FacebookTwitter
Hatrack River Forum Post New Topic  Post A Reply
my profile login | register | search | faq | forum home

  next oldest topic   next newest topic
» Hatrack River Forum » Active Forums » Books, Films, Food and Culture » Yo stupidly simple Calculus question

   
Author Topic: Yo stupidly simple Calculus question
Blayne Bradley
unregistered


 - posted            Edit/Delete Post   Reply With Quote 
okay so if I have f(x)=(x+2)^2

then f'(x) = 2*(x+2)* (dx/dy) x+2 which is 1

so f'(x) = 2x+4

However if I want to find the integral of 2x+4

so: f(x) = 2x+4 (dx)

x^2+4x+C

Which I imagine if you factor it probably equals to (x+2)^2 if you solve for c

x^2+2x+2x+4
x^2+4x+4

Which seems a little wierd, is the answer good enough at the polynomial x^2+4x+C or do I have to get it back to (x+2)^2?

Next if I have (x+2)^2, do I find its integral? Foil it out and do f(x)= x^2+4x+4 (dx)?

In which case it should be (1/3)(x^3) + 2x^2 + 4x + C?

Or is there a reverse chain rule I can use?

(1/3)[(x+2)^3](x^2+2x)+C?

//one third times x+2 cubed times the intergral of the inside plus a constant?

IP: Logged | Report this post to a Moderator
Phanto
Member
Member # 5897

 - posted      Profile for Phanto           Edit/Delete Post   Reply With Quote 
Your question doesn't really make sense to me.

quote:

is the answer good enough at the polynomial x^2+4x+C or do I have to get it back to (x+2)^2?

The integral of 2x + 4 is x^2 + 4x + C... where are you getting the whole weird (x+2)^2 stuff from?

The fact that you can rewrite an equation in another way doesn't mean you should, and unless you know what C is for some reason then it is unknown.

Posts: 3060 | Registered: Nov 2003  |  IP: Logged | Report this post to a Moderator
Blayne Bradley
unregistered


 - posted            Edit/Delete Post   Reply With Quote 
The point is say I have f(x) = (x+2)^2

The idea is figuring out how to integrate it and how to integrate back from its derivative.

IP: Logged | Report this post to a Moderator
fugu13
Member
Member # 2859

 - posted      Profile for fugu13   Email fugu13         Edit/Delete Post   Reply With Quote 
You can't, perfectly. Information is lost. Taking the derivative of any constant is 0, so it is impossible to integrate and work out any constant terms that were part of the original.

(x+2)^2 has a constant term -- writing it another way, as x^2 + 4x + 4, you can see that clearly. After taking the derivative, all information about that constant term is lost. It was a 4, but from the derivative you can't tell if it was a 23 or a 7 or 2.1 billion. So no, you can't get back to (x+2)^2 from knowing the derivative is 2x + 4. Because it could have been x^2 + 4x + 3.14159 instead of x^2 + 4x + 4.

Now, if you have additional information, you can compute the constant factor. For instance, sometimes you are given the intercept or somesuch.

Posts: 15770 | Registered: Dec 2001  |  IP: Logged | Report this post to a Moderator
The Rabbit
Member
Member # 671

 - posted      Profile for The Rabbit   Email The Rabbit         Edit/Delete Post   Reply With Quote 
quote:
Originally posted by Blayne Bradley:
okay so if I have f(x)=(x+2)^2

then f'(x) = 2*(x+2)* (dx/dy) x+2 which is 1

so f'(x) = 2x+4


You've lost me completely.

What is y? What variable are you differentiating with respect to?

what is 1?

Aside from that, the answer fugu and Phanto have given you is accurate.

You loose information when you take a derivative, that's why you have to add in an arbitrary constant when you integrate. In order to recover the original function by integrating a derivative, you have to have more information. Typically, you have to know the value of the initial function at some point.

Posts: 12591 | Registered: Jan 2000  |  IP: Logged | Report this post to a Moderator
The White Whale
Member
Member # 6594

 - posted      Profile for The White Whale           Edit/Delete Post   Reply With Quote 
I think he means f'(x) = 2*(x+2)*d/dx(x+2), and that d/dx(x+2) = 1.
Posts: 1710 | Registered: Jun 2004  |  IP: Logged | Report this post to a Moderator
Mike
Member
Member # 55

 - posted      Profile for Mike   Email Mike         Edit/Delete Post   Reply With Quote 
quote:
You loose information...
Speaking of which, I can't wait 'til my information bow arrives in the mail. [Wink]
Posts: 1810 | Registered: Jan 1999  |  IP: Logged | Report this post to a Moderator
willthesane
Member
Member # 11754

 - posted      Profile for willthesane           Edit/Delete Post   Reply With Quote 
the integral from a to b of f(g(x))*g'(x) is the same as the integral from g(a) to g(b) of f(x).

so, the integral of f(x)=(x+2)^2 dx from 0 to 2 is the same as the integral of f(u)=u^2 from 2 to 4.

http://en.wikipedia.org/wiki/Integration_by_substitution

if this is what you were looking for i wish you luck.

Posts: 7 | Registered: Sep 2008  |  IP: Logged | Report this post to a Moderator
   

Quick Reply
Message:

HTML is not enabled.
UBB Code™ is enabled.
UBB Code™ Images not permitted.
Instant Graemlins
   


Post New Topic  Post A Reply Close Topic   Feature Topic   Move Topic   Delete Topic next oldest topic   next newest topic
 - Printer-friendly view of this topic
Hop To:


Contact Us | Hatrack River Home Page

Copyright © 2008 Hatrack River Enterprises Inc. All rights reserved.
Reproduction in whole or in part without permission is prohibited.


Powered by Infopop Corporation
UBB.classic™ 6.7.2