Integrals with syms over the variable x

Hi!

I want to solve the following integrals but i want it to integrate over the variable x while c1, c2 and c3 are constants that are not known beforehand but the plan is to get my 3 equations so i can solve the equationsystem after the integrations for c1,c2 and c3. Can someone help me, please? The code i got so far is the one below.

My 3 equations:

% Galerkin method
A0=6e-4;
E=70e9;
L=0.5;
P=5000;
syms c1 c2 c3

e1=E*A0*x * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P) 
e2=E*A0*x^2 * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P)
e3=E*A0*x^3 * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P)

F1=int(e1, L ,0);
F2=int(e2, L, 0);
F3=int(e3, L, 0);

[Walter Roberson]

f1 = round(randn*100);
f2 = round(randn*100);
f3 = round(randn*100);

% Galerkin method
A0=6e-4;
E=70e9;
L=0.5;
P=5000;

syms c1 c2 c3 x
e1=E*A0*x * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P) 

e1 = 

e2=E*A0*x^2 * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P)

e2 = 

e3=E*A0*x^3 * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P)

e3 = 

F1=int(e1, x, L ,0)

F1 = 

F2=int(e2, x, L, 0)

F2 = 

F3=int(e3, x, L, 0)

F3 = 

sol = solve(F1==f1, F2==f2, F3==f3, [c1, c2, c3]);
[sol.c1; sol.c2; sol.c3]

ans = 

[Daniel Arvidsson]

Thanks, i sorted it out =) Somehow it chooses automatically to integrate over x if you also defines x as sym.

[Bjorn Gustavsson]

Good!

Walter's advice to explicitly state the variable of integration is good. Matlab use a reasonably clever procedure to decide that, but if one are explicit about it things will not go wrong.

[Daniel Arvidsson]

Yes, very true, thanks for input Björn! :)

Answer 1

[Bjorn Gustavsson]

If you have the symbolic toolbox this can be done:

syms E A0 L P c1 c2 c3 P x
e1=E*A0*x * (c1*(2 - (x/L)) + c2*(4*x - 2* (x^2) *(1/L)) + c3*(6*(x^2) - 3*(x^3)*(1/L)) - P);
F1=int(e1, L ,0) 
% Returns: 
% F1 =
% 
% -(A0*E*L^2*(27*c3*L^2 + 25*c2*L - 15*P + 20*c1))/30

The symbolic toolbox should also be able to solve the 3 eqs for c1, c2, and c3.

But considering that you have 3 polynomials in x, you should be able to calculate the integrals by hand easily.

HTH

[Walter Roberson]

I recommend specifying the variable of integration explicitly, as there are four variables in e1 and it is clearer to specify the variable of integration instead of requiring that the person reading the code be completely certain about the procedure for chosing the default variable.

[Daniel Arvidsson]

Very true, Walter, thanks for input :)