Given that $\int_a^b \abs{x}dx=\frac{1}{2}$ satisfies $a\leq0$ and $b\geq 0$, find the maximum and minimum areas of the plane region enclosed by the curve $y=x^2+ax$ and the line $y=bx$
#help-38
fresh pendant
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riyobi
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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
1
burnt mulch
This is two unrelated things clubbed together, so I’d start by breaking down the conditions
Let’s start with $\int^{b}_{a} |x| \dd{x}$
solid kilnBOT
Civil Service Pigeon
burnt mulch
So how would you set up the integral?
fresh pendant
$\int_0^{b-a} (-x^2 + (b - a)x)dx$
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riyobi
burnt mulch
```
What step are you on?
1. I don't know where to begin.
2. I have begun but go...
Ngl it seems like you weren’t 1 here
But anyway
burnt mulch
**riyobi**
Computing this explicitly, what’s the result?
fresh pendant
$\int_0^{b-a} -x^2 dx + \int_0^{b-a} (b - a)x dx=\left(\frac{2}{6} - \frac{1}{6}\right)(b - a)^3 = \frac{(b - a)^3}{6}$
solid kilnBOT
riyobi
burnt mulch
Yeah, so now we just need to find the max and min of $\frac{1}{6}(b-a)^3$ for $a^2+b^2=1$
solid kilnBOT
Civil Service Pigeon
burnt mulch
There’s a few ways you could go about this
But I’d think about it this way - if (…) is maximized, then so is ((b-a)^3)/6
What’s a “nice” expression you could put in the blank?
burnt mulch
Not rlly tbh
Yk how if you have distance, it’s often easier to find the max/min of the square of the distance instead?
So you can avoid the square root
burnt mulch
(b-a), but yeah
So we’ll just focus on (b-a)
However, I’m lazy, so I’m going to avoid dealing with the square roots if I can avoid it
There’s two main ways you can do this:
- Discriminants (hint: ||consider the line b-a=k and the circle a^2+b^2=1||)
- Consider (b-a)^2 = 1 - 2ab, then optimize ab using similar logic to what we did earlier
I probably have to go soon, but here’s two paths you can go on from here
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proper kernel
Can anyone else explains the second way here, how to optimize ab
might be wrong on this, but since a^2 + b^2 = 1, "ab" is usually at its most extreme when a = b
at those two corners are when ab is highest
at lower values of ab, you can see it can still be better
when ab is lowest, its just flipped
fresh pendant
Where can I learn these techniques systematically
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rugged canyon
I have been stuck on these clue things for at least two days because I have no idea what I'm doing or what I'm supposed to do
Completely forgot how these work over the summer so I've been using a bunch of websites and whatnot but it's kinda inefficient
rugged canyon
I got notes on how to solve these but I don't know where they are and my memory is really bad
I'm a very slow learner when it comes to math
rugged hull
(x+a)(x+b)=x^2+(a+b)x+ab
so for q 1
ab=30
a+b=-11
a=-5
b=-6
so (x-5)(x-6)=0
x=5,6
or you could also just put them directly into the quadratic formula
q4 uses the difference of squares a^2-b^2 = (a-b)(a+b)
@rugged canyon
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wraith hinge
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molten comet
Uh
vagrant prism
yes
molten comet
Why not leave the numerator as 1
wraith hinge
Why not leave the numerator as 1
i dont want it
molten comet
Okay
vagrant prism
in the end it's still x1y1 tho
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pearl lotus
The hint says
By fermats little theorem p(2ⁿ) is congruent p(2) mod n where n is prime
And so p(2)= 0
Can some one explain this to me
pearl lotus
.close
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pearl lotus
what was the solution?
Idk come to mathematical olympiad server there you can get it
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main sigil
How does the 2nd underlined inequality follow from the first one?
amber python
move denom over to the left
amber python
goes under the name of titu's lemma or sedraykan's inequality
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unique compass
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unique compass
How do I find asymptotes of F(K,L) but isoqusnt form
In implicit form
For the graph to the left
Which is written for a constant value Q= F(K,L)
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dreamy spire
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dreamy spire
i can find 3.384 using ASTC in the 3rd quardrant but cant find the other 3 points
wooden remnant
What is ASTC? And what are you searching for?
dreamy spire
im trying to find the points on the coordinate where y=0 the 4 points i ticked
dreamy spire
What is ASTC? And what are you searching for?
all sine tangent cosine?
wooden remnant
You solving this with a calculator? Or are you trying something exact?
dreamy spire
i can use calculator to solve this
but im still not sure how to solve with calculator
i can only find 1 value what if i want to find the other 3 values
tight drift
+π no?
dreamy spire
well the x value i get from calculator is -0.243 rad when i +pi it doesnt give me the ans
tight drift
Wait let me try
wooden remnant
You can visualize it for yourself with a unit circle
Ive written the angle theta here because this is the value your calculator is giving you (the negative value)
tight drift
Yeah sorry I was being dumb it’s +2πk cause periodic
wooden remnant
The other possible angle is the one on the other side
tight drift
Sine is a periodic function
Yeah 2pik is a full revolution around the unit circle which gives the same answer
The k is just how many solutions you want
dreamy spire
so a possivle value here wld be pi+theta?
tight drift
K=3 4th sol
dreamy spire
Yeah sorry I was being dumb it’s +2πk cause periodic
im not sure i understand?
tight drift
Basically
tight drift
If you look on the unit circle this person did
Say you have an angle of like 45° which is the answer to a trig question
Another solution to the same equation will be 45°+360°
Cause it’s the same angle just with an extra revolution around the unit circle
dreamy spire
oh
but lets say
i want the first + value of the qn here
its a whole revolution?
tight drift
So any solution to the equation would be 45°+k*360°
wooden remnant
tight drift
Wait did I say something wrong
wooden remnant
This is what you were looking for
dreamy spire
but wldnt theta +pi give 3.38 which yea
is the 3rd point
but if i wanted to find the 1st point
how wld i do it?
cause i cant seem to solve it no matter what i did
wooden remnant
Well trig sucks so you would have to fill in the formula for some points to find all points on your interval
Here wait a minute
wooden remnant
You would have to do this and then take the points located in the interval you were searching for
dreamy spire
ohh
dreamy spire
but lets say if they want u to prove
the point 1.29044 exist on the graph
where y=0
is it possible?
wooden remnant
Oh you could just put x=1.29044
wooden remnant
And if you fill in your oroginal gormula you would have to get y=0 then
If you would get a value different from 0 when filling in the formula the point is obviously somewhere else and it wouldnt be on the graph
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keen void
Could someone help me determine what my bounds should be ?
keen void
"Determine the value to the generelised integral ... where D is the area in the planes first quadrant limited by the lines y=x, y=2x and x=1"
"planes first quadrant"
What does this exactly mean? Are we in 3d space?
opal breach
This is a double integral so it's over a plane meaning it's 2D. First quadrant is the one where both x>0 and y>0. (The top right one)
keen void
ooh okay
that makes more sense...
i was trying to draw in 3d space..
so integrals and double integrals are in 2D and tripple integrals are in 3D space?
bleak birch
Yes, but I am not very familiar with the topic
It's covered in calc 3 (which I am just starting)
keen void
Okay so i'm trying to calculate the area D which is that triangle in the middle compact with orange, red and green
How do i determine the bounds?
bleak birch
Okay, so the bounds of integration must be from 0 to 1
And the area of the triagle
Is going to be determined by the integral of 2x-x
I can't use LaTeX well
keen void
Hmm oki
I'll see if we got any pros
<@&286206848099549185> We got any Calc 3 geniuses online? 
ancient flicker
yo
keen void
I need some quick help in understanding how to determine what bounds i will use on my integral
wooden remnant
Yup x bound 0 to 1
y bound x to 2x
keen void
Aaah crap okay that makes sense
wooden remnant
If I am not mistaken calc 3 was a while ago for me
wooden remnant
I assumed this was your area D
First the y bound so you get a eq with only x
Then you have just a standard integral from 0 to 1 wrt x
bleak birch
Could someone help me determine what my bounds should be ?
The entire double integral seems like a calc 3 problem
wooden remnant
If you do other way around youll see soon enough it wont work because youll get variables in your final eq
bleak birch
But this part is just integral calculus
keen void
$\int_0^1 \int_{x}^{2x} \frac{1}{\sqrt[3]{xy}}dydx$
wooden remnant
No
keen void
This is not allowed?
keen void
Ok. Does the boundary for x where it goes from 0 to 1 correlate to my dx?
Which one is correct? Left or right?
wooden remnant
The int 0^1 …dx is one part and the int x^2x …dy is another
You need to see them as separate operations
bleak birch
$\int_{0}^{1} (2x-x)dx$
wooden remnant
Which one is correct? Left or right?
Right
solid kilnBOT
Merineth
wooden remnant
**Merineth**
This means youll fiest integrate wirh respect to y, and then fill in the boundsss y=2x and y=x
keen void
$\int_{0}^{1} (2x-x)dx$
Can you remove this pls
keen void
This means youll fiest integrate wirh respect to y, and then fill in the boundss...
Yeah that makes sense
another quick question tho
wooden remnant
Afterwards youll get new function which will then be integrated from just 0 to 1 wrt x
keen void
$\int_0^1 \int_{x}^{2x} \frac{1}{\sqrt[3]{xy}}dydx = \int_{x}^{2x} \int_0^1 \frac{1}{\sqrt[3]{xy}}dxdy$
fuck me
solid kilnBOT
Merineth
wooden remnant
Yesssssss
keen void
There we go
wooden remnant
That is correct
keen void
That is fine?
wooden remnant
But when youll integrate now youll see it wont work
Because your final integral will have x variables in it
Just try both versions
keen void
Ah right so the trick is to determine which version i should work with first?
wooden remnant
Soonest way too see what i mean and youll learn so you wont make the mistake later
wooden remnant
Ah right so the trick is to determine which version i should work with first?
Yes
keen void
Okay hold on i'll try and integrate
oufh this is a tough one to integrate
I start with
$\int_0^1 (xy)^{-1/3}dx$
solid kilnBOT
Merineth
keen void
so i have to first add 3/3 to the exponent and determine what should be infront of my 
wooden remnant
When you are integrating wrt x just look at other variables at constants
So you can take y out
nova spire
wait there's a small problem
nova spire
**Merineth**
you're not allowed to do that
the bounds of integration on y depend on x
so you can't swap them
nova spire
wait when
keen void
So the bounds of x which is 0 to 1 should be integerated with dy?
bleak birch
**Merineth**
I got lost when this started happening
wooden remnant
It is fine @nova spire it just wont work
nova spire
It is fine <@246787399991033858> it just wont work
then it's not fine
nova spire
**Merineth**
because this isn't true
the final result shouldn't be depending on either x or y
wooden remnant
Thats what i mean with it wont work
keen void
So what is actually correct then?
nova spire
so don't let Merineth writing something untrue without saying it xd
nova spire
btw quick side talk, how did your test go Merineth?
keen void
It went okay ish, i try not to think about it since i don't have time to cry
I'll inform of how it went when it's graded
nova spire
alr alr
keen void
**Merineth**
So what exactly is wrong with this?
nova spire
So what exactly is wrong with this?
if you integrate with respect to dx first
then you should lose the dependence in x
keen void
nova spire
the problem is the bounds in y still depend on x
so you can't start with integrating dx
maiden hare
If you wanted to integrate wrt to x first, you'd need to write something like this:
$\int_0^2 \int_{\frac{y}{2}}^{min(1,y)} \frac{1}{\sqrt[3]{xy}} , dx , dy$
solid kilnBOT
Nel
maiden hare
Just swapping things doesn't work
nova spire
you're integrating with respect to a very particular domain in R^2
it's a fixed domain
maiden hare
Computing the thing on the right here yields a function of x, not a value, and certainly not the value you're looking for
keen void
Okay so when i'm given a function, i shouldn't swap the order of the dx dy?
wooden remnant
This is what i meant
nova spire
This is what i meant
a,b,c,d are constants
maiden hare
This is what i meant
This is not the case here
wooden remnant
See the note
nova spire
sure
nova spire
That is correct
but um
wooden remnant
You would have to change the bounds that is why it wont work but yes this is the case here
keen void
Isn't that what Nel did?
nova spire
don't go say untrue things xdd
wooden remnant
Isn't that what Nel did?
Yes
maiden hare
Saying this is fine to do without mentioning that the bounds need to change is just not helpful
nova spire
Isn't that what Nel did?
here, the domain on which we're integrating is 0<=x <= 1, x<= y <= 2x
if you wanted to integrate with respect to x first
wooden remnant
I tried to say it by telling him it wont work and he’ll see when he tries to integrate it but yeah sry
nova spire
you need to do the opposite instead
keen void
Just to clarify. The function and bounds would look something like this:
$\int_{x}^{2x} \int_0^1 \frac{1}{\sqrt[3]{xy}}dxdy$
solid kilnBOT
Merineth
nova spire
$\int_{x}^{2x} \int_0^1 \frac{1}{\sqrt[3]{xy}}dxdy$
if you're integrating with respect to x first, it's the bounds of x that are a function of y
and the bounds of y must be constant
wooden remnant
**Merineth**
Other way around
nova spire
which means you would need to rewrite "0<=x <= 1, x<= y <= 2x"
into "constant <= y <= constant, f(y) <= x <= g(y)"
keen void
$\int \int$ function $dydx$
I'm completely new to double integrals however.
Should integrals normally be like this:
solid kilnBOT
Merineth
keen void
Does the order of the dydx / dxdy matter?
wooden remnant
This is what i meant
.
nova spire
Does the order of the dydx / dxdy matter?
depends if you wrote it as $\int_{x=...}^{x=...} \int_{y=...}^{y=...} $ function $dydx$ or the opposite
solid kilnBOT
rafilou2003
nova spire
if the "outer" integral is bounds of x, you integrate with dx LAST
$\int_{x=...}^{x=...} (\int_{y=...}^{y=...} $ function $dy)dx$
solid kilnBOT
rafilou2003
wooden remnant
Have you ever done like double integrals over a square or something? So just constants not functions as bounds? @keen void
keen void
Interesting. So you are saying that if i'm integrating wrt dx then it should be x = ...
But if i'm doing integrating wrt dy then it should be y = ...
maiden hare
This is your region.
If you start with x, you go from 0 to 1 on the x axis, and compute the corresponding y values for the inner bounds (so y=x as lower bound and y=2x as upper bound). Then the dy comes first because the inner bounds are for y, and after that comes the dx.
If you start with y, you go from 0 to 2 on the y axis, and compute the corresponding x values for the inner bounds (so x=y/2 as lower bound and x=min(y,2) as upper bound). Then the dx comes first because the inner bounds are for x.
nova spire
$D = {(x,y) | 0\leq x \leq 1, x\leq y \leq 2x}$
solid kilnBOT
rafilou2003
keen void
Interesting. So you are saying that if i'm integrating wrt dx then it should be ...
So based on that Nel, this should be reversed?
maiden hare
Note that without the min in x=min(y,2) (so if you just wrote x=y as upper bound), you would be integrating over this region instead:
maiden hare
So based on that Nel, this should be reversed?
No, what you said is right, if the inner integral has bounds like x=..., you want the dx first
keen void
Ahh okay so a good rule of thumb is when integrating wrt dx i do x = ... but when integrating wrt dy i do y = ...
Since i want their appropriate values?
maiden hare
Imagine parentheses around the inner integral:
$\int_{x=0}^1 \left( \int_{y=x}^{2x} \frac{1}{\sqrt[3]{xy}} , dy \right) , dx$
solid kilnBOT
Nel
nova spire
How you compute an integral depends on how you defined your domain.
You can either first find the range of all possible values of x, and then find the possible values of y IN TERMS of x.
Or doing the opposite, find the range of all possible values of y, and then find the possible values of x IN TERMS of y
keen void
Wait hold on this is too much haha
wooden remnant
Just try integrating it both ways it is the fastest way to see what is right and wrong
maiden hare
The other way around would be
$\int_{y=0}^2 \left( \int_{x=\frac{y}{2}}^{min(1,y)} \frac{1}{\sqrt[3]{xy}} , dx \right) , dy$
solid kilnBOT
Nel
maiden hare
(what I wrote earlier, just with parentheses and specifying the variable in the bounds)
solid kilnBOT
Merineth
keen void
I was given this function
and i was given this graph (by drawing out the lines)
So i have to determine the bounds of my integrals before i can proceed
so i have to determine the dx one first
since that's my innermost integral
nova spire
it's the opposite
wooden remnant
so i have to determine the dx one first
You need to determine whichever you find easiest and your integral is based on what you chose
maiden hare
You can determine either one first, it's your choice
keen void
Ok
maiden hare
But one way is easier than the other
keen void
Well if i wanted to determine the bounds for x
i can clearly see that it goes from 0 to 1
nova spire
so you determined that first
maiden hare
Ok yes, but that would be the outer integral
nova spire
meaning that it's gonna be the outer integral
keen void
for dy?
maiden hare
No, for dx
keen void
**Merineth**
But dx is my inner integral
How can it be my outer integral and apply to dx?
maiden hare
It isn't
maiden hare
I mean it can be, but for the easier way, it isn't
keen void
$\int_0^1 \int \frac{1}{\sqrt[3]{xy}}dxdy$
solid kilnBOT
Merineth
keen void
This is what you are saying?
maiden hare
**Merineth**
This is not accurate
keen void
This is not accurate
That is literally the integral i was given in the assignment
maiden hare
$\iint_R \frac{1}{\sqrt[3]{xy}} dR$
solid kilnBOT
Nel
keen void
maiden hare
Yes the main difference is the D below the integrals
It means it's not two integrals, it's one double integral over a region
maiden hare
No
keen void
huh
wooden remnant
Can we start over from a easier integral and bound for practice? because this isn’t going anywhere
Like just an area integral for a square
maiden hare
That dxdy in this case is really just dD
keen void
So why does he write dxdy?
maiden hare
dD can be dxdy or dydx, depending on which way you choose
maiden hare
So why does he write dxdy?
Because it's not an issue when writing it as one double integral
and maybe he didn't want to write dD 
keen void
okay so i should always just ignore the dxdy when given double integrals
and assume it's dD
wooden remnant
and assume it's dD
It is the same
keen void
Clearly not?
maiden hare
and assume it's dD
Pretty much yes
wooden remnant
Writing convention
maiden hare
It's the same in a double integral, it's not the same after you split it into two integrals
keen void
Ok so i determined that the x bounds for my function goes from 0 to 1 so i have :
$\int_0^1 \int \frac{1}{\sqrt[3]{xy}}dydx$
solid kilnBOT
Merineth
keen void
Is there a reason it gets placed as the outermode bound?
Or does every bound that i find first become outerbound?
maiden hare
Or does every bound that i find first become outerbound?
I mean kinda
nova spire
kinda yea
maiden hare
If the bounds of the outer integral are for x, the dx is at the end
maiden hare
Again, imagine parentheses around the inner one
wooden remnant
You want the bound not dependent on variables on the outside
keen void
so we want it in terms of y = ...
And we have two functions that "hold" the triangle together
y = 2x and y = x, right?
$\int_0^1 \int_{x}^{2x} \frac{1}{\sqrt[3]{xy}}dydx$
So i'm left with this?
wooden remnant
You switched the lower and upper bound
nova spire
the lower bound for y is x
solid kilnBOT
Merineth
keen void
Yeah of course, my bad
nova spire
there we go great
keen void
NIce
<3
Okay so this is jsut solved normally
like Nel explained with the
and start with dy?
nova spire
so you can see this as
$\int_0^1 \left(\int_{x}^{2x} \frac{1}{\sqrt[3]{xy}}dydx\right)$
solid kilnBOT
rafilou2003
keen void
$\int_{x}^{2x} \frac{1}{\sqrt[3]{xy}}dy$
solid kilnBOT
Merineth
keen void
So i start here with this integral
nova spire
great
solid kilnBOT
Merineth
nova spire
yeah you can notice the primitive more easily
but this is an integral with respect to y
so using some power rule, there's an even easier way to look at this integral
keen void
$\frac{3(xy)^{2/3}}{2}\frac{xy^2}{2}$
solid kilnBOT
Merineth
keen void
Seems ok?
nova spire
well y^2 = 2y when derived
where is y^2 mentioned?
keen void
well
wooden remnant
where is y^2 mentioned?
Also my question
wooden remnant
Well the first part yes
And nvm first part also raises question
keen void
the integral of y = y^2/2
nova spire
the integral of `y` = `y^2/2`
I think you're taking the wrong approach
nova spire
so using some power rule, there's an even easier way to look at this integral
.
(xy)^a = ?
wooden remnant
Just take x out of the integral if it makes it too confusing
keen void
a increases by 1
wooden remnant
x can be looked at as a constant
keen void
and then i have to make sure that when i derive the original problem that when a "falls down" it disappears
$x^2 = \frac{x^3}{3}$
solid kilnBOT
Merineth
nova spire
the integral of `y` = `y^2/2`
just because (f o g)' = g' * (f' o g)
doesn't mean that when you're trying to integrate f(g(y))
you have to integrate g
and integrate f
wooden remnant
I think she means rhe integral of rafilou
keen void
I am confident i know integrals
wooden remnant
The integrak of x^2 = x^3/3
nova spire
tho let's go for an extreme case
integral of (2x)^2
"well i have to integrate the outer function"
(2x)^3/3"
and then integrate the inner function
"x^2"
wooden remnant
I am confident i know integrals
I’m not confident you do tho haha
nova spire
"so x^2 * (2x)^3/3"
nova spire
integral of (2x)^2
btw this 4x^2
nova spire
"so x^2 * (2x)^3/3"
which basically got integrated into ... x^5 :(
wooden remnant
I’m not confident you do tho haha
Haha sry
nova spire
btw this 4x^2
what is it Nel?
maiden hare
Integral of (2x)^2 is 4x^3/3
nova spire
Integral of (2x)^2 is 4x^3/3
... that's what I was saying
the way of doing it is wrong xdd
keen void
$(xy)^{-1/3}$
solid kilnBOT
Merineth
nova spire
yes
maiden hare
Oh ok, didn't understand what you meant
keen void
First i add 3/3 to my exponent
this is how i do it
ok
Since when i derive something like that instead of doing -1 to exponent in integrals i add +1
this results in
$(xy)^{2/3}$
Ok?
solid kilnBOT
Merineth
keen void
HOWEVER
wooden remnant
Okay I am completely finished I’ll integrate it myself
wooden remnant
keen void
BRUH
STO
P
PSOTING
LIKE
I
DON
TWANT
YOU
ANSWER
Mygod it's actually so infuriating
nova spire
let's let merineth finish first pls
keen void
$3\frac{(xy)^{2/3}}{2}$
solid kilnBOT
Merineth
keen void
Such that i'm left with this
When the 2/3 fall down, it will cancel out the 3/2 leaving 1
nova spire
is that the final answer?
nova spire
alr
keen void
Lemme think, one moment
Shit i just remembered there is another way to do it
Can't i factor out x first?
nova spire
(xy)^a = ?
alright so
keen void
x^a*y^a
keen void
$x^{-1/3} \int_x^{2x} y^{-1/3}dy$
nova spire
wrong bounds
solid kilnBOT
Merineth
nova spire
alr great
keen void
$x^{-1/3} \int_x^{2x} y^{-1/3}dy = [x^{-1/3} * \frac{3y^{2/3}}{2}]_{y=x}^{y = 2x}$
maiden hare
No the square brackets are around the integrated part, not the constant x^{-1/3}
wooden remnant
No the square brackets are around the integrated part, not the constant x^{-1/3}
Is the same (as long as she remembers the bounds are for the variable y)
Could be written down more clearly by specifying the bounds as y = 2x and y = x
solid kilnBOT
Merineth
maiden hare
Yeah ok, I'm just getting tired
nova spire
it's this way around
keen void
Ok will calc
one moment
$=x^{-1/3}\frac{3(2x)^{2/3}}{2} - (x^{-1/3}\frac{3x^{2/3}}{2}) =$
solid kilnBOT
Merineth
keen void
x^(-1/3) cancel out
maiden hare
cancel out?
wooden remnant
x^(-1/3) cancel out
Uhh can you specify how you would do that?
nova spire
she maybe meant cancel out/get absorbed with x^(2/3)
keen void
$=x^{-1/3}\frac{3(2x)^{2/3}}{2} - x^{-1/3}\frac{3x^{2/3}}{2} =$
solid kilnBOT
Merineth
nova spire
she maybe meant cancel out/get absorbed with x^(2/3)
so yeah you can either do that first
or factor
keen void
Yeah i'm thinking of factoring it immediately instead of expanding
maiden hare
Whatever works for you
keen void
$=\frac{3x^{2/3}}{2} (2x^{-1/3}-x^{-1/3})$
solid kilnBOT
Merineth
keen void
Uhm is it the -1/3 that's the problem?
nova spire
it's the 2 in 2x^(-1/3) that's the problem
also wait you kinda got the powers the other way around right?
didn't you mean to write $=\frac{3x^{-1/3}}{2} ((2x)^{2/3}-x^{2/3})$?
solid kilnBOT
rafilou2003
keen void
Nah i'm doinglike nel said and expaninddg first..
$=x^{-1/3}\frac{3(2x)^{2/3}}{2} - x^{-1/3}\frac{3x^{2/3}}{2} =$
solid kilnBOT
Merineth
keen void
both the 2 and x gets the 2/3. right?
nova spire
yeah
keen void
$=x^{-1/3}\frac{3*2^{2/3}x^{2/3}}{2} - x^{-1/3}\frac{3x^{2/3}}{2} =$
solid kilnBOT
Merineth
solid kilnBOT
Merineth
maiden hare
That's fine but the x^{-1/3} can also be factored out
solid kilnBOT
Merineth
nova spire
alr and one last simplification step
keen void
change back the x^-1/3 into 1/cuberootx?
wooden remnant
x^a * x^b = ???
keen void
$=\frac{3x^{1/3}}{2}(2^{2/3}-1)$
solid kilnBOT
Merineth
solid kilnBOT
rafilou2003
nova spire
I'll let you try out the rest
nova spire
holy crap integrals are exhausting xD
ikr, a lot of detail
keen void
Me personally if i wanted to integrate that
I would expand again
and make it into two different integrals
Does that seem fair?
wooden remnant
Can be done but a lot of work
keen void
Seems easier than trying to work with product rule?
nova spire
and make it into two different integrals
you can, but it requires change of variables first
wooden remnant
Easier to just take out a constant term
nova spire
as the two integrals are not independent yet
(the bounds of the inner are dependent of the outer)
keen void
the variable being 3/2?
wooden remnant
What do you mean @nova spire?
nova spire
**rafilou2003**
Me personally if i wanted to integrate that
wait are you talking about this integral?
or the original one
keen void
okay no i see what you mean raf
nova spire
I was talking about the original dxdy form
wooden remnant
I was talking about the original dxdy form
Aaah i thought the final one we arrived at
That clears it up at least for me
keen void
$\frac{3}{2}(2^{2/3}-1) \int_0^1 x^{1/3}dx$
solid kilnBOT
Merineth
nova spire
$\frac{3}{2}(2^{2/3}-1) \int_0^1 x^{1/3}dx$
yeah perfect
keen void
$\frac{3}{2}(2^{2/3}-1) \int_0^1 x^{1/3}dx = [\frac{3}{2}(2^{2/3}-1) * \frac{3x^{4/3}}{4}]_0^1$
solid kilnBOT
Merineth
wooden remnant
Yesss
Now I personally would take the bounds just over the part with x in it and leave the constant term for later to avoid a lot of algebra but both will work
Or you could immediatly see the result from the 1 and 0 bound but that also depends on you
nova spire
$\frac{3}{2}(2^{2/3}-1) \int_0^1 x^{1/3}dx = [\frac{3}{2}(2^{2/3}-1) * \frac{3x^...
whenever you put something in factor of the integral, you can keep it in factor of the [...] version
don't feel obligated to put it back in everytime as sometimes the factored form is easier
so something like $\frac{3}{2}(2^{2/3}-1) * [\frac{3x^{4/3}}{4}]_0^1$
solid kilnBOT
rafilou2003
nova spire
if you see what I mean
sleek canopy
i see
keen void
I did it on paper and got
sleek canopy
where did you come from xd
ops
keen void
$\frac{3^{7/3}(2^{2/3}-1)}{8}$
solid kilnBOT
Merineth
nova spire
$\frac{3^{7/3}(2^{2/3}-1)}{8}$
where did this fractional power of 3 come from?
keen void
Hm
nova spire
i think you thought (3x)^4/3
keen void
$\frac{3}{2}(2^{2/3}-1)\frac{3^{4/3}}{4}$
solid kilnBOT
Merineth
nova spire
yeah ok that's what happened
keen void
is that wrong?
nova spire
recall
keen void
4*2 = 8
keen void
nova spire
alr
keen void
$\frac{3}{2}(2^{2/3}-1)\frac{3(1)^{4/3}}{4}$
solid kilnBOT
Merineth
nova spire
yep great
keen void
So it's just 3/4?
nova spire
yeah the term on the right is 3/4
solid kilnBOT
Merineth
nova spire
congrats!
wooden remnant
🥳
nova spire
ok wait do you want to know how else we can do this?
keen void
I remember @whole coral saying that she learned calc 3 with great difficulty. I can soooooooooooooooooo relate to that
nova spire
alr enough for today huh xd
keen void
Shower + food + rest
My calc 3 exam is tomorrow, i wont pass but i thought i could aswell learn double integrals
Since i'd assume tripple integrals works fairly similarly
keen void
What about 8 and 10
nova spire
(when the bounds are already specified)
nova spire
oh that looks like a line integral and surface integral
great
well I always transform this into double and triple integral respectively
keen void
"determine counterclockwise circulation"
Do you think i have enough time tonight to learn it? xD
nova spire
oh certainly it's simple
nova spire
well now you know double integrals
wooden remnant
Do you think i have enough time tonight to learn it? xD
Youre cooked
nova spire
(triple integral is the same)
nova spire
Will you be on in like an hour raf?
yeah I'm chronically online don't worry xd
DM ping me if I don't respond
keen void
Alright i'm gonna take a break for an hour roughly to process what i jsut did
will be back soon
@maiden hare @nova spire @wooden remnant
thanks for the help
.close
dreamy loom
A TV with an effect of 160 W is on for 3 hours. Calculate the usage of electric energy in both kWh and J.
dreamy loom
My book did it in an earlier one
vagrant marsh
can you show that one pkease
dreamy loom
They did 100W * 24 * 3600s = 8,64MJ
Oh that’s the joules
So I just do 160W * 3h = 480kWh
No
vagrant marsh
remember that 1Ws = 1J
dreamy loom
0,48kWh
vagrant marsh
yeah
vagrant marsh
looks good
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keen void
Determine the counterclockwise circulation where r is the edge to the circle sector
keen void
Counld someone teach me how these circulations work? C:
My assumption is that it has something to do with integration
@nova spire hi :D
Does it have something to do with this?
nova spire
so from that you just have to find Q'x and P'y
and if you remember double integrals you can try to parametrize "dxdy"
here polar coordinates are most appropriate
keen void
I'm not sure what P'_y and Q'_x is
mighty prism
,, \int_{\gamma} F(r) \cdot dr = \int^b_a F(\varphi(t)) \cdot \varphi'(t) dt
solid kilnBOT
Bair
keen void
ok um not sure what that is
keep in mind i know nothing
I've tried looking up videos on it for an hour but their all dogshit per usual
:(
nova spire
I'm not sure what P'_y and Q'_x is
F = [P Q]
mighty prism
dude you don't need to apply green's theorem
nova spire
dude you don't need to apply green's theorem
different possibilities
mighty prism
you would need to compute the integral over the entire boundary, and over the region inside
keen void
So P and Q are just functions derived wrt x and y?
nova spire
you would need to compute the integral over the entire boundary, and over the re...
it's an integral over the domain
that's stokes theorem
keen void
Makes sense
nova spire
you can guess what P and Q are in your exercise?
mighty prism
that's stokes theorem
what are you talking about? this is a 2d line integral
keen void
would that mean our x boundary is from 0 to 1?
nova spire
what are you talking about? this is a 2d line integral
closed curve flux integral
by stokes theorem
we can write this as a double integral
mighty prism
to apply stokes theorem you need to F to be a field in R3
nova spire
this theorem
keen void
The solution provided by the teacher is solved using Greens formula
mighty prism
yeah you can do that, but it's extra work
nova spire
exactly this is used
mighty prism
idk what this guy is talking about with stokes theorem
nova spire
yeah you can do that, but it's extra work
not really
nova spire
idk what this guy is talking about with stokes theorem
green is kinda a consequence of stokes
anyways
keen void
So P is the function derived wrt y and Q is derived wrt x?
nova spire
So P is the function derived wrt y and Q is derived wrt x?
P is the first coordinate function
P'_y is that function derived wrt y
nova spire
$\textbf{F}(x,y) = \begin{pmatrix}P(x,y)\Q(x,y)\end{pmatrix}$
solid kilnBOT
rafilou2003
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rafilou2003
mighty prism
here F = (-y, x) so P = -y, and Q = x
keen void
oooh ok
nova spire
$P'_y = \frac{\partial P}{\partial y}$
solid kilnBOT
rafilou2003
nova spire
yes
keen void
Ok so i just apply the formula?
and determine the bounds
Well first of all lets determine Q'_x and P'_y
nova spire
Does it have something to do with this?
alright first write that last double integral
we still don't know exactly what Omega is
but we'll find out in due time
keen void
Um
nova spire
Well first of all lets determine Q'_x and P'_y
yes so great
nova spire
that's not what we're asking for tho
nova spire
Well first of all lets determine Q'_x and P'_y
no we're doing this
nova spire
Shouldn't P'_x and Q'_y just be [0,0] ?
it's just I think you got confused a bit here
nova spire
yeah
keen void
Oh ok
I was looking at the third one
So i assumed Q was first and P was second
If that's the case
P = -y
Q = x
P'_y = -1
Q'_x = 1
$(Q'_x-P'_y) = (1 -(-1)) = (1+1) = 2$
solid kilnBOT
Merineth
nova spire
great
mighty prism
yeah so now you integrate that over omega
nova spire
so you're integrating 2 over Omega
keen void
Oki so our function is just 2 ? xD
keen void
Fair enough 
mighty prism
ok that was indeed easier, i thought gamma was just the circular part
also you were referring to generalized stokes' earlier, not regular stokes'
keen void
Well my guess would be
nova spire
The boundary has to be 0 to 1 for dx
ah we don't even have to worry about that
keen void
not entirely sure what dy is tho
keen void
oh
I don't really get that sentence
We are computing 2 omega?
I thought we were creating our double integral?
nova spire
I thought we were creating our double integral?
We could
But the easy trick is
integral(constant) over a 2D region is constant*area
keen void
Nooo this is not easy at all
There's no way i'll learn it tonight
It's too much new stuff
mighty prism
the only new thing is green's theorem
just remember to swap P and Q when differentiating
solid kilnBOT
Merineth
keen void
So we have this? (idk how to make omega under double integral)
mighty prism
what is omega
keen void
huh
mighty prism
so the region bounded by gamma is omega
nova spire
partial Omega is boundary
keen void
is omega
mighty prism
in the original question
nova spire
keen void
oh
mighty prism
,, \partial \Omega = \gamma
solid kilnBOT
Bair
mighty prism
yeah green's theorem relates an integral on the boundary of a region to an integral on the region itself
so we're going from $\partial \Omega$ to $\Omega$
solid kilnBOT
Bair
mighty prism
mighty prism
the integral on the left is equal to the integral on the right
mighty prism
one is an integral on the boundary of a region and the other is an integral on the region itself
hence the double integral
keen void
I can't understand what you mean
"one is an integral on the boundary of a region"
what is that
mighty prism
do you know what a line integral is
keen void
I know what an integral is
mighty prism
ok well intuitively what it is is we are splitting up a curve into a bunch of small pieces and evaluating the function F at each of them and then adding all of those small pieces together
like an integral does for a function
mighty prism
yeah gamma
keen void
I assumed the green here is a normal integral?
dusty sleet
Wdym?
nova spire
the original integral is over the red part
keen void
HAHAHA
nova spire
we transform it into an integral over the green domain
keen void
I don't think i've been more confused since i joined xD
dusty sleet
How the heck are you supposed to solve this exercise if you don't know the very basics?😭
Did your teacher by chance explained how to solve this kind of exercises?
dusty sleet
Ahnn now I see
keen void
Got re-exam tomorrow and raf said it was easy to learn since i learnt how to do double integrals relatively quickly. So we assumed i'd be able to handle this also.
But that clearly isn't the case.
But yeah anyway i'll just drop the calc 3 for now
and retry during the semester
thanks tho.
.close
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nova spire
thanks tho.
sorry I wasn't that much available for now
nova spire
I can talk more about this in a couple of hours sry
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