Jump to content
Join Talk Chelsea and log in to join the CFC discussion! ×

The Pub - Discuss Anything

Manuf

By Manuf,
in General Chat

 Share

Recommended Posts

  • Replies 29.3k
  • Created
  • Last Reply

Top Posters In This Topic

  • Rmpr

    2515

  • Stingray

    2479

  • Spike

    2472

  • CHOULO19

    1921

Popular Days

Top Posters In This Topic

Popular Days

Popular Posts

We Hate Scouse
We Hate Scouse

Finally have internet so I'll be posting more. In other news here is some photos of my son

Muzchap
Muzchap

Talking of marriage and seeing as the thread is already de-railed - I got married at the weekend It was amazing - hired a marquee and had Dodgems and a carousel along with a double decker bus convert

We Hate Scouse
We Hate Scouse

Posted Images

Amblève.

Amblève. 4,995

Posted
Posted

Dude you sound like me, (praying for a 4) there was no way I was getting a 5 cuz I never went to the teacher for extra help, always did it on my own. Hahaha

And lmaooo nah I'm only a sophomore,, im just acting like a senior (in it that I'm doing absolutely nothing cuz we're gonna be done soon) but i still got 2 more painful years left till im off to college.

2 more years?! Jeeeez. I can't even wait till these 12 years at school are over in a month. Good luck mate, you're gonna need it I think :D

It's been a while since I last took calculus, but I don't think you can find the antiderivative for that without going into power series expansion or finite expansion (Ugh, I hated this stuff!). You can find it's integral over some period using a calculator, though.

Mate, it was 3am when I posted that, I don't have a clue what I was writing :lol:

2e^(x-3) should be the correct answer, according to integrals.wolfram.com.

I thought about that :ph34r: but don't really see it as a feasible option tbh. I thought about doing my master's in the US perhaps but 1) it's too expensive 2) it's too restrictive since I don't want to get an MBA. Plus since my current bachelor's degree is from the University of London, it makes it much much easier for me to either get a job in the UK or get accepted into a UK master's programme!

Maybe if things go well I might be able to work there someday? Not sure tbh. My family isn't loaded lol so I'll just have to see what happens. Anyway Singapore's financial sector is doing really well atm. But like I said I definitely need to get out of here for a few years.

Right! If you go overseas, you'll have to take their own examinations right? To qualify as a lawyer there?

Have you thought about what type of law you want to practice?

You'll do fine wherever you go ;)

Getting experience at as many places as possible would be ideal I think. Not just because of the working experience, mainly because of the lifetime experiences you could gain from visiting many parts of the world.

Yeah, most likely. Law's really stupid regarding getting permissions in other countries, but there's nothing really you can do about.

I'm planning on specializing in human rights. Obviously I don't know if that's gonna be my final decision as well but right now I feel like this part of law would be brilliant for me. But yeah, who knows what' gonna happen...

Link to comment
Share on other sites
CHOULO19

CHOULO19 24,332

Posted
Posted
Mate, it was 3am when I posted that, I don't have a clue what I was writing 2e^(x-3) should be the correct answer, according to integrals.wolfram.com.

Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

Link to comment
Share on other sites
Amblève.

Amblève. 4,995

Posted
Posted

Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

None of them.

(e^(x-3))*2

Link to comment
Share on other sites
Madmax

Madmax 9,219

Posted
Posted

Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

None of them.

(e^(x-3))*2

That's easy, it stays exactly the same (plus a certain constant C).

Yep, I definitely understood all of that.

Link to comment
Share on other sites
Stingray

Stingray 9,441

Posted
Posted

Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

Wtf !!!

post-6387-0-92889000-1369328469_thumb.jp

:Goober:

Link to comment
Share on other sites

Please sign in to comment

You will be able to leave a comment after signing in



Sign In Now
 Share
Go to topic listing

  • 0 members are here!

    • No registered users viewing this page.
×
×
  • Create New...

talk chelse forums

We get it, advertisements are annoying!
Talk Chelsea relies on revenue to pay for hosting and upgrades. While we try to keep adverts as unobtrusive as possible, we need to run ad's to make sure we can stay online because over the years costs have become very high.

Could you please allow adverts on this website and help us by switching your ad blocker off.

KTBFFH
Thank You

Sure, let me in!
Enter