help-27

[jaunty mantle]

Oh is it?

[vital edge]

Yes

[jaunty mantle]

Oh yeah it is

[vital edge]

I am pretty sure

So let $L_1 \cong L_2 \cong k$

[woven radish]

Zavier 🌺

[vital edge]

Then $L \cong k^2$

[woven radish]

Zavier 🌺

[jaunty mantle]

Yeah sure

[vital edge]

with the trivial bracket

You can now define an ideal for each $\lambda \in k$

[woven radish]

Zavier 🌺

[jaunty mantle]

That’s where you just multiply the elements of k together right?

[vital edge]

That gives you infinitely many ideals

[jaunty mantle]

Using that k is a field

[vital edge]

No

It is the one which sends every pair to zero

As trivial as it gets

[jaunty mantle]

Okay, what’s this then

[vital edge]

The abelian bracket is ab - ba right

[jaunty mantle]

Is it?

Isn’t abelian brackets [x, y] = [y, x]

[vital edge]

Not in the context of lie algs iirc

Hold on lemme find Humphreys

[jaunty mantle]

Oh you’re using no gap to represent the brackets

[vital edge]

Oh nvm the abelian bracket is just [x,y] = 0

[jaunty mantle]

Yeah

[vital edge]

I was thinking of the commutator bracket

[jaunty mantle]

This with the minus sign says it’s all 0

[vital edge]

(Note that they are identical for a 1d lie alg)

[jaunty mantle]

With [x, y] = -[y, x] implies [x, y] = 0 for all x, y

Which I’m happy with

[vital edge]

Anyway

[jaunty mantle]

So first I look at k² with the trivial bracket?

[vital edge]

Yes

And then look at the ideals in it

[jaunty mantle]

Okay

[vital edge]

Intuitively, the ideals are "lines"

[jaunty mantle]

Huh?

[abstract plover]

<@&268886789983436800>

[faint hearth]

i mean just think of R^2. The ideals are the linear subspaces

which are literally lines through the origin

[jaunty mantle]

I need to think about that

[faint hearth]

and each line has a slope

so each lambda in k "defines an ideal"

[jaunty mantle]

So let’s say y = 2x, if I pick 2 points (1, 2) and (2, 4) and bracket them I get (2, 8)?

No I bracket them I get 0?

[faint hearth]

are you sure? the bracket here just sends everything to 0

yeah and 0 is in the ideal

[jaunty mantle]

Okay great

What forces me to have straight lines then

[faint hearth]

i mean an ideal is defined to be a linear subspace + closed under bracketing

so the linear subspace condition alone forces lines, {0} or the whole space

[jaunty mantle]

Hmm I’m forgetting the linear part

I was thinking ring ideals

So what’s wrong with my solution then

[faint hearth]

you should check yourself if what you have defined are indeed ideals, im having a tough time reading it

[jaunty mantle]

Oh obviously they aren’t closed under scaling

I’d need something like l¹ = 2 l² and then this is an actual linear subspace

Then go S_a is when l¹ = al² but then this is the same as the lines interpretation

Aha

Thank you!

Now do I really need that dim L1 = 1? Or that L1 iso L2? Or is that just convenient?

[faint hearth]

You need it to make the bracket trivial

[jaunty mantle]

Do I have to have the trivial bracket tho

Or does this just make it trivially true that it’s an ideal so we don’t need to check that part

[faint hearth]

I mean here we needed it. Idk what happens in general

[jaunty mantle]

Mhmm okay thanks!

.close

[devout snow]

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[devout snow]

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[desert seal]

Not really sure how to approach this question

[devout snow]

@desert seal Has your question been resolved?

[shy osprey]

in part a, the impulse recieved sets the particle in a direction perpendicular to its motion so using this try to draw a rough figure of all impulses and momentum of particle

[desert seal]

yeah

gimme a sec ill pull up what i had drawn first

this is what u thought

but not sure if this is right or where to go from this

[shy osprey]

Yeah thats right!

See now the impulse imparted wont alter the initial velocity right?

As in the particle will keep moving in the same direction with the same speed but it will also simulataneoulsy move up

[desert seal]

yeah kinda so would it be moving in the direction of where i drew V?

[shy osprey]

So the velocities (left and up) when added together in vector notation will give the magnititude of the resultant velocity right?

[shy osprey]

Yeah

So first lets calc its velocity upwards

it would be I/m right?

where m is mass (2kg)

[desert seal]

is this like rearranging the impulse formula

[shy osprey]

Yeah

Impulse (I) = m* v

[desert seal]

yeah

so I/2 like you said

[shy osprey]

Yeah

Now we have velocities upwards and left

[desert seal]

so pythagoras?

[shy osprey]

The resultant velocity (mag) will be

[shy osprey]

yeah

[desert seal]

i see

i kind of get it now

lemme try explain what i was thinking intially

[shy osprey]

Okey

[desert seal]

so i was thinking of using the other equation for imupulse I = m(v-u) , and then reaarange to get I +mu = mv

but not sure where to go from this

[devout snow]

Channel closed

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[shy osprey]

.reopen

[devout snow]

✅ Original question: #help-27 message

[shy osprey]

see the thing here is that

Impulse is a vector

So inital velocity in the upward direction is zero

so we can treat I = m(v-0)

[desert seal]

okay

lemme upload 1 image and if u can, can u try explain it please

[shy osprey]

Oh you mean vector wise

Yeah would be better to understand using vectors

[desert seal]

is this questions kind of saying mv is the resultant vector right?

[shy osprey]

Yeah

[desert seal]

okaoakao

[shy osprey]

you mean momentum no?

[desert seal]

oh yeah mb

yeah

[shy osprey]

mv is final momentum, v is final velocity

[desert seal]

yeah

i kind of get it now

[shy osprey]

okey!

lmk the final ans!

[desert seal]

and is this also sayign the same thing then?

[shy osprey]

yeah

[desert seal]

Okay

Thank you once again 😭 😉

[shy osprey]

np

[shy osprey]

.

[desert seal]

oh yeah lemme quickly do it

[shy osprey]

we still have part b btw

[desert seal]

😉

Got the value of I to be 9.6Ns

[shy osprey]

,w 100 = (2.8)^2 + x^2

oh oops

[woven radish]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Apr 21, 2026, 5:31 PM

[shy osprey]

,w 100 = (2.8)^2 + x^2

[woven radish]

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[shy osprey]

Yeah thats correct!

[desert seal]

yessirrr

[shy osprey]

-# (But the negative one is also correct btw)

doesnt matter anyways since they ask mag

welp now for Q2 what do you have

as in diag?

[desert seal]

you mean for part b right?

[shy osprey]

yeah

mb

[desert seal]

essentially i think this part im pretty chill at - what im thinking is we know there gonna be the wall above it, and so like this

and then we can find the angles from our triangle from previou answer

[shy osprey]

Yeah

But

becareful tho when you apply restitution formulae

They are only valid for velocities normal to their plane of striking

[desert seal]

lemme see what i get 😉

i got 7/32

[shy osprey]

oh uh one sec

wait the mass of particle is 2kg?

like same as prev one?

[desert seal]

yeah i think so

[shy osprey]

o one sec

,w 4.8/1.75

[woven radish]

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[shy osprey]

,w 7/32

[woven radish]

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Upgrade to WolframAlpha Pro!

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[shy osprey]

uhm?

whats the formula u used for coeff of restitution?

[desert seal]

v=eu right

[shy osprey]

yeah

oh shoot

,w 1.75/4.8

[woven radish]

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Upgrade to WolframAlpha Pro!

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[shy osprey]

uhm now where did i go wrong

[desert seal]

hmm

[shy osprey]

what did you take v as btw

[desert seal]

i took v as 5

[shy osprey]

5?

why

[desert seal]

then resolved that into up and right speeds

[desert seal]

since it was the speed the ball was moving at in part a

[shy osprey]

wait

we have to combine part a and b?

[desert seal]

yeah

like they're linked together

[shy osprey]

oh sorry then i didnt really get the qsn

one sec

[shy osprey]

oh i see

[desert seal]

np

[shy osprey]

,w (1.75)^2 = (1.4)^2 + x^2

[woven radish]

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[desert seal]

hmm

[shy osprey]

,w 1.05/4.8

Yeah its correct

good job!

tho i dont know how you got exact 7/32 fraction wise

[desert seal]

i put into calc

[woven radish]

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[shy osprey]

oh i missed 5 as 4

am tweaking today😭

Anyways good job!!

[desert seal]

so like 1.4^2 + 4.8^2*e^2 = 1.75^2 then solve for e is what i did

[desert seal]

😭 dw bro

[desert seal]

thanks for the help as always 😉

[shy osprey]

Anytime bro!

Anymore?

[desert seal]

Nah not right now lol

Imma finish the rest of that worksheet today/tonight 😉

[shy osprey]

Aight sure❤️

so if u dont have one rn js

!done

[devout snow]

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

[desert seal]

.close

[devout snow]

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[shy osprey]

Your wlcm!
Have a nice day!

[devout snow]

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[vestal nebula]

sorry team but i need help in math it's my first year in High school , sorry for the malfunction

[trail eagle]

You've claimed this channel now. If you need help for a specific exercise or clarifications, do post your question. Otherwise, please close the channel with .close.

[vestal nebula]

It would be more about one or more chapters that my mom introduced me to this channel, which is great for math.close.

[neat solstice]

elaborate please

any specifics?

[devout snow]

@vestal nebula Has your question been resolved?

[devout snow]

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[serene root]

Hi!! This is for my quantitative methods in social sciences class, we’re using SPSS as a computing system but the textbook says that the learning program doesn’t help with calculating the critical value for the confidence interval, nor the margin of error.

Can someone help me calculate both? I’m not sure what to do, I have filled out the confidence levels from my SPSS output so I have a start!

[serene root]

computer pic is the problem, and the table on paper is my work

[devout snow]

@serene root Has your question been resolved?

[serene root]

<@&286206848099549185>

[tranquil ferry]

How i may help you?

[devout snow]

@serene root Has your question been resolved?

[devout snow]

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[serene root]

Hey!! So sorry

[serene root]

Can you help me find the confidence interval critical value? And the margin of error?

using the info from the table

[tranquil ferry]

interval critical value

interval critical value

[devout snow]

@serene root Has your question been resolved?

[devout snow]

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[hollow meteor]

hello me and meefs have a question

[karmic atlas]

anybody know why the conic gets interrupted by the circle / another way to construct a conic as the inversions of the envelope of a circle?
basically rn, the conic is the locus of points of the inversions from the tangents of A over circle C

[karmic atlas]

violin teacher !!!

[hollow meteor]

booooo

[karmic atlas]

brahhmmsmmsmsmms

[hollow meteor]

suht up meefs

[karmic atlas]

<@&286206848099549185> hlep pls

[hollow meteor]

hello helpers we would like helps

[karmic atlas]

ok nvm figured it out just being a dumby

.close

/.close

huh

.close

@hollow meteor i think u have to type .close

[hollow meteor]

.close

[devout snow]

Channel closed

Closed by @hollow meteor

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[hollow meteor]

i did it

[karmic atlas]

do it agian

[tranquil ferry]

.reopen

[devout snow]

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[carmine remnant]

Hi i would need help with doing differential homogenous equation bc im stuck at this one part on what to do with this

[patent loom]

wihch part??

[carmine remnant]

How do i integrate the fraction on left?

[runic prawn]

let t = v-1

[carmine remnant]

This is the answer but idk how it turned into -2 with 1/ v-1

[patent loom]

$\int \frac{2}{(v-1)^2} , dv = -\frac{2}{v-1} + C

how to do ts??

[runic prawn]

forget about the answer key

[patent loom]

$\int \frac{2}{(v-1)^2} , dv = -\frac{2}{v-1} + C$

[woven radish]

Chetan???

[runic prawn]

power rule

$t = v-1$ so $\dd{t} = \dd{v}$

[woven radish]

#whatdatmean

[runic prawn]

$\int \frac{2}{(v-1)^2} \dd{v} = \int \frac{2}{t^2} \dd{t}$

[woven radish]

#whatdatmean

[runic prawn]

$=\int 2t^{-2}\dd{t}$

[woven radish]

#whatdatmean

[patent loom]

quick tip : if the function has a degree of 1, u can treat it like .. x

basically sin(2x-a) integration would be -cos(2x-a)/2 or
i guess this is where u were stuck??

[runic prawn]

$= 2\int t^{-2}\dd{t}$

[woven radish]

#whatdatmean

[runic prawn]

$=2 \frac{t^{-2+1}}{-2+1}$

[woven radish]

#whatdatmean

[runic prawn]

$=\frac{-2}{t} = \frac{-2}{v-1}$

[woven radish]

#whatdatmean

[runic prawn]

!done

[devout snow]

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

[carmine remnant]

thank you im still trying to process this 😭

.close

[devout snow]

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[runic prawn]

welcome lol

[devout snow]

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[analog wing]

20. Two polynomials f(x) and g(x) with a leading coefficient of 1 satisfy the following conditions. When f(x) is divided by x^2 - 1, the quotient is Q(x) and the remainder is R(x), where R(1) = 5. Find the value of Q(-2) + R(2).

[ Conditions ]
(a) g(x) is a cubic polynomial.
(b) The quotient and the remainder of f(x) divided by g(x) are the same.
(c) g(1) = g(-1) = 0, g(2) = 6
(d) f(1) = f(-1) + 4

① 2 ② 4 ③ 6 ④ 8 ⑤ 10

[analog wing]

Uh need 1

[arctic jacinth]

!status

[devout snow]

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

[glossy dew]

<@&268886789983436800> ?

[arctic jacinth]

@obsidian bridge I am scared

[glossy dew]

did you try defining the quotient and remainder functions in question

say g(x) = ax³+bx²+cx+d and the quotient/remainder function mentioned in b) is h(x)

try writing all conditions mentioned as equations now

[devout snow]

@analog wing Has your question been resolved?

[devout snow]

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[analog wing]

Tried but idk after making some simple things

[arctic jacinth]

@analog wing were you able to find g(x)

[hasty cargo]

Next time please use .reopen

[devout snow]

@analog wing Has your question been resolved?

[analog wing]

g(x) = x(x^-1)

[dark sable]

🥀 sorry bro I tried helping I'm kinda stupid

[dark sable]

with c, if g(1)=g(-1)=0 then they are roots and note that since the leading coefficient is 1, then there is only one possible thing to multiply to the factors (x-1)(x+1) to get a cubic polynomial with a leading coefficient of 1 leading to g(x)=x(x-1)(x+1)=x^3-x

with the question itself, reconfiguring the given gives us f(x)=Q(x) * (x^2-1)+R(x)
substituting x=1 gives us f(1)=Q(1) * 0+R(1)=5, so now we know f(-1)=1 since f(1)-4=f(-1)
with that, we can get f(-1)=1=Q(-1) * 0+R(-1) => R(-1)=1

with b, let q(x) be the quotient and remainder of f(x)/g(x)
you can reconfigure it to f(x)=q(x)g(x)+q(x)
notice how q(x)g(x) is divisible by x^2-1, Q(x)=q(x) * x and R(x)=q(x)
(note that R(x) doesn't change due to: f(x)=Q(x)g(x)/x+R(x) )
so we can understand that Q(x) and R(x) differ by only one degree

from here

not entirely sure how to continue

[analog wing]

One of the hardest problem fr

[dark sable]

I mean I do know that you can now express f(x) as a function of q(x)(x+1) so that gives us that that since it is also equal to q(x)(g(x)+1) there is some bsery going on in the last paragraph I made up

probably in the equivalence of f(x)=Q(x)g(x)/x+R(x)=q(x)g(x)+q(x)

at the very least we do know R(x) has a degree of at least one

[analog wing]

f(x) = g(x) h(x) + h(x) and idk h(x)'s value

[dark sable]

do u mean x(x^2-1)

[analog wing]

X^ ax+ b is so

[analog wing]

Yeah alr solved that

[dark sable]

you can make assumptions around Q(x) and R(x) based from h(x) but I'm iffy on that

you can see it from the thing I said

[dark sable]

notice how g(x)/x=x^2-1

f(x)/(g(x)/x)=Q(x)+R(x)/(x^2-1)=>f(x)/g(x)=Q(x)/x+R(x)/x(x^2-1)=Q(x)/x+R(x)/g(x) wait a minute

wait is the assumption true

[analog wing]

Nahh (x+1) (x-1) (x-k) g(2)=6..

I used this thought

[dark sable]

r u not gonna read the wall of text I sent

it would help partly

this is probably some factor theorem application I'm too dumb to understand

[analog wing]

Hmm not the idea

I'm just not good at this part.. it's OK tho

[glossy dew]

were you able to write some set of equations like this at least

try working with c) first

[dark sable]

they got c

[glossy dew]

oh, ok

oh right

[analog wing]

f(1)=R(1)??

[arctic jacinth]

Guys, I think I have a easier fact to point out, I dont know if you got it already or no

[analog wing]

Yeah 5 ig

[arctic jacinth]

May I ?

[dark sable]

wait ok so w/ b, q(x) has a degree less than the degree of g(x) which means q(x) is 2nd degree and with f(x) having a leading coefficient of 1, q must also have a leading coefficient of 1 so q(x)=x^2+ax+b and as we did get earlier f(x)=q(x)g(x)+q(x)=q(x)(g(x)+1), now since f(1)=5=q(1)(0+1)=q(1) and f(-1)=1=q(-1)(0+1)=5
so um you can get a series of equations to then get f(x)

[arctic jacinth]

We can find the g(x) by given information, it's a cubic polynomial with 2 of its roots given to us, you can find the whole of g(x) as g(2) is given

After that we can find Q(x)

[dark sable]

f(1)=R(1) because f(x)=Q(x) * (x^2-1)+R(x)=Q(1) * 0+R(1) for x=1 or x=-1

[analog wing]

Oh got it

[dark sable]

so like 5=1+a+b, 1=1-a+b so a and b=2, so q(x)=x^2+2x+2 so ||f(x)=(x^2+2x+2)(x^3-x+1)|| now you just have to get Q(x) from there and R(x) by dividing by x^2-1 (hint: you can separate the function into ||(x^2+2x+2)(x^3-x)+(x^2+2x+2)||

[analog wing]

I think I solved

.close

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[dry nest]

<@&268886789983436800>

mb

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[muted folio]

how to prove this for a subjective exam

[muted folio]

the solution allows for a "crude approximation"

but i'm looking for something watertight

[faint gorge]

water tight

[sand quarry]

subjective exam?

what does that mean

[muted folio]

lol-

[muted folio]

like theres an objective portion

and a portion where you write down stuff

[sand quarry]

how can maths be subjective ok whatever. this is besides the point

[muted folio]

huh?

[faint gorge]

Prove n!>3^n for n>8 via induction

[arctic jacinth]

Formal mathematics where you need to proof writing & stuff

[gloomy aurora]

logarithms

[gloomy aurora]

well that dont help

[muted folio]

alright lemme try that out

[muted folio]

what is a log gonna do for factorials htough

[arctic jacinth]

On LHS, we will have to pull out every power of 2, which isn't that much of a big task to be very honest..

[faint gorge]

Oh right

[gloomy aurora]

Prove $n! > 4^n$ via induction. It works for $n \ge 9$ i think.

[woven radish]

Annie Maqionde

[arctic jacinth]

You just do the box & keep diving by 2 ?

[faint gorge]

Yeah that ought to work

[muted folio]

ah

[viscid root]

out of curiosity, is this CMI 2021

[muted folio]

will i have to state it working when and where?

[muted folio]

yeah

[viscid root]

i recognize this question

awesome

[muted folio]

omg

[gloomy aurora]

i mean yes............

[muted folio]

like the n>9 condition?

ohwait

i think yeah

alright i think this is good enough

[faint gorge]

It's water tight

[slow vigil]

this is a fun problem

[sand quarry]

you can pair up every product of the factorial which gives you ten different products

[muted folio]

no make fun of me T_T

[sand quarry]

it should be easy to prove that each of those products are all at least equal to 20

[faint gorge]

Sorry it was srs

[muted folio]

phew

[sand quarry]

which is going to be greater than 16 each

[muted folio]

yeahhh

[slow vigil]

so many people talking in this thread lol
as @sand quarry said, try rewriting the factorial. think like n! = 1 * 2 * 3 * … * n. then try pairing each term from the ends. for example, pair (1 * 20), (2 * 19) etc

one you do that, look at where the largest value happens (aka, which multiplication pair gives the largest value)

[sand quarry]

its basically what this is all about anyway

[viscid root]

my favorite solution to this is to ||write out the prime factorization of 20! and replace all the primes with 2||

[sand quarry]

but you dont need induction

since its just one case

you dont need to generalise for all n

[muted folio]

true i guess that could be overkill

[muted folio]

yeah i guess thats the general consensus

like

manipulating the values

n stuff

[faint gorge]

If you took it as a fact I meant you could argue easily why it's false

[muted folio]

wdym

[faint gorge]

Since n!>4^n for n>8 then 2^20>20! is false

[muted folio]

wait

did i miss something

hold on

whoops

uhmm

WAIT

yeah

sorry i got a bit confused

.close

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[olive pivot]

.

[muted folio]

.close

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[coarse flume]

<@&268886789983436800>

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[willow helm]

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[crisp arch]

Hi, I would like some help solving a series problem. I've tried everything I can think of but I can't even get in the right direction, I don't want the full answer, just to find out what the correct direction is. The question is as in the attached picture.

I've tried to rearrange e part into different forms of sin and cos, but that doesn't help. I also tried treating it as the sum of a geometric series with common ratio re^inφ. I don't even understand how the sum is convergent, because even though I have to find the absolute value squared (which reduces the e section in every part of the sum to 1) the r^n is divergent unless 0<r<1, but that wasn't specified in the question. I also checked the answer in the back of the book in case the answer was that the sum was divergent, but it gives a finite expression that also includes φ which disproves my thought that the e part is reduced to 1 in the absolute.

I'm just not sure where I'm going wrong.

Any help is much appreciated!

[devout snow]

@crisp arch Has your question been resolved?

[lunar harbor]

You’re correct that this is a geometric series, but the common ratio isn’t just $r$. Note that
$$r^n e^{in \varphi}=\left(re^{i\varphi} \right)^n.$$

[woven radish]

Civil Service Pigeon

[crisp arch]

Oh sorry, I must have mistyped. I have tried that as the common ratio but I couldn't get it to work.

[lunar harbor]

!show

[devout snow]

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

[lunar harbor]

Show your actual calculations

[crisp arch]

Okay, give me a second to take a picture

Sorry the picture didn't work. I tried to use the formula for the sum of a geometric series, which is (R^n-1)/(R-1), in this case substituting R for re^iφ. when I did that and set the limit for n->infinity I got it to be lim(n->inf) -e^inφ

I finally got a picture. Sorry for messy working

[lunar harbor]

How did you swap the limits and absolute values like that

And why not just use the formula for the sum of an infinite geometric series directly

$$\sum \cdots=\frac{1}{1-re^{i\varphi}}$$

[woven radish]

Civil Service Pigeon

[crisp arch]

That looks correct, how did you get that?

[lunar harbor]

And why not just use the formula for the sum of an infinite geometric series directly

$$|r|<1 \implies \sum^{\infty}_{n=0} ar^n=\frac{a}{1-r}$$

[woven radish]

Civil Service Pigeon

[lunar harbor]

The proof goes as you did to find the sum of the finite and take the limit as n -> inf

You just did some very confusing stuff after that which I’m not in a position to parse atm

[crisp arch]

Oh right, I was using the formula for the sum when not infinite, a-r^n/1-r. But how can we be sure that re^iφ is less than 1?

[lunar harbor]

You implicitly assume the series converges otherwise you can’t find a closed form answer. Of course, this requires |r|<1.

Note that the magnitude of the exponential is 1

[crisp arch]

Okay, that makes a lot more sense now. Thank you so much for your help

[lunar harbor]

[devout snow]

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[crisp arch]

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[vital edge]

A translation would help

Alternatively, you could wait for someone who knows the language to show up

[icy flower]

Yes, these are just for the graphs, I will translate momentarily

The first screenshot reads “function f has a primitive function F. Below is the graph to F”

The question I’m having trouble with is :

“Another function f has a primitive function G. One of the graphs A-F shows primitive function G. Which of the graphs show G if g(x)dx=3?”

So, the question essentially wants to know which graph shows G if it has the area of 3 from between 0 and 1 on the x axis

Specifically, as far as I know, integrals show the total area within a function, but the total area of all the graphs shown is more than 3

<@&286206848099549185> pinging due to inactivity

[sand quarry]

You are confusing what the integral $\ds\int_0^1g(x)\dd x$ is calculating. Yes, it calculates the area under the curve $g(x)$, but the graphs shown in A-F are that of $G(x)$ (the primitive function) NOT $g(x)$

[icy flower]

I see

That part makes sense

[sand quarry]

You have to use the fundamental theorem of calculus to connect them

Do you know that

[icy flower]

I am not familiar with that, no

[sand quarry]

,, \int_a^b g(x) \dd x = G(b) - G(a)

[sand quarry]

This, basically

[icy flower]

Ah, I see, so in order to find the correct G it’s the difference between 1 and 0

[sand quarry]

Yes.

Exactly

G(1) - G(0) = 3, to be specific

Just to clarify

Can there be more than one correct answer

[icy flower]

Given we aren’t given the equation, I’m unsure how to solve this?

I would assume it’s B graph, since it’s the only point where y=3

[sand quarry]

Youre given the graphs, though

You just have to see which one(s) give you a difference of 3

[icy flower]

I am unsure how that translates into a graph, though

y=3 is the only info we’re given

[sand quarry]

Look at B

[icy flower]

Yeah

[sand quarry]

Can you evaluate G(0) and G(1)

[icy flower]

0,2 and 1,3

[sand quarry]

What is G(1) - G(0)

[icy flower]

1

3-2 =1

[sand quarry]

Is it 3

[icy flower]

No

[sand quarry]

So it is not the correct option

Repeat the process for the rest

[icy flower]

1,2 and 0,3 gives -1

[sand quarry]

Not that either, then

[icy flower]

-3,8— 3.8 = 0 for f

Which is noteworthy

Wait that’s wrong

Nv

-3,8-3 gives -0,8

[sand quarry]

Not 0 but it's too small of a difference to be 3 anyway

[icy flower]

1– - 2 gives 3

So it’s E

Ok, I understand

[sand quarry]

Yay :D

[icy flower]

While I’m here

How do I properly write out an integral? I know you have to find the primitive

Actually, I’ll just write it and take a pic

[sand quarry]

Oki

[icy flower]

It reads “find area between 5 and 0”

Is this the correct notation?

[sand quarry]

Yeah

Seems perfect

[icy flower]

Ok awesome

Thanks for the help 👍

[sand quarry]

Of course

Nice Hornet pfp btw

[icy flower]

Shaw!

Thank you stranger, this appearance has been a source of strength during my travels

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[ocean isle]

I'm not able to solve this question.
Let $A = \left{\frac{1}{n} : n \in \mathbb{N}\right}$, $B = A \cup {0}$, $I = [0,1]$ and let:
$$X = A \times I, Y = B \times I, Z = Y \cup {(x,0) : 0 \leq x \leq 1}$$
and gave $X, Y, Z$ the relative topology to $\mathbb{R}^2$.

Determinate if $X,Y,Z$ are connected, local connected, path-connected or local path-connected.

I have solved for $X$ and $Y$, but not for $Z$.

[woven radish]

Martu

[devout snow]

@ocean isle Has your question been resolved?

[ocean isle]

.close

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[golden isle]

can someone help me get started on this problem

[supple knot]

use parallelepiped / triple scalar product formula

[golden isle]

sorry idk what that formula is

[supple knot]

https://en.wikipedia.org/wiki/Parallelepiped

[faint gorge]

You can try to construct two vectors that span this parallelogram, then the magnitude of their cross product returns the area iirc

[supple knot]

and or https://en.wikipedia.org/wiki/Triple_product

[golden isle]

oh

wait there are four points

and the triple product only has three

[supple knot]

the triple product has 3 vectors

[dapper tiger]

it's normal, in a parallelogram one point among four is given by the other 3

you only have 3 vectors

[supple knot]

do you know the difference between a vector and a point?

[golden isle]

yea

[supple knot]

so why do you think the triple product has only 3 points

[golden isle]

so make one point zero and translate all the remaining points

[golden isle]

it said it requires three vectors

[supple knot]

i don't think you understand

[golden isle]

what is it then

im bad at learning so i forget these stuff easily

[supple knot]

vertices (the points you're given) don't have direction. vectors do

to form a vector from two points, you take their difference

[golden isle]

okay

[supple knot]

e.g. in two dimensions:

[golden isle]

so do i make one point of the 3d verticies 0

and transalate the remaining three points

[supple knot]

you're given the equivalent of OA (point from origin to A) and OB (point from origin to B)

try it and see

[golden isle]

okay

ty

[dapper tiger]

star deleted his/her method and I don't know why, it worked and was the easiest too

[golden isle]

its not deleted

[dapper tiger]

woah, huh, yeah

I'm very tired

my bad

[faint gorge]

It's still here but I get as usually ignored

[dapper tiger]

nah I read and liked the answer : )

just couldn't see it again when I read again

the reason it works is elegant too, the magnitude of cross product is geometrically the product of the lengths x sin(angle)

so base x height = area of the parallelogram

[golden isle]

(-1, 0, -2), and (-1, 0, -2)?

[dapper tiger]

you need to span the parallelogram, so find two vectors that are not the same, and not the diagonal

you want to cross product these

[golden isle]

(-1, 0, -2), (-2, -1, 0)

(-2, 4, 1)?

[dapper tiger]

I agree with the choice of (-2, -1, 0)

but your (-1, 0, -2) is a sign error to me

it should be (-1, 0, 2) or (1, 0, -2)

[golden isle]

oh

[dapper tiger]

depending on the direction you took

[golden isle]

so the cross product is (2, -4, 1)?

and answer is sqrt21?

[dapper tiger]

yep

[golden isle]

Yayy

Ty

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[blissful ocean]

Can anyone help me with understanding this concept? I just can't begin to understand how these are equal.

[muted reef]

the 1 can be rewritten

[blissful ocean]

Oh, I didnt see that. That makes complete sense thank you.

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[stark spruce]

is this a local maximum?

[stark spruce]

first derivative test tells me that function is increasing for its entire domain, that means no local maxima or minima i assume?

[spiral raptor]

Generally a local maximum is the maximum in a certain range

But, the point 0,-1 is not a maximum of the function

[stark spruce]

gotcha

but it is an inflection point? or no?

[spiral raptor]

It looks like it is an inflection point. Try checking the second derivative

[devout snow]

@stark spruce Has your question been resolved?

[dark sable]

if I'm remembering this right, if you're considering entire domains you would be checking for absolute maxima/minima and local is for a certain interval of x

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[rose cloud]

How did the piecewise function of his make this graph? Isn't 3x x>1 meant to make a straight line?

[rose cloud]

I get the "greater than" and how it's cut off until the 1 x axis, but I don't understand how it made a 45* angle.

[hardy trail]

Hi! So you think y = 3x is a straight line? Could you please explain why :)

Btw the angle is not really 45 degrees, but I will touch on that a bit later.

[rose cloud]

Is my blunder being that only numbers without coefficients can qualify as a straight line? Since it's 3x, it should be in a 45* angle?

Either that, or I've got my x and y axis all wrong and the 3 is at the y axis, and still counts as an x (Which is normally viewed as a 45* angle)

[hardy trail]

Mmmm not exactly

Before anything,

Can I ask you what
"y = 3x" means?

Like if I were to give you this equation, what would this mean to you?

What is it saying?

[rose cloud]

It means 3 is on the y axis, so it would be (0,3), meaning 0 in the x axis, which would put it at the region (or like the starter place i forgor the name) and 3 would be going 3 up

OH WAIT

Is it because of rise over run, and given the fact the x value is there makes it qualified?? So It would be 3/1 and it would go 1 right and 3 up, which is the right part for the point to be

[hardy trail]

Yeah exactly! I mean to simplify this even more,

The equation "y = 3x" is saying:
"For every x value, the y coordinate is 3 times that x coordinate"

So, by following this logic - when x = 1, y = 3. When x = 2, y = 6, when x = 3, y = 9, and so on so forth.

When you look at it this way, indeed - the line is not horizontal or vertical, it is going diagonally!

[rose cloud]

I see

[hardy trail]

That brings me to my next point - the angle is not 45 degrees. If it were, the line would be increasing by the same amount each time. I.e going up 1 whenever you go right 1. The angle will therefore be a little higher than 45 since the graph is going up. (As a bonus exercise, how would you find this angle? Hint: Use trig).

But anyway, that should cover your main question.

[rose cloud]

Thank you, I'll close the channel now!

.close

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[last parrot]

Hi, just a quick question, we know a function is integrable in $[a,b]$ when $U(f,P) - L(f,P) \to 0$, then, by using Darboux sum, how can I show $\int_0^1 2x$ exists using $\varepsilon-\delta$?

[woven radish]

Minλ

[round raptor]

A minh spotted

[vagrant skiff]

you need to show that for any ε > 0, there exists a δ > 0, such that for every partition P of [0,1] with mesh size ||P|| < δ, the difference between the upper and lower Darboux sums is strictly less than ε

[last parrot]

Hmm what could this mean? Do I fix ε > 0 then pick a P?

[sand quarry]

Yes

Also it probably helps that your function is strictly increasing

[last parrot]

Yes I can see it

[vagrant skiff]

you can do that too

[sand quarry]

The sup and inf will strictly lie on the endpoints of any subinterval you may consider

[last parrot]

I recognise the idea but I don't know where to start with, do I just pick a P first, then state ε > 0, or other way around

[vagrant skiff]

it's the standard Darboux criterion

[vagrant skiff]

you always want ε > 0

Let epsilon > 0 be given. We must show there exists a partition P of [0,1] such that U(f,P) - L(f,P) < epsilon

that's what you want

then you construct such P

[last parrot]

Right, then I state épsilon first

Then I want each sections have same length then their length is $\frac{a-b}{n}$ for $n\in \mathbb{N}$?

will do this

[woven radish]

Minλ

[supple knot]

sounds fine. you know a and b in this case

[supple knot]

did you typo the question and mean [0, 1] instead of [a, b]

[last parrot]

Ah yes

If a function is integrable in $[a,b]$ when $U(f,P) - L(f,P) \to 0$ then what $ U(f,P) - L(f,P) \to ?$ for $y=2x$? ;-;

[woven radish]

Minλ

[last parrot]

Is it 2?

[sand dove]

Well no, it should be 0

we're guessing that f(x) = 2x is integrable on [0,1]

So we must have what you said

[last parrot]

Right so we want to show it equal 0

This rule apply to all functions right?

[sand dove]

not that it equals 0 for every P, it should go to 0

[last parrot]

yea to 0

[sand dove]

so, you said to try with uniform partition size

go for it, try P = {uniform partition with n+1 points}

[last parrot]

Yes, I want to find an epsilon in term of n

[sand dove]

It's more the other way around?

[last parrot]

Hmm, I asked my professor for a hint and she said this is 2/n

[sand dove]

Start by computing U(f,P) - L(f,P) for this P

[last parrot]

She says the subtraction of U - L = 2/n

[sand dove]

ok, did you check? or do you just want to admit it?

[last parrot]

How can I do this? ;-;

[sand dove]

? Apply the definition?

[last parrot]

No I haven't checked it yet but when I asked my professor just then she said it is 2/n

Ah yes sorry this is quite confusing

[sand dove]

$P = \left{\left[\frac kn,\frac{k+1}{n}\right] :: k\in {0,...,n-1}\right}$

[woven radish]

Rafilouyear2026

[devout snow]

@last parrot Has your question been resolved?

[last parrot]

Just a quick check, is $\sum_{i=1}^n\frac{2}{n^2}$ is $\frac{2}{n^2}$ add itself $n$ times right?

[woven radish]

Minλ

[gloomy aurora]

yup.

[last parrot]

Yeah i worked out like this

Alright thanks guys, much âppreciate!

.close

[devout snow]

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[jaunty moat]

I don't understand how to do c. I've gotten 0.03 but apparently that's wrong

[jaunty moat]

For context x = 0.14

[hardy trail]

Hi! If Vincent got a total greater than 6 points, what scores might he have gotten?

[jaunty moat]

P ( 4 n 3 ) + P ( 4 n 4 ) + P (3 n 4)

Which sums to 0.05

[hardy trail]

Good. Now, what is the question actually asking for?

[jaunty moat]

The probablity that he got a score of 4 first in the two games he played

Which is P ( 4 n 4 ) and P ( 4 n 3 )

[hardy trail]

Good, but remember specifically - it's the probability he got a score in the first game given that he got a score higher than 6.

[hardy trail]

So what do you think you need to do with this probability?

[jaunty moat]

Mutiply them together, that's what I did

Which got me 0.03

[hardy trail]

Right, but

P(4 n 4) + P(4 n 3) = 0.03
And
P( 4 n 3 ) + P( 4 n 4 ) + P(3 n 4) = 0.05,

[jaunty moat]

Yes

[hardy trail]

I guess to reframe my question in a different way; how do you find the probability of A given that B happens?

[topaz axle]

so it's 3/5=0.6

[jaunty moat]

Ohhhh, this is conditional probablity isn't it? I kinda forgot what to do

[hardy trail]

Yeahh exactly.

[jaunty moat]

It's been awhile since I've done probablity

[hardy trail]

Fair enough

But yeah - you are given that the total score is greater than 6; the word "given" should remind you of conditional probability.

[jaunty moat]

Alr tysm

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[wispy warren]

yo I'm stuck on this problem

[wispy warren]

we were taught to do something similar to this

but don't know what the point of that is

[last parrot]

Integration by part ?

Or partial fractions

[wispy warren]

partial fractions

[last parrot]

Yup either way

Factorise out denominator

[wispy warren]

okey one moment

x*2 + x - 2 = 0

[faint gorge]

no

D=b²-4ac=1-4(1)(-2)=9

[wispy warren]

what's that

[faint gorge]

The discriminant

Your sqrt(3) should be sqrt(9)

[wispy warren]

ah yes true

got it

thank you so much

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[silk panther]

hi, is this correct?

[silk panther]

when I wanna show lim f(x) implies lim f(1/x) thing

[trail eagle]

Note that provided $x\to\infty$ then $t=\frac{1}{x}$ already goes to $0^+$.

[woven radish]

Azyrashacorki

[trail eagle]

So in particular you don't really know that $\lim_{t\to 0} f\left(\frac{1}{t}\right) = L$.

[woven radish]

Azyrashacorki

[silk panther]

shouldn't t tend to 0 without the plus sign

[trail eagle]

It does but from the right side

[silk panther]

x -> inf then t = 1/x -> 0

why is it only 0+

[stone stump]

1/x is always positive

never negative

so it says nothing about approaching 0 from the left

[silk panther]

ah I get it

[trail eagle]

If anything it just means that the second half of you proof is unnecessary (and wrong).

[silk panther]

wait so do I need to prove anything now

do I just let t = 1/x

[stone stump]

well have you proved that you can just substitute values like that?

[silk panther]

wdym

why that needs proofing

[mint nebula]

An argument for this is always nice

[stone stump]

why can you just let t=1/x and plug in

why should that work

dont say obvious

I'm not sure how rigorous your course is. but this is something that you would have to prove in general

[silk panther]

well as long as x is not zero, I can safely let t = 1/x?

is it like that? Idk how else to prove

[faint gorge]

Just use εδ ig should be immediate

[silk panther]

I tried this but its confusing ngl

lemme send it

[faint gorge]

Yeah

[silk panther]

I am sure this needs some improvements

[faint gorge]

That should be it

[silk panther]

<@&268886789983436800>

[silk panther]

erm alr

but then idk if I really showed the t -> 0+ thing

[faint gorge]

Well you showed that lim x->0+ f(1/x)=L

[silk panther]

wat

[silk panther]

well showed?

[faint gorge]

bruh

[mint nebula]

You have showed that when t is sufficiently close to 0 from the right, the value f(1/t) can be made as close to L as we want (This holds for every epsilon > 0, so we can make epsilon arbitrarily small). This is the definition of the right-hand limit at 0.

[silk panther]

oooo is it because of the 0 < t?

keeping the t positive

[mint nebula]

Say, if you had 0<= t instead, then one could argue that 1/t can't hold, since you could end up with 1/0.

[silk panther]

ok ok understood

and I have another question actually, how many forms of limit definition are there? I am confused

[mint nebula]

Anyhow, the real reason for doing so it because we are approaching 0 from the right. t>0 means we stay on the positive side of 0, and t< delta means t get arbitratily close to 0.

[silk panther]

like there's the epsilon-delta ones, and infinite limit definition

[mint nebula]

I know this is not the exact answer you are looking for, but for calc, my best advice is to get comfortable with the epsilon-delta definition of the limit.

[silk panther]

yes I can tell im not used to it

because there are questions where the delta is the minimum of two things

that is confusing

why can't just pick one

and also, even after looking at the illustration, 5 mins later i will be damned

[mint nebula]

Sometimes, plotting this in geogebra or making a plot on paper helps with the intuition. Even my professor (supervisor) does this for some of the simple stuff, as it is easier to make sense of it that way

[silk panther]

yea thats why I said after visualizing it, I will get confused with the mathematical term again

and do u also understand why some delta is assigned to be the minimmum of two things?

[mint nebula]

Do you have something in mind, then we can talk about it?

[silk panther]

for the minimum thingy?

[mint nebula]

Yessir

[silk panther]

I think I remembered

lim x->3 of x^2 - 9

= 0

prove with epsilon delta

[mint nebula]

Can you show where you are stuck, then I can help you better understand

[silk panther]

alr ill write my thoughts

pls wait

[silk panther]

yes

[mint nebula]

One cannot simply set delta = epsilon / delta+6, since this is circular.

From |x^2 - 9| < delta (detlta+6), force delta <= 1, such that delta + 6 <= 7, so |x^2 - 9| < delta(delta+6) <= 7 delta.

[silk panther]

why u force delta to be 1

[mint nebula]

We need to control the factor |x+3|. After factoring, |x^2 - 9| = |x-3| |x+3|.

Here, we know that |x-3| becomes small when x->3, but |x+3| is still variable. So we force x to stay close to 3 by requiring |x-3| < 1

This is done by delta <= 1

[silk panther]

hmm fair

but what if delta is huge like 100

[mint nebula]

If you allowed delta = 100, then 0 < |x-3| < 100, would allow x anywhere in -97 < x < 103, and then |x+3| could be large, you we "loose control" over the factor x+3

Numerically, for delta = 100,

| x^2 - 9| < 100(100+6) = 10600

which is useless, if we choose a small epsilon, say epsilon = 0,01. Here we would need |x^2 - 9| < 0,01, and not < 10600

The whole idea is to be able to choose an epsilon arbitrarily small enough

[silk panther]

but it says for every epsion in the definition no? So epsilon can be big

[mint nebula]

But we want to show the distance is small between the terms

Okay, maybe not so ideal on a black background, but look. It would make no analysis sense to choose epsilon so far away

[silk panther]

why must it be small, as long as the actual limit and f(x) is inside the epsilon range, its true right

[mint nebula]

Yes, epsilon can be large or small, in fact the definition only requires epsilon > 0. But once epsilon is given, we get to choose a delta that works. Here, we choose a delta that is small on purpose; Not because epsilon is small, but because a small delta keeps x close to the 3, and makes the estimate work

[silk panther]

but why we need x close to 3

this might be dumb lol

[mint nebula]

The need x close to 3 because the definition of

lim_(x->3) (x^2 - 9) = 0, is about what happens when x approaches 3. Not what happens when we are far away from 3

[silk panther]

oooooooooooooooooooooo

[mint nebula]

If |x+3| = 10003, when |x^2-9| is far away from 0, which is not what we want to do analysis on, thus we restrict x to stay close to 3

[silk panther]

I understand it now

[mint nebula]

Great!

[silk panther]

ok so back to the x+3

what made you think that u should force delta to be 1? Why not smaller like 0.1

we want delta to be as small as possible here I believe

[mint nebula]

Just for convenience. We had |x^2 - 9| = |x-3| |x+3|, and need to control |x+3|.

My thought was to force x to lie in some small neighborhood of 3, where x+3 is bounded.

To do so, we choose a fixed radious around 3, i just choose 1, so |x-3|<1, then
2 < x < 4 => |x+3| < 7.

If we choose delta <= 0,1 then

|x-3| < 0,1
=> 2,9 < x < 3,1
=> 5,9 < x+3 < 6,1
=> |x+3| < 6,1

[silk panther]

u thought of forcing x to lie near 3 because of the |x+3| right? Else, do not need to do this

[mint nebula]

Yes! I only forced x near 3 because of |x+3|. If the factor was not there, then no need for that extra restriction

[silk panther]

ok ok, so then we will have delta = epsilon/7

so we choose delta as that?

[mint nebula]

That would make |x^2 - 9| < 7 delta = epsilon

But this gives us some trouble - can you spot why?

[silk panther]

erm

can I answer this if I visualize

[mint nebula]

Let me give you a clue:

We came to |x^2 - 9| < 7|x-3| under the assumption that |x-3|<1. This came from forcing x near 3 so that |x+3| <7.

So if you only choose delta = epsilon / 7, it only works when epsilon/7 <= 1, i.e.

espilon <= 7

[silk panther]

oooo so it only works for some epsilon

[mint nebula]

Right!

[silk panther]

so if epsilon is bigger than 7, everything falls

[mint nebula]

Yes - if we only choose that delta

So we have to get clever now, in choosing our delta

[silk panther]

wait a min, when u say epsilon/7 <= 1, why not epsilon/7 = 1

I mean the reason we got that expression is by equating 7*delta with epsilon

[mint nebula]

We don't need 7delta = epsilon, we only need 7delta <= epsilon.

Equality is unnecessary since anything smaller also works.

[silk panther]

ohh so when we get 7delta, we are actually doing 7delta <= epsilon?

[mint nebula]

Yup

[silk panther]

understandable, alr lets get back to the problem again

[silk panther]

how should we approach this

we know 7*delta <= epsilon

[mint nebula]

Well, for starters, let's list all the restrictions we have on delta as for now

[silk panther]

delta = 1

[mint nebula]

Not quite

[silk panther]

vro we forced delta to be 1

[mint nebula]

To be <= 1

So we have delta <= 1 and delta <= epsilon/7

[silk panther]

is the reason to capture every delta smaller than 1

[mint nebula]

Not to capture every smaller delta.

It is because we need our chosen delta to be a most 1, so that |x-3|<delta automatically implies |x-3|<1, which gives the bound |x+3|<7

[silk panther]

ahh i get it

u da goat

[silk panther]

yes

[mint nebula]

Here, I think that you can end the proof easily by yourself 🙂

[silk panther]

so delta can be at most epsilon/7 for evey epsilon <= 7

and for epsilon > 7, delta <= 1

oh so is that why they write min{1, epsilon/7}

[mint nebula]

Almost - it is not only for the epsilon <= 7.

delta <= epsilon / 7 holds for every epsilon > 0. This comes from wanting 7 delta <= epsilon, which is always true.

The extra condition was delta <= 1, so be need both delta <=1 and delta <= epsilon/7.

And yes, this is why we choose

delta = min(1, epsilon/7)

[silk panther]

why not just delta = 1

oh

im tripping

is it bcuz it wont capture 7*delta ,<= epsilon when epsilon < 1?

[mint nebula]

Yes

[silk panther]

i mean the inequality wont be satisfied

holy

i understand it now

i think need more practice for this one

to get used to this

[mint nebula]

We all started somewhere

[silk panther]

ok my last q for u

[mint nebula]

A fast one, I need to get on with my own work

[silk panther]

do u have any channel recomnendation for learning epsilon delta

thx for helping me all this time bro 🫡

[mint nebula]

Calculus

[silk panther]

what

[mint nebula]

Oh - like youtube?

[silk panther]

yee

or any source actually

website etc.

[mint nebula]

I think 3blue1brown has a good video about it

Else, i'd suggest googling some lecture notes on that

[silk panther]

alr tysm man

cya :blobcry

[mint nebula]

cheers

[silk panther]

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[gritty saddle]

Ive this question: im reading buragos book a course in metric geometry and im trying to find some length structure in R² which induce the Euclidean metric and there are some differences in some lengths of curves, but i couldn't find something. I tried the length structure induced by different metrics in R²,but nothing worked

[devout snow]

@gritty saddle Has your question been resolved?

[devout snow]

@gritty saddle Has your question been resolved?

[hasty cargo]

You can ping helpers now actually

[devout snow]

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[wispy warren]

hey I need help with this problem

[summer summit]

what have you tried so far?

[wispy warren]

but I'm stuck

[summer summit]

second line looks good so far

try factoring in the integral on the third line

bring the constant outside and then use polynomial division

[summer summit]

last sign should be a -

[wispy warren]

oh yes right

it looks a bit messy though

[summer summit]

very close

integration of that last term is not ln

think of your trig formulas

[wispy warren]

oh damn right

arctan(x^2 + 1)

I think

no

arctan(x)

[summer summit]

it is arctan but not x^2 + 1 in the inside

[summer summit]

yes

[wispy warren]

tysm for the help

that is the correct answer

.close

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[hollow laurel]

So I don't understand how to find f(0) as that would be to take the deriviative and I'm not sure how to take the deriviative of just 0 I thought it had to be on bounds.

[hollow laurel]

U freaking kidding? Someone ban this hacked account bro

[hasty cargo]

They did.

[hollow laurel]

ok....

[supple knot]

did you try to find the equation for f'(x) for x in [0, 4]

[hollow laurel]

Like point-slope form?

HOLY I'M GOING TO CRASH OUT

<@&268886789983436800>

[fair juniper]

point slope form only works for lines

[fair juniper]

you don't need to find the equation for part (a)

[hollow laurel]

Am i just setting up a integral and doing area under the curve from [0,5]?

[fair juniper]

do you remember FTC?

(fundamental theorem of calculus)

[hollow laurel]

Yeah when you do a deriviative of a anti-deriviative

I think

[fair juniper]

its a little more than that

specifically,

$\int_a^b f'(x) dx=f(b)-f(a)$