- INT is a Symbolic Math Toolbox command, so it will try to do the integration symbolically. In this case, since the function is quite complicated, it's having a hard time doing so.
- Instead, you probably want to try INTEGRAL if you have a recent enough MATLAB version, or QUAD otherwise.
- To handle the extra parameters, check out this doc page: Parameterizing Functions
How can I compute numerical integration from a multivariate function ( f(t,C) ) where C'=[c_1,c_2,...,c_m] with respect to " t "
1 回表示 (過去 30 日間)
古いコメントを表示
I want to find the numerical integration from a multivariate function ( f(t,C) ) where C'=[c_1,c_2,...,c_m] with respect to " t ",
what command is the most accurate and efficient Matlab command for doing this?
for example this is my function:
f(t,c_1,c_2,c_3,c_4) =
(t^2 + 1)*((5040315906112595*t^(3/5)*(156*c_1 - 156*3^(1/2)*c_2 + 156*5^(1/2)*c_3 - 156*7^(1/2)*c_4 + 195*3^(1/2)*c_2*t - 585*5^(1/2)*c_3*t + 1170*7^(1/2)*c_4*t + 450*5^(1/2)*c_3*t^2 - 2250*7^(1/2)*c_4*t^2 + 1250*7^(1/2)*c_4*t^3))/702561541869797376 + 1)^2 + (sin(t)*(c_1 - (exp(t)*((5040315906112595*t^(3/5)*(156*c_1 - 156*3^(1/2)*c_2 + 156*5^(1/2)*c_3 - 156*7^(1/2)*c_4 + 195*3^(1/2)*c_2*t - 585*5^(1/2)*c_3*t + 1170*7^(1/2)*c_4*t + 450*5^(1/2)*c_3*t^2 - 2250*7^(1/2)*c_4*t^2 + 1250*7^(1/2)*c_4*t^3))/702561541869797376 + 1))/2 + 3^(1/2)*c_2*(2*t - 1) + 5^(1/2)*c_3*(6*t^2 - 6*t + 1) + 7^(1/2)*c_4*(20*t^3 - 30*t^2 + 12*t - 1))^2)/cos(t)^2
note that the command " int ( f,t,0,1 ) " didn't work and it's said: Warning Explicit integral could not be found.
I'll be very appreciate if you help me with this.
Best regards,
Siavash.
0 件のコメント
回答 (1 件)
Yannick
2013 年 10 月 16 日
編集済み: Yannick
2013 年 10 月 16 日
Hi - a few things:
Hope this helps!
2 件のコメント
Yannick
2013 年 10 月 17 日
Hi Siavash,
Could you show your code here, as well as how you execute it and what result/warning/error you get? Then I can help figure out with you why it isn't working as expected.
Yannick
参考
カテゴリ
Help Center および File Exchange で Symbolic Math Toolbox についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!