Numerical Integration: Find x to 3 Decimal Points

[Albert1]

\[2.37=\frac{1}{\sqrt{6}} \int_{0}^{x} \sqrt{\frac{e^x}{e^x-1}}dx\]

please find x to three decimal point

 

[MarkFL]

Using technology and some guesswork, I find:

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

$x=4.425$

 

[Albert1]

I get x $\approx$ 4.428

may be your answer is more accurate

 

[MarkFL]

wolframalpha.com gave the result I cited as being exact, but my TI-89 gives:

$x\approx4.42501043622$

 

[Nono713]

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]

W|A also says the value my TI-89 gave results in the same exact value as well. (Tmi)

 

[Mathwebster]

Do you really want the upper limit and dummy variable (the integration variable) to both be $x$?

 

[Mathwebster]

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]

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!

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

 

[Mathwebster]

Nice!

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.