필터 지우기
필터 지우기

Numerical Integration on Matlab

조회 수: 2 (최근 30일)
Emrah Can Ucar
Emrah Can Ucar 2022년 4월 7일
편집: Torsten 2022년 4월 8일
There are 4 integrals I need to write, but I can't get correct results from them. Did I write it right?
Equations that end in c are constants. they don't change
xi_0 = r(1)/Rprop;
xi_1 = 1;
I1_c = 4*G*(1-epsilon*tan(phi));
I1 = integral(@(xint) xint*I1_c, xi_0, xi_1);
I2_c = lambda*(1+epsilon/tan(phi))*sin(phi)*cos(phi);
I2 = integral(@(xint) ((xint*I1_c)/2.*xint),xi_0,xi_1);
J1_c = 4*G*(1+epsilon/tan(phi));
J1 = integral(@(xint) xint*J1_c ,xi_0, xi_1);
J2_c = (1-epsilon*tan(phi))*((1+cos(2*phi))/(2));
J2 = integral(@(xint) (xint*J1_c)/2 ,xi_0,xi_1);

답변 (2개)

Riccardo Scorretti
Riccardo Scorretti 2022년 4월 8일
Hi. I would say no: you forget to multiply I2 and J2 by the respective constants I2_c and J2_c. That being said, these functions are polynomials with respect to xi: at your place I would go analytically.

Torsten
Torsten 2022년 4월 8일
편집: Torsten 2022년 4월 8일
I2 = integral(@(xint) ((xint*I1_c*I2_c)/2.*xint),xi_0,xi_1);
J2_c = (1-epsilon*tan(phi))*(cos(phi))^2;
J2 = integral(@(xint) (xint*J1_c)/2*J2_c ,xi_0,xi_1);
Not sure whether
I2' = lambda*(I1'/2 * zeta ...
or
I2' = lambda*(I1'/(2*zeta) ...
Some authors write
1/2*zeta
and mean
1/(2*zeta)

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

제품


릴리스

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by