Symbolic Integral Giving answer as infinity

Hi,

I am writing a function and need some help. I want to solve a symbolic integral, but it keeps outputting the answer as infinity. The integral is a function of x.

syms Es Ib Ic H L P x R2 R1

Vin5=@(x) (1/(E*Ic))

int(Vin5,0,H)

Answer 1

[Star Strider]

Note that the RHS of ‘Vin5’ is not a function of ‘x’. You have also not defined ‘E’.

The correct symbolic expression is:

syms Es Ib Ic H L P x R2 R1
Vin5(x) =  (1/(E*Ic))
int(Vin5,0,H)

however ‘E’ remains undefined, and will throw an error.

If you instead use:

Vin5(x) =  (1/(Es*Ic))

the integral is:

H/(Es*Ic)

That still does not solve the problem of ‘Vin5’ having no expressions in terms of ‘x’ on the RHS. However it will produce an appropriate result, if you are integrating a constant.

[Socheat Tun]

Hey Star Strider,

So I tried typing exactly what you put in and got this error:

Error using sym/subsindex (line 855). It seems that whenever I try to integral 1/(sym), it always inputs an error or inf as the answer. Do you know how to fix this?

[Star Strider]

Please post the complete code you used.

[Walter Roberson]

Check if you accidentally assigned to a variable named int

[Socheat Tun]

syms E a H G L P R2 R1 x

Ib=(2*a*L)/E

Ic=(a*H)/E

%Integration of virtual work equations for released structure

Vin1=@(x) (P/(4*E*Ic))*(-H+x);

dhreleased=int(Vin1,0,H)

Vin2=@(x) (P*L/4)*(-1/(E*Ic))

thetareleased=int(Vin2,0,H)

Output:

Ib =

(2*L*a)/E

Ic =

(H*a)/E

dhreleased =

-(H*P)/(8*a)

Vin2 =

function_handle with value:

@(x)(P*L/4)*(-1/(E*Ic))+0*x

thetareleased =

-Inf*sign(L*P)

[Daniel M]

If you implement the actual suggestion, you should not receive an error.

clear
clc

syms E a H G L P R2 R1 x
Ib=(2*a*L)/E;
Ic=(a*H)/E;
%Integration of virtual work equations for released structure
% Vin1=@(x) (P/(4*E*Ic))*(-H+x); % Don't do this.
Vin1(x) = (P/(4*E*Ic))*(-H+x);
dhreleased=int(Vin1,0,H)
% Vin2=@(x) (P*L/4)*(-1/(E*Ic)) % Don't do this.
Vin2(x) = (P*L/4)*(-1/(E*Ic));
thetareleased=int(Vin2,0,H)

[Star Strider]

I just now saw this.

I agree with Daniel M.

It is best to use symbolic functions — not anonymous functions — with symbolic operations (such as int).

Your (slightly edited) code, replacing the anonymous functions with symbolic functions:

syms E a H G L P R2 R1 x
Ib=(2*a*L)/E;
Ic=(a*H)/E;
%Integration of virtual work equations for released structure
Vin1(x) = (P/(4*E*Ic))*(-H+x);
dhreleased=int(Vin1,0,H)
Vin2(x) = (P*L/4)*(-1/(E*Ic))
thetareleased=int(Vin2,0,H)

with the functions defined as:

Vin1(x) =
-(P*(H - x))/(4*H*a)

Vin2(x) =
-(L*P)/(4*H*a)

and the integrations produce these results when I run your code:

dhreleased = 
-(H*P)/(8*a)

thetareleased =
-(L*P)/(4*a)

Note that ‘Vin2’ is a constant, and not a funciton of ‘x’.