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Numerical integration

Albert

Well-known member
Jan 25, 2013
1,225
\[2.37=\frac{1}{\sqrt{6}} \int_{0}^{x} \sqrt{\frac{e^x}{e^x-1}}dx\]

please find x to three decimal point
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Using technology and some guesswork, I find:

$\displaystyle \frac{1}{\sqrt{6}}\int_0^{4.425}\sqrt{\frac{e^t}{e^t-1}}\,dt=2.37$

$x=4.425$
 

Albert

Well-known member
Jan 25, 2013
1,225
I get x $\approx$ 4.428

may be your answer is more accurate
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
wolframalpha.com gave the result I cited as being exact, but my TI-89 gives:

$x\approx4.42501043622$
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
wolframalpha.com gave the result I cited as being exact, but my TI-89 gives:

$x\approx4.42501043622$
The TI is probably having floating-point accuracy problems - it's not exactly a trivial approximation - I doubt W|A is wrong. I would use Mathematica to confirm but I don't have it installed right now :(
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
W|A also says the value my TI-89 gave results in the same exact value as well. (Tmi)
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Do you really want the upper limit and dummy variable (the integration variable) to both be $x$?
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
If the integration variable is in fact $t$ (like MarkFL notes), then we can solve exactly for $x$ giving

$x = 2 \ln \dfrac{k^2+1}{2k}$ where $k = e^{2.37\sqrt{6}/2}$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If the integration variable is in fact $t$ (like MarkFL notes), then we can solve exactly for $x$ giving

$x = 2 \ln \dfrac{k^2+1}{2k}$ where $k = e^{2.37\sqrt{6}/2}$.
Nice! :cool:

Did you find this by directly computing the improper integral, or did you exploit the FTOC in some other way?
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Nice! :cool:

Did you find this by directly computing the improper integral, or did you exploit the FTOC in some other way?
I integrated directly. Once you let $u = e^{t/2}$, you get an integral very manageable.