integration

1 次查看(过去 30 天)
Edward
Edward 2012-5-25
please can someone help me to solve this problem: I want to integrate this function:
exp(0.5*(1-((l-hm)/(2*H))-exp(-((l-hm)/(2*H)))))
I make this program:
hm=400;
H=50;
syms l;
H0=int(exp(0.5*(1-((l-hm)/(2*H))-exp(-((l-hm)/(2*H))))),l,hm,700)
but the answer is:
100*2^(1/2)*pi^(1/2)*exp(1)^(1/2)*(erf(2^(1/2)/2) - erf(2^(1/2)/(2*exp(3/2))))
how to get the right answer???

回答(3 个)

Alexander
Alexander 2012-5-25
Why do you think that the answer is incorrect? If you just want a number, use double on the result:
>> double(H0)
ans =
209.1673
Or use INTEGRAL as Mike Hosea suggests.
  2 个评论
Mike Hosea
Mike Hosea 2012-5-25
:lol: I didn't even bother with the symbolic attempt and completely misinterpreted the question. I'm blaming this on Walter. :)
Walter Roberson
Walter Roberson 2012-5-25
It's a fair cop, gov!

请先登录,再进行评论。


Mike Hosea
Mike Hosea 2012-5-25
I never quite know what to say when questions like this come up. There's nothing wrong with wanting to do integrals analytically, to be sure, and some times there are legitimate reasons, I guess, but I wonder what people need the analytical answer for. This problem is really no problem for INTEGRAL (or QUADGK)
>> integral(@(l)exp(0.5*(1-((l-hm)/(2*H))-exp(-((l-hm)/(2*H))))),hm,700)
ans =
2.091673274741571e+02
>> 100*2^(1/2)*pi^(1/2)*exp(1)^(1/2)*(erf(2^(1/2)/2) - erf(2^(1/2)/(2*exp(3/2))))
ans =
2.091673274741571e+02

Walter Roberson
Walter Roberson 2012-5-25
MuPAD is not able to complete that integral. (Maple is not able to do it either.)
Perhaps there is an appropriate change of variables, but I have not been able to locate one.
  7 个评论
Christopher Creutzig
If it helps to get some confidence in the result: To get integrals.com to integrate the expression (indefinitely), substitute x=exp(-l/200). http://integrals.wolfram.com/index.jsp?expr=-200*exp%285%2F2+-+%281%2F2%29*x^2*exp%284%29%29&random=false
Alexander
Alexander 2012-5-25
Okay, that explains my confusion.

请先登录,再进行评论。

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by