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School work help

posted
Total Posts
52

Elementary, middle, or high school? (For the Australians and anyone else who uses this system)

Elementary School (year lvl prep-6)
0
0.00%
Middle school (year level 7-9)
3
8.57%
High school(year level 10-12)
16
45.71%
Something else
16
45.71%
Total votes: 35
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1
For anyone needing help with school work post what they need help with in here and hopefully someone will help.
RULES

1. Don't make fun of people for their trouble, everyone is at various degrees of challenges.

2. Only post if you genuinely having trouble, don't just post all of your work and expect people to give you answers.

3. When helping someone try to explain the resolution instead of just giving short responses,(basically, be helpful)

I hope this gets used, but i don't really mind if it doesn't, just trying to give people help if they are struggling, mostly because me, as ab Australian, really don't get much homework, but ik student over seas get bombarded with school work, and can lead to many mental issues and self doubt, anyway hope this helps:)
Patatitta
1,244 posts
Joined February 2020
Patatitta
i'm breaking all the rules just to say, if someone votes that they're in elementary school and drop-kicking them
Reyalp51
962 posts
Joined April 2021
Reyalp51
we had a 9 yo on the forums once but never a 5 yo, lets try to break the record
Dementedjet
236 posts
Joined December 2020
Dementedjet

Reyalp51 wrote:

we had a 9 yo on the forums once but never a 5 yo, lets try to break the record
Solution = 🚽😀

But on a thread, middle school was best for me and probably will remain that way
Corne2Plum3
Global Moderator
1,766 posts
Joined November 2019
Corne2Plum3
There isn't my school grade here lol
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

Corne2Plum3 wrote:

There isn't my school grade here lol
What grade u in?
Jangsoodlor
Jangsoodlor
Can anyone help me find ∂f/∂x of f(x,y,z) = (z(x^2))(ycos(z) + (x^2)y) ?
Last edited by Jangsoodlor , edited 3 times in total.
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1
To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:

\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]

Using the product rule, we have:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]

Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:

\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]

\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]

Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]

This would be the expression for \( \frac{\partial f}{\partial x} \).
sametdze
339 posts
Joined July 2021
sametdze

Jebus1 wrote:

To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:

\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]

Using the product rule, we have:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]

Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:

\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]

\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]

Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]

This would be the expression for \( \frac{\partial f}{\partial x} \).
did brother use chatgpt for this or did he actually work it out
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

sametdze wrote:

Jebus1 wrote:

To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:

\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]

Using the product rule, we have:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]

Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:

\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]

\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]

Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]

This would be the expression for \( \frac{\partial f}{\partial x} \).
did brother use chatgpt for this or did he actually work it out
I started working it out, got like halfway, and then just thought it'd be easier to get chatgpt to explain it 🤣

And tbh it did a pretty good job of explaining 😅
Last edited by Jebus1 , edited 1 time in total.
Jangsoodlor
Jangsoodlor

Jebus1 wrote:

sametdze wrote:

Jebus1 wrote:

To find ∂f/∂x of the function \( f(x,y,z) = z(x^2)(ycos(z) + (x^2)y) \), we'll use the product rule and the chain rule. Let's break it down:

\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} [z(x^2)(ycos(z) + (x^2)y)] \]

Using the product rule, we have:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(\frac{\partial}{\partial x}[ycos(z) + (x^2)y]) \]

Now, let's differentiate \( ycos(z) + (x^2)y \) with respect to x using the chain rule:

\[ \frac{\partial}{\partial x} [ycos(z) + (x^2)y] = y\frac{\partial}{\partial x}[cos(z)] + y\frac{\partial}{\partial x}[(x^2)] + (2xy) \]

\[ = -ysin(z)\frac{\partial z}{\partial x} + 2xy \]

Now, substituting this back into the expression for \( \frac{\partial f}{\partial x} \), we get:

\[ \frac{\partial f}{\partial x} = z(2x)(ycos(z) + (x^2)y) + z(x^2)(-ysin(z)\frac{\partial z}{\partial x} + 2xy) \]

This would be the expression for \( \frac{\partial f}{\partial x} \).
did brother use chatgpt for this or did he actually work it out
I started working it out, got like halfway, and then just thought it'd be easier to get chatgpt to explain it 🤣

And tbh it did a pretty good job of explaining 😅

ChatGPT usually failed to calculate maths correctly. Here is the example and Anoter example. Notice the inconsistencies of its answer.

the actual answer (if my calculation is correct) is zx^2(2xy) + (ycosz+yx^2)2xz
Last edited by Jangsoodlor , edited 1 time in total.
[-Omni-]
960 posts
Joined July 2021
[-Omni-]
for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
anaxii
15,000 posts
Joined December 2022
anaxii
apparently if i base myself on the australian school system, i'm in "senior high school" (i'm in 11th grade)
MistressRemilia
157 posts
Joined January 2024
MistressRemilia
I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
anaxii
15,000 posts
Joined December 2022
anaxii

MistressRemilia wrote:

I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at it
MistressRemilia
157 posts
Joined January 2024
MistressRemilia

Anaxii wrote:

MistressRemilia wrote:

I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at it
Definitely not physics, but maybe some biology yeah
anaxii
15,000 posts
Joined December 2022
anaxii

MistressRemilia wrote:

Anaxii wrote:

MistressRemilia wrote:

I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at it
Definitely not physics, but maybe some biology yeah
i would like you to help me differentiate between the different leukocytes with your words 🙏
B0ii
591 posts
Joined November 2021
B0ii

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
american moment
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

B0ii wrote:

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
american moment
💀💀, in Australia (ik I put middle school) we don't have middle school we just have primary and high, there is also kindy or kindergarten but they aren't really with the education system they're all independent plus I doubt any kindy kids are gonna be playing OSU!💀
Patatitta
1,244 posts
Joined February 2020
Patatitta

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
different countries use different systems because education is not a solved science. It's not something like metric where you can get every country (except america for some reason) to agree to measure stuff on meters
MistressRemilia
157 posts
Joined January 2024
MistressRemilia

Anaxii wrote:

MistressRemilia wrote:

Anaxii wrote:

MistressRemilia wrote:

I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at it
Definitely not physics, but maybe some biology yeah
i would like you to help me differentiate between the different leukocytes with your words 🙏
So there's like lymphocytes which hang out mainly in the lymphatic system until they're needed. They include T-cells, B Cells, and the killer cells (I don't remember if they have a name). Part of what makes yersinia pestis (aka, the plague) so deadly is that it attacks the lymphatic system, hence why you get bubonic plague, where your lymph nodes swell up like nuts (there's other forms of plague, too).

B Cells are what makes your antibodies. They're part of your adaptive immune system and get activated after receiving signals. There's a bit of time between when your innate immune system notices a pathogen, when it activates the adaptive immune system through antigens, and when the B Cells have time to start making antibodies to fight an infection.

I forget exactly what T-Cells do... but they get hit by HIV. They're obviously important. Kurzgesagt has a very good video on the different bits of the immune system, so look that up and it'll probably tell you what T-Cells do.

So those above are the lymphocytes and are part of the adaptive immune system parts. T-Cells, B Cells, and natural killer cells.

There's also eosinophil, which attack parasites, and are also why you get allergies and such.

Basophils are responsible for inflammation, asthma, allergies, anaphylaxis... all the fun stuff like that. They're not well understood, and are the least common leukocytes in the blood stream. They release histamines.

Then there's monocytes, which are the largest white cells and go around eating shit up similar to neutrophils. They go around looking for invaders or trauma, and then go nom-nom-nom on infected cells. They'll call in neutrophils if they need help.

Neutrophils are also part of the innate immune system. iirc, they're the most common leukocyte in the blood, and for some reason, my body in particular tends to enjoy producing more neutrophils than normal. They go around killing things as well. When they need help, they release dendrytic cells. Those go find your adaptive immune system and kick it into action.

So you've got lymphocytes (B Cells that produce antibodies, T-Cells, and natural killer cells), eosinophils (kills parasites, does some stuff with inflammation and allergies), basophils (releases histamines to cause inflammation), neutrophils (KILL EVERYTHING EVEN WHEN IT MEANS KILLING MYSELF), and monocytes (nom nom nom dudes looking for trouble).

I think I have those correct... double-check with your textbooks and such in case I got something wrong.
Last edited by MistressRemilia , edited 1 time in total.
Jangsoodlor
Jangsoodlor

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
Here it's
Primary : 6 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
Last edited by Jangsoodlor , edited 1 time in total.
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

Jangsoodlor wrote:

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
Here it's
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
so you've only got 9 years of education?
Jangsoodlor
Jangsoodlor

Jebus1 wrote:

Jangsoodlor wrote:

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
Here it's
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
so you've only got 9 years of education?
*Primary 6 Years I mis-typed kek.
Last edited by Jangsoodlor , edited 3 times in total.
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

Jangsoodlor wrote:

Jebus1 wrote:

Jangsoodlor wrote:

[-Omni-] wrote:

for me, its

elementary: prek-5
middle: 6-8
high: 9-12


smh everywhere should use the same things it would be much easier😞
Here it's
Primary : 3 years
Junior High:3 years
And then you can go either Senior High (3 years) or go get vocational certificate (which is also 3 years)
so you've only got 9 years of education?
*Primary 6 Years I mis-typed kek.
Ah gotchu
anaxii
15,000 posts
Joined December 2022
anaxii

MistressRemilia wrote:

Anaxii wrote:

MistressRemilia wrote:

Anaxii wrote:

MistressRemilia wrote:

I mean... I'm 40, so I might be able to help someone with some homework. Just not math, I absolutely suck at math.
are you okay with physic and biology? because i suck at it
Definitely not physics, but maybe some biology yeah
i would like you to help me differentiate between the different leukocytes with your words 🙏
So there's like lymphocytes which hang out mainly in the lymphatic system until they're needed. They include T-cells, B Cells, and the killer cells (I don't remember if they have a name). Part of what makes yersinia pestis (aka, the plague) so deadly is that it attacks the lymphatic system, hence why you get bubonic plague, where your lymph nodes swell up like nuts (there's other forms of plague, too).

B Cells are what makes your antibodies. They're part of your adaptive immune system and get activated after receiving signals. There's a bit of time between when your innate immune system notices a pathogen, when it activates the adaptive immune system through antigens, and when the B Cells have time to start making antibodies to fight an infection.

I forget exactly what T-Cells do... but they get hit by HIV. They're obviously important. Kurzgesagt has a very good video on the different bits of the immune system, so look that up and it'll probably tell you what T-Cells do.

So those above are the lymphocytes and are part of the adaptive immune system parts. T-Cells, B Cells, and natural killer cells.

There's also eosinophil, which attack parasites, and are also why you get allergies and such.

Basophils are responsible for inflammation, asthma, allergies, anaphylaxis... all the fun stuff like that. They're not well understood, and are the least common leukocytes in the blood stream. They release histamines.

Then there's monocytes, which are the largest white cells and go around eating shit up similar to neutrophils. They go around looking for invaders or trauma, and then go nom-nom-nom on infected cells. They'll call in neutrophils if they need help.

Neutrophils are also part of the innate immune system. iirc, they're the most common leukocyte in the blood, and for some reason, my body in particular tends to enjoy producing more neutrophils than normal. They go around killing things as well. When they need help, they release dendrytic cells. Those go find your adaptive immune system and kick it into action.

So you've got lymphocytes (B Cells that produce antibodies, T-Cells, and natural killer cells), eosinophils (kills parasites, does some stuff with inflammation and allergies), basophils (releases histamines to cause inflammation), neutrophils (KILL EVERYTHING EVEN WHEN IT MEANS KILLING MYSELF), and monocytes (nom nom nom dudes looking for trouble).

I think I have those correct... double-check with your textbooks and such in case I got something wrong.
i think i understand a bit now
thank you for your help
Jangsoodlor
Jangsoodlor
Can anyone help me solve this double integral?
anaxii
15,000 posts
Joined December 2022
anaxii

Jangsoodlor wrote:

Can anyone help me solve this double integral?
Nekathe
52 posts
Joined March 2021
Nekathe

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta
Nekathe's solution

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8

u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.

I am going to edit this according to what i've solved:

1. According to my old notes and the book: [1] N. Piskunov, "Capitulo XI: integral definida", Cálculo diferencial e integral, tercera edición, Editorial Mir, URSS, Editorial Mir Moscú, 1977, pp. 436-437. (IEEE style reference)

Here's the english version of my book: N. Piskunov differetial and integral calculus, i think it is the first edition but it has the information we use here. The page is 404-405 of the PDF, 403-404 of the book.

I'll first take out the 6 out of the 2 integrals to simplify the operation, thats what we used to do in math class.

2.I'll solve the exponential minus one in the power. -(2x+3y)=-2x-3y

now we have: e^(-2x-3y)

3. Well i comfirmed this with my notes and this vid:
Double Integral: e^(x + y) dy dx , y = 2 to 3 , x = 1 to 2


Basically if you have for example: e^(a+b) you can separate into a product (e^a)(e^b)
Therefore: e^(-2x-3y)=(e^-2x)(e^-3y) and the first integral it is respect to y so we treat (e^-2x) as in the vid.

at this point i'll take (e^-2x) out multiplying the dy integral.

then acoding to [1] solve the definite integral from o to 1-x of (e^-3y) dy
this is wrong xd
leaving us with: this expresion: 6 ∫0_1 (e^-2x)∫0_1-x (e^-3y)dy dx

evaluate the integral from 0 to 1-x:
6 ∫0_1 (e^-2x) [(e^-3y)]0_1-xdx

evaluating:
6 ∫0_1 (e^-2x) [(e^-3⟨1-x⟩-e^-3⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-e^⟨0⟩]dx
6 ∫0_1 (e^-2x) [(e^⟨-3+3x⟩-1]dx
Multiplying (e^-2x) with the evaluated integral:
here we apply the power rule of the product a^m(a^n)=a^m+n.
6 ∫0_1 e^⟨-3+x⟩ - e^⟨-2x⟩dx
Here's almost done just solve this integral using the previous rules in [1]

6[e^⟨-3+x⟩ - e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨x⟩- e^⟨-2x⟩]0_1
6[e^⟨-3⟩.e^⟨1⟩- e^⟨-2⟩-(e^⟨-3⟩.e^⟨0⟩- e^⟨-2.0⟩)]
6[e^⟨-2⟩-e^⟨-2⟩-(e^⟨-3⟩.1-1)]
6[e^⟨-2⟩-e^⟨-2⟩-e^⟨-3⟩+1)]
I am struggling here with the signs and the algebra maybe you can solve it to get the answer. I am going to sleep, rn heres midnight. hope u understand this. tomorrow hopefully i'll get the answer, it was really fun to solve this, thanks for this challenge :D, i re-learned a few things. Salutations o/

26/02/24

i can't figure out the last part but i've remembered another few rules to make it easier

27/02/24

I figured this out i think, thanks to Joanna Angel at DC Mathematics server. With her help i figured out that i skipped a rule, a basic rule while integrating e powered to constants and variables greater than 1.

here is what i did:




Rules applied:

Last edited by Hachiman Ryouta , edited 34 times in total.
Nekathe
52 posts
Joined March 2021
Nekathe

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk
Jangsoodlor
Jangsoodlor

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk

Well, here's the actual answer to that question


I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta

Jangsoodlor wrote:

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk

Well, here's the actual answer to that question


I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.
Nekathe
52 posts
Joined March 2021
Nekathe

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk

Welp, i did embarrass myself. I solved it with “x-1” instead of “1-x”

I cant believe i made a mistake like that
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta

Nekathe wrote:

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk


Welp, i did embarrass myself. I solved it with “x-1” instead of “1-x”

I cant believe i made a mistake like that
yeaaah it be like that, but hey, don't be discouraged by that, you made all the process, just mechanize and solve it twice, like in a test, you not only need to be able to solve it, but to solve it fast and check it again. Salutations o/.
Last edited by Hachiman Ryouta , edited 1 time in total.
Jangsoodlor
Jangsoodlor

Hachiman Ryouta wrote:

Jangsoodlor wrote:

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk

Well, here's the actual answer to that question


I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.
Well, I just got introduced to multi-variable integral in Calculus II course last week so I don't know any techniques yet (except basic integration techniques from Calculus I like trigonometric substitution and stuffs). This problem is from another course I took, which technically requires only Calculus I, but the instructor mysteriously gave a problem that requires Calculus II knowledge.

As for the deadline, it's already passed lol. Again, thanks for your help and maybe I'll post my attempt once I'm equipped with appropriate knownledge to solve this.

Here's the full problem btw


In order to find the Probability Distribution Function of f, given that event A happened, I need to solve for Probability of Event A happened first. Which is the area under that graph where x+y <=1. That's where Integration comes in. But I didn't expected it to be multi-variable lol

And here's the full solution


As for the book name, it's "Probability and Stochastic Process: A friendly introduction for Electrical and Computer Engineers" Second Edition by Roy D. Yates and David J. Goodman. It's a great book to read if you want to calculate the probability of you getting a certain character in Gacha games, then have literal mental breakdown and then questions your existence and sanity.
Last edited by Jangsoodlor , edited 3 times in total.
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta

Jangsoodlor wrote:

Hachiman Ryouta wrote:

Jangsoodlor wrote:

Nekathe wrote:

Hachiman Ryouta wrote:

Nekathe wrote:

Jangsoodlor wrote:

Can anyone help me solve this double integral?
I have not learned this... but if I don't solve this, it'll live rent free in my head for the next week.

Time to embarrass myself.

first lets start with the inside integral.
  1. u sub, let u = 2x+3y
  2. since we are working with dy, we treat x as a constant when finding du. du/3 = dy
  3. change the bounds,
    1. top bounds: y=x-1, u then equals 2x+3(x-1); 5x-3
    2. bottom bounds: y=0, u then equals 2x
  4. the integral is now /int_2x^5x-3 (2e^-u)du
  5. solve the simple integral, -2e^(-5x+3)+2e^(-2x)
now the outside integral
  1. /int_0^1 (-2e^(-5x+3)+2e^-2x)dx, finally...something i know how to do
  2. separate the integral, /int_0^1 (2e^-2x)dx - /int_0^1 (2e^(-5x+3))dx
  3. solve the first integral
    1. find indeterminant form, -e^-2x. i guess you could u sub but i dont think you need to
    2. evaluate between the bounds, -(e^(-2)-e^-0)
    3. -e^(-2)+1
  4. now solve the second integral
    1. this one is similar to the previous one, indeterminant form yada yada
    2. indeterminant form, (-2/5)e^(-5x+3)
    3. evaluate between the bounds, (-2/5)(e^(-2)-e^3)
FINALLY combine everything together to get:-e^(-2)+1-(-2/5)(e^(-2)-e^3) ~= -7.11541593922
(consistent with mathway)

KEEP IN MIND I HAVE NOT COMPLETED A YEAR OF ANY CALCULUS COURSE.

goated video: https://www.youtube.com/watch?v=rZQ5DI3dvn8
u is not -(2x+3y)? then du would be -3? That will change everything, idk, i am rusty i haven't solved one since 2019 so..

well i tried to solve this, but i got another start, different from yours Nekathe.
I dont think it should make a difference since the integral is evaluated anyways, it just has a negative power which is accounted for. But if you get a different answer, lmk

Well, here's the actual answer to that question


I just got introduced to the concept of double integral on calculus course last week. Wtih easy example like how to solve for a 1x1 square. But I also took another cute and funny course called "Probability & Statistics" which requires me integrating that not-so-cute-and-funny function to get the answer. I saw the answer and don't quite understand how to solve for the questions. Thanks for your help
Can you provide more info, like the books you are using and the methods used by the teacher, also the deadline for this homework, I am working in the solution provided by symbolab but idk if I can make it if it's for monday, anyways I'll let u know how i would solve this, later, i think... Thanks for providing more info btw.
Well, I just got introduced to multi-variable integral in Calculus II course last week so I don't know any techniques yet (except basic integration techniques from Calculus I like trigonometric substitution and stuffs). This problem is from another course I took, which technically requires only Calculus I, but the instructor mysteriously gave a problem that requires Calculus II knowledge.

As for the deadline, it's already passed lol. Again, thanks for your help and maybe I'll post my attempt once I'm equipped with appropriate knownledge to solve this.

Here's the full problem btw


In order to find the Probability Distribution Function of f, given that event A happened, I need to solve for Probability of Event A happened first. Which is the area under that graph where x+y <=1. That's where Integration comes in. But I didn't expected it to be multi-variable lol

And here's the full solution

As for the book name, it's "Probability and Stochastic Process: A friendly introduction for Electrical and Computer Engineers" Second Edition by Roy D. Yates and David J. Goodman. It's a great book to read if you want to calculate the probability of you getting a certain character in Gacha games, then have literal mental breakdown and then questions your existence and sanity.
OMG probability and statistics wasn't the greatest subject back then when i took it in college the teacher was kind of inexperienced a i did not undertand it deep enough. anyways don't worry about the deadline, just solve the problem, because if you don't understand the first one, it is likely that you don't understand the next ones. Salutations o/

PD: aah i can't figure out the last part of the problem, i got it with an amplitude of 3 times the size of the solution and sings changed :S
Last edited by Hachiman Ryouta , edited 1 time in total.
NitroZz
140 posts
Joined November 2022
NitroZz
Do you have any apps or techniques to improve in English?
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta

NitroZz wrote:

Do you have any apps or techniques to improve in English?
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.
anaxii
15,000 posts
Joined December 2022
anaxii

NitroZz wrote:

Do you have any apps or techniques to improve in English?
honestly i learned english thanks to youtube videos and translators lol
Karmine
890 posts
Joined November 2015
Karmine

Hachiman Ryouta wrote:

NitroZz wrote:

Do you have any apps or techniques to improve in English?
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.
That's how I learned, playing games in english (sometimes because there was no translation), watching videos, movies etc.
Also being active on forums and chats (vc/text) helps a lot since you not only listen/read but also speak/write.
Last edited by Karmine , edited 1 time in total.
NitroZz
140 posts
Joined November 2022
NitroZz

Hachiman Ryouta wrote:

NitroZz wrote:

Do you have any apps or techniques to improve in English?
Salutations o/. find something to do regularly in english, like watching a youtuber, do little changes that make you think in english, for example, change your phone to english. Idk i did that, kinda worked for me.
Thank you for the advice
anaxii
15,000 posts
Joined December 2022
anaxii
fortunately, no one is in elementary school which is GOOD
Gsun
Gsun
https://education12.quora.com/This-is-my-answer-to-how-to-help-teenagers-in-high-school-and-anyone-in-school-get-through-learning-when-theyre-slow Hopefully this helps people.
NitroZz
140 posts
Joined November 2022
NitroZz

Anaxii wrote:

fortunately, no one is in elementary school which is GOOD
Maybe I'm there
anaxii
15,000 posts
Joined December 2022
anaxii

NitroZz wrote:

Anaxii wrote:

fortunately, no one is in elementary school which is GOOD
Maybe I'm there
i can help you kiddo :)
[-Omni-]
960 posts
Joined July 2021
[-Omni-]
test in 30 mins wish me luck 😔
MistressRemilia
157 posts
Joined January 2024
MistressRemilia

[-Omni-] wrote:

test in 30 mins wish me luck 😔
Good luck!! 🤞
Hachiman Ryouta
50 posts
Joined October 2014
Hachiman Ryouta

[-Omni-] wrote:

test in 30 mins wish me luck 😔
Best of luck
Jebus1
215 posts
Joined June 2022
Topic Starter
Jebus1

[-Omni-] wrote:

test in 30 mins wish me luck 😔
How'd the test go? 🫣

I didn't know what emoji to use😭😭😭😭
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