My Math Forum Integral u sub problem

 Calculus Calculus Math Forum

 December 26th, 2009, 01:07 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=u-2 So... ?(u-2)(u^1/2)du =?u^3/2-2u^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](3X-4) + 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, 03:04 PM #2 Global Moderator   Joined: May 2007 Posts: 6,586 Thanks: 612 Re: Integral u sub problem Your answer and the book answer are equal. Just play with the algebra a little.
December 26th, 2009, 05:02 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Integral u sub problem

Hello, austen951!

Quote:
 $\int x(x\,+\,2)^{\frac{1}{2}}\,dx$

When given the square root of a linear expression,
[color=beige]. . [/color]I often let $u$ equal the entire radical.

$\text{Let: }\,u \:=\:\sqrt{x\,+\,2} \qquad\Rightarrow\qquad x\,+\,2 \:=\:u^2 \qquad\Rightarrow\qquad x \:=\:u^2\,-\,2 \qquad\Rightarrow\qquad dx \:=\:2u\,du$

$\text{Substitute: }\;\int(u^2\,-\,2)\cdot\,u\,\cdot(2u\,du) \;=\;2\int\left(u^4\,-\,2u^2\right)\,du$

[color=beige]. . . . . . . . [/color]$=\;2\left(\frac{u^5}{5}\,-\,\frac{2u^3}{3}\right)\,+\,C \;=\; \frac{2}{15}u^3\,\left(3u^2\,-\,10\right)\,+\,C$

$\text{Back-substitute: }\:\frac{2}{15}\,\left(\sqrt{x\,+\,2}\right)^3\,\b igg[3(x\,+\,2)\,-\,10\bigg] \;=\;\frac{2}{15}\,(x\,+\,2)^{\frac{3}{2}}\,(3x\,-\,4)\,+\,C$

December 26th, 2009, 06:42 PM   #4
Global Moderator

Joined: Dec 2006

Posts: 19,547
Thanks: 1752

Quote:
 Originally Posted by austen951 I think in the past I checked by differentiation and it didn't seem to work out anyway.
Such a check should work, but a considerable amount of manipulation is sometimes required.

How would you tackle finding ??(x³ + 2x²)dx (where x > -2)?

 December 31st, 2009, 12:04 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, 12:09 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, 12:40 PM #7 Global Moderator   Joined: May 2007 Posts: 6,586 Thanks: 612 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](3X-4) + 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, 10:43 PM   #8
Global Moderator

Joined: Dec 2006

Posts: 19,547
Thanks: 1752

Quote:
 Originally Posted by austen951 it becomes the same problem.
Are you sure?

 January 1st, 2010, 12:37 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^3-2x^2) and x sqrt(x-2). 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, 02:49 AM #10 Global Moderator   Joined: Dec 2006 Posts: 19,547 Thanks: 1752 Do you have a textbook that explains this topic?

 Tags integral, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post arthurduh1 Calculus 0 June 28th, 2012 05:13 PM jackson_wang Calculus 5 September 13th, 2011 02:19 AM Tiome_nguyen Calculus 3 March 25th, 2011 07:54 PM good_phy Calculus 3 October 16th, 2008 07:38 AM Saeid Applied Math 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top