February 16th, 2015, 09:07 AM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Checking mistake
Where did I make mistake ? My answer is $\displaystyle \frac{1}{4}\pi^2+2$. The correct answer is $\displaystyle \frac{1}{4}\pi^22$. Last edited by jiasyuen; February 16th, 2015 at 09:14 AM. 
February 16th, 2015, 09:19 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs 
The second time you applied integration by parts, you did not distribute the negative sign to the resulting integral. 
February 16th, 2015, 09:31 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
obviously a sign error ... $\displaystyle \int x^2\cos{x} \, dx$ $\displaystyle u = x^2$ ; $\displaystyle du = 2x \, dx$ $\displaystyle dv = \cos{x} \, dx$ ; $\displaystyle v = \sin{x}$ $\displaystyle \int x^2\cos{x} \, dx = x^2\sin{x}  \int 2x\sin{x} \, dx$ $\displaystyle u = 2x$ ; $\displaystyle du = 2 \, dx$ $\displaystyle dv = \sin{x}$ ; $\displaystyle v = \cos{x} \, dx$ $\displaystyle \int x^2\cos{x} \, dx = x^2\sin{x}  \left[2x\cos{x}  \int 2\cos{x} \, dx\right]$ ... from this step to the next is where I think you made a sign error $\displaystyle \int x^2\cos{x} \, dx = x^2\sin{x} + 2x\cos{x}  \int 2\cos{x} \, dx$ 
February 16th, 2015, 03:07 PM  #4 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
Thanks a lot guys. I wonder when will the time I be one of you guys.

February 16th, 2015, 04:03 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  
February 16th, 2015, 06:56 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
As Jesus once said, let he who has not made a sign error throw the first stone. Something like that, anyway.


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