - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \
[ \ln(1+x^2)x ] - \int_0^1 \frac{x}{u}du$

[glossy valve]

@devout valley

$[ \ln(1+x^2)x ] - \int_0^1 \frac{2x}{1+x^2}x dx \\
[ \ln(1+x^2)x ] - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \\
[ \ln(1+x^2)x ] - \int_0^1 \frac{x}{u}du$
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   \xparse function is not expandable 
l.50 [ \ln(1+x^2)x ]
                     - \int_0^1 \frac{2x^2}{u} \frac{du}{2x} \\
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

[queen crater]

I don't understand either lmao

[devout valley]

Welp ah well

[next basalt]

Does it like fine atleast?

[thick hedge]

Try breaking it into multiple texit commands

[next basalt]

$\int_0^1 \frac{2x}{1+x^2}x dx\
\int_0^1 \frac{2x^2}{u} \frac{du}{2x}\
\int_0^1 \frac{x}{u}du$

[glossy valve]

Merineth

[next basalt]

[thick hedge]

How do you have both x and u in the same integral, that's wrong

[next basalt]

I was told that is ok?

[queen crater]

It's not wrong, it's just that you can't find a result for it

[next basalt]

Yeah i got stuck at that last part

$\int_0^1 \frac{2x^2}{1+x^2}dx$

got any advice for this?

I see that the numerators deriviate is 2x

oh wait

[glossy valve]

Merineth

[next basalt]

this is what i have

But how do you know that?

[devout valley]

Top heavy

[next basalt]

Top heavy?

[queen crater]

The denominator is a sum, you want it in the numerator and then split the fraction

(because you already have x^2 in both)

[devout valley]

[and you ideally do not want to be dealing with "polynomial fractions" when the degree of the numerator is not strictly less than that of the denominator...]

[next basalt]

Ooh i see

could i potentiall do

polynomial division here?

[queen crater]

Yea

Though it's a pretty easy one

[next basalt]

$2-\frac{2}{x^2+1}$

[glossy valve]

Merineth

[next basalt]

Easy for you!

[queen crater]

Nice, but when you have a factor like this, you could just put it aside

[next basalt]

I do that step when i integrate them both

factoring out that is

[queen crater]

$\int_0^1 \frac{2x^2}{1+x^2}dx = 2 \int_0^1 \frac{x^2 + 1 - 1}{1+x^2}dx = 2 \int_0^1 1 - \frac{1}{1+x^2}dx$

[glossy valve]

Nel

[queen crater]

Do what you feel is easier, that's just an example

[next basalt]

Yeah i'd probably not od that

$\int_0^1 2dx - 2\int_0^1 \frac{1}{x^2+1}dx$

[glossy valve]

Merineth

[next basalt]

This server should 100% get a "dx" emote

the same way +C

[devout valley]

plusc isn't from here, but it should be too

Might add it to my emote server

[next basalt]

Okay my integral became

$[xln(1+x^2) -2x-2arctan(x)]_0^1$

[glossy valve]

Merineth

[next basalt]

Do i include +C inside the []?

[queen crater]

Technically it's there inside the brackets but it cancels out when you compute the difference

[next basalt]

Does it seem ok?

,w integrate ln(1+x^2)dx

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by merineth

[devout valley]

You probably forgot somewhere but otherwise fine

[next basalt]

[queen crater]

Yeah it's +arctan

[next basalt]

integrals is superhard

one minor mistake and the entire thing is ruined

Can the chain rule be applied on integrals?

But in reverse?

[devout valley]

That's what integration by substitution is

[next basalt]

i keep forgetting

ILATE

What's the difference between

A and E?

E is e^x ?

while A is poly?

[devout valley]

Algebraic is polynomials, or fractions of

Yep, and also stuff like (x + 1)/(3x^2 + 6x + 9)

[devout valley]

and e being exponentials, e^x, 2^x etc etc

[next basalt]

$\int_1^e (ln(x))^2 dx = \int_1^e u^2xdu = [\frac{(ln(x))^2x^2}{2}-\frac{2ln(x)x^3}{6}]_1^e$

Does this seem reasonable?

[glossy valve]

Merineth

[devout valley]

Hmmm

May want to justify the steps you took

[thick hedge]

Nope, you haven't performed the usub properly

Hint ln(x) ' derivative is 1/x

Not 1

[devout valley]

They have noted that but didn't complete the changing of variables, it seems, see the extra x that wasn't undone

[thick hedge]

Ah,ok

[devout valley]

But before you do any integrating, make sure you've gotten rid of all appearances of x, if you didn't do that(!)

[next basalt]

why? and how

[devout valley]

The whole point of substitution is that the resulting integral you get, in terms of the new variable, is easier, but you cannot ignore x because it still depends on u

Oh, and also, limits too

[thick hedge]

Think of it this way dx is a unit, for now you can only have one unit in your integral

So you want everything to be in terms of that unit

If this analogy is wrong, someone please correct me

[next basalt]

$\frac{du}{dx} = \frac{1}{x} \implies dx = du*x$

[glossy valve]

Merineth

[next basalt]

So now i have to change my x to the unit u before integrating?

[thick hedge]

Yes

[next basalt]

And i set up the equation:
$u = ln(x)$ and solve for x?

[glossy valve]

Merineth

[thick hedge]

Yes, so you now express x in terms of a function of u

[next basalt]

Ok so $x = e^u$

[glossy valve]

Merineth

[next basalt]

$\int_1^e (ln(x))^2 dx = \int_1^e u^2 e^udu$

[glossy valve]

Merineth

[next basalt]

And now apply IBP

where i deriviate u^2 and integrate e^u

[devout valley]

A note: be very careful about the limits: those apply to the values of x, not of u, there are two ways to deal with that

[next basalt]

Can't i just integrate and change u back to terms of x?

and then utilize the limits?

[devout valley]

Either change them to be in terms of u, then you can always forget about them and just put them back in

[thick hedge]

Yes, you can

[next basalt]

hahah chart

which one do you prefer?

[devout valley]

Or make it clear that they aren't in terms of u, so like $\int_{x=1}^{x=e} u^2 e^u \dd u$

[glossy valve]

@devout valley

[next basalt]

If i wanted to use u

and integrate with u

how would i change them?

u = ln(1) ?

solve for 1?

[devout valley]

Well solve for u, then replace that as the lower limit

Similar replace the upper limit with ln(e), simplifying that down

[next basalt]

$\int_1^e (ln(x))^2 dx = \int_{e^1}^{e^{e}} u^2 e^udu$

[devout valley]

,calc log(1)

[glossy valve]

Result:

0

Merineth

[next basalt]

Like that?

[queen crater]

It's ln, not e^

[next basalt]

u = ln(e)

[queen crater]

u = ln(x), so if you have x, you put it in ln to make u

[devout valley]

What's ln(e) equal to?

[next basalt]

1

Okay so we just replace x with the limits in the equation and then replace our limits with that result?

u = ln(1) = 0
u = ln(e) = 1

[devout valley]

Yep, and in the corresponding order

[next basalt]

So our actual limits change from 0 to 1

in terms of u

[queen crater]

$\int_{x=1}^{x=e} \ldots dx = $\int_{u=0}^{u=1} \ldots du$

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After i have integrated with the new limits and gotten myself an answer, think that is fine? I don't have to solve based on dx?

[glossy valve]

Nel
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[queen crater]

Meh, screw you compile error

[devout valley]

You don't

[next basalt]

[devout valley]

$\int_{x=1}^{x=e} \ldots dx = \int_{u=0}^{u=1} \ldots du$

[next basalt]

Me everytime compilerror appears

[devout valley]

the dollar sign in the middle haah

[next basalt]

Chartbit:s latex > Nel:s latex
Confirmed

[queen crater]

Too late to change it

[glossy valve]

@devout valley

[devout valley]

Why does the markdown have to obscure characters for

[queen crater]

I mean yeah, I don't pretend to be good at either LaTeX or maths

[next basalt]

but you are

[next basalt]

I'm gonna start using this

[queen crater]

It's all relative

[next basalt]

Way easier to keep track

Okay

so

$[u^2e^u-2ue^u+2e^u]_0^1 = e-2$

[glossy valve]

Merineth

[next basalt]

That is my final answer

,w integrate from 1 to e (ln(x))^2

[glossy valve]

Results provided by WolframAlpha

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Requested by merineth

[next basalt]

YAY

[devout valley]

Beat me to it

[queen crater]

WA understands int_1^e

[next basalt]

aaah nice

Wasn't sure

[queen crater]

It also understands frac{}{}, and probably more

Sometimes it's a bit random

[next basalt]

Don't give me the answer but

$\int_0^{pi/4} \frac{x}{cos^2(x)}dx = \int_0^{pi/4} x(cos(x))^{-2}dx$

Can i treat it as such?

[thick hedge]

yes, but 1/cos^2(x)=sec^2(x), that's a more convenient way of expressing it

[glossy valve]

Merineth

[next basalt]

Does the cosx)^-2 count as a poly or trig?

Ok, so i deriviate x

[thick hedge]

Trig, which is to avoid confusion I suggest you express it as $sec^2(x)$

[glossy valve]

Why am. I here

[next basalt]

I haven't learned sec cot sadly

so i try to avoid it

[devout valley]

Do you know what the integral $\int \frac1{\cos^2(x)} \dd x$ is at least?

[glossy valve]

@devout valley

[next basalt]

tan