My Math Forum numerical integration

 Calculus Calculus Math Forum

 July 1st, 2012, 08:22 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 numerical integration $2.37=\frac{1}{\sqrt{6}}\int_{0}^{x}\sqrt{\frac{e^x }{e^x-1}}\,\,\,dx, \,\,please \,\,\,find\,\,\, x$ $Ans :\,\, x\approx\,\, 4.428$
 July 1st, 2012, 09:41 PM #2 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: numerical integration Let $t=e^x\in[0,e^{x_0}]$ $x=\ln t\in [0,x_0]$ $dx=\frac{dt}{t}$ $I=\int_1^{e^{x_0}} \frac{dt}{sqrt{t^2-t}}=\int_1^{e^{x_0}} \frac{dt}{sqrt{(t-1/2)^2-1/4}}$ $=\ln(t-1/2+sqrt{t^2-t})\|_{t=e^{x_0}}-\lim_{t\to 1}\ln(t-1/2+sqrt{t^2-t})$ $=\ln(t-1/2+sqrt{t^2-t})\|_{t=e^{x_0}}+\ln2$ $=\ln$2(t-1/2+sqrt{t^2-t})$\|_{t=e^{x_0}}$ $=\ln$2e^{x_0}-1+2sqrt{e^{2x_0}-e^{x_0}}$$ We have $e^I=2e^{x_0}-1+2sqrt{e^{2x_0}-e^{x_0}}$ Rearrange and simplify, $e^{x_0}=\frac{(e^I+1)^2}{4e^I}$, So, $x_0=2\ln(e^I+1)-I-2\ln2$ Let $I=2.37sqrt 6$ I think you may get the answer..
 July 1st, 2012, 09:46 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: numerical integration I get: $x=\ln$$\frac{2+e^{-\frac{237\sqrt{3}}{50\sqrt{2}}}+e^{\frac{237\sqrt{ 3}}{50\sqrt{2}}}}{4}$$\approx4.425$
July 1st, 2012, 10:13 PM   #4
Senior Member

Joined: Oct 2010
From: Changchun, China

Posts: 492
Thanks: 14

Re: numerical integration

Quote:
 Originally Posted by MarkFL I get: $x=\approx4.425$
Me too.

I use Matlab to execute this integral,

Code:
syms x;int(sqrt(exp(x)/(exp(x)-1))/sqrt(6),x,0,x_0)
while x_0 in [4.425, 4.428], I got the same approximate value 2.37....

 Tags integration, numerical

### numerical integration my maths

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post AnomanderRakeSoD Applied Math 0 June 21st, 2013 02:11 AM dallairius Calculus 1 April 6th, 2012 09:07 AM PageUp Applied Math 1 March 19th, 2012 05:58 AM VFernandes Applied Math 1 January 5th, 2012 12:38 PM papna Applied Math 0 June 19th, 2009 01:08 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top