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        riddles >> medium >> Integral
        
(Message started by: mathstudent155 on Jan 27^th, 2013, 12:31pm)

Title: Integral
Post by mathstudent155 on Jan 27^th, 2013, 12:31pm

Hello everybody!

May I ask you a big favor?
I need help.
What I have is:
(2x-3)/ sqrt(4x-1)
What is the integral of it??

I've tried to use Integral Calculator here:
numberempire.com/integral...var=x&answers=

And I've got : (x-4)*sqrt(4*x-1)/3

Seems like it is the right answer.
But I want to understand the way of solving it.
Please show me step by step solution.

Thanks in advance!

Title: Re: Integral
Post by towr on Jan 27^th, 2013, 1:08pm

You can use integration by parts. http://en.wikipedia.org/wiki/Integration_by_parts

We can take (2x-3)/ sqrt(4x-1) = (2x-3) d[1/2 sqrt(4x-1)]
Then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif(2x-3) d[1/2 sqrt(4x-1)]
= 1/2 sqrt(4x-1) * (2x-3) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1/2 sqrt(4x-1) d[2x-3]
= 1/2 sqrt(4x-1) * (2x-3) - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifsqrt(4x-1) d x
= 1/2 sqrt(4x-1) * (2x-3) - 1/4 * 2/3 * (4x-1)^(3/2)
= ((2x-3)*3-(4x-1))/6 * sqrt(4x-1)
= (x-4)/3 * sqrt(4x-1)

(I think it's literally been years since I've last done this..)

Title: Re: Integral
Post by tsitut on Feb 20^th, 2013, 12:12pm

Thanks, I actually needed it too :P