December 26th, 2009, 02:07 PM  #1 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Integral u sub problem
Hi, I can't seem to get this problem to work out right... ? x(x+2)^1/2 dx I let... u=x+2 du=dx x=u2 So... ?(u2)(u^1/2)du =?u^3/22u^1/2 du = (2/5)u^5/2  4/3u^3/2 +C = 2/5(x+2)^5/2  4/3(x+2)^3/2 + C But the book has it as... [2/15(X+2)^3/2](3X4) + C It might just be a simplification issue, but I'm not so sure because the example in the book, which in the practice set referred me to, had an answer similar to mine. I think in the past I checked by differentiation and it didn't seem to work out anyway. Thanks. 
December 26th, 2009, 04:04 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562  Re: Integral u sub problem
Your answer and the book answer are equal. Just play with the algebra a little.

December 26th, 2009, 06:02 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Integral u sub problem Hello, austen951! Quote: When given the square root of a linear expression, [color=beige]. . [/color]I often let equal the entire radical. [color=beige]. . . . . . . . [/color]  
December 26th, 2009, 07:42 PM  #4  
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493  Quote:
How would you tackle finding ??(x³ + 2x²)dx (where x > 2)?  
December 31st, 2009, 01:04 PM  #5 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Re: Integral u sub problem
Thanks all. I got it now...I didn't find any errors in my work and didn't see how to manipulate to the books answer. I guess it's just a different approach. And Skip I've never done an integral like that. I've only just finished a semester of Calc AB, and it seems over my head based on the integrals we've done. 
December 31st, 2009, 01:09 PM  #6 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Re: Integral u sub problem
Wait, I believe you can just factor out an x^2 and it becomes the same problem. hah

December 31st, 2009, 01:40 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562  Re: Integral u sub problem
[quote 2/5(x+2)^5/2  4/3(x+2)^3/2 + C But the book has it as... [2/15(X+2)^3/2](3X4) + C[/quote] 2/5(X+2)^5/2 = [2/5(X+2)^3/2](X+2) therefore 2/5(x+2)^5/2  4/3(x+2)^3/2 + C = (X+2)^3/2[2/5(X+2) 4/3] + C however [2/5(X+2) 4/3] = 2/15[3(X+2)  10] = 2/15(3X  4) therefore your answer and book answer are the same!!!!! 
December 31st, 2009, 11:43 PM  #8  
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493  Quote:
 
January 1st, 2010, 01:37 AM  #9 
Newbie Joined: Dec 2009 Posts: 6 Thanks: 0  Re: Integral u sub problem
Actually no. I checked the graphs of both and I noticed an ever slight difference between sqrt (x^32x^2) and x sqrt(x2). Isn't it the same principle of say, reducing sqrt 28 to 2 sqrt 7? It seems like it, but I don't remember the rules for this at the moment... 
January 2nd, 2010, 03:49 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
Do you have a textbook that explains this topic?


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