expanded vs unexpanded polynomials in symbolic integrals

I was doing some simple kinematics equations in matlab and I was getting the wrong answer for a long time until I expanded my function before integration using expand(). Is there a reason for this?

This was my orginial code which outputs position as 144.3144 meters

time = 2; %sec
m = 0.183; %kg
g = 9.81; %m/s^2
syms t
T = 15 - 15 * (t-1)^2; %N
aT = T/m;
a = aT - g;
v = int(a);
x = int(v);
position = double(subs(x,t,time))

position = 144.3144

This is the code I ended up using with the added expand() function which outputs position as 89.6696 meters

time = 2; %sec
m = 0.183; %kg
g = 9.81; %m/s^2
syms t
T = 15 - 15 * expand((t-1)^2); %N
aT = T/m;
a = aT - g;
v = int(a);
x = int(v);
position = double(subs(x,t,time))

position = 89.6696

Answer 1

[Torsten]

Without giving the limits of integration, "int" returns an arbitrary antiderivative for the function you want to integrate. But you want a special one in your application.

time = 2; %sec
m = 0.183; %kg
g = 9.81; %m/s^2
syms t
T = 15 - 15 * (t-1)^2; %N
aT = T/m;
a = aT - g;
v = int(a,0,t);
x = int(v,0,t);
position = double(subs(x,t,time))

position = 89.6696

time = 2; %sec
m = 0.183; %kg
g = 9.81; %m/s^2
syms t
T = 15 - 15 * expand((t-1)^2); %N
aT = T/m;
a = aT - g;
v = int(a,0,t);
x = int(v,0,t);
position = double(subs(x,t,time))

position = 89.6696

Maybe even these results for position are wrong because they implicitly assume that you want x(0) = v(0) = 0.