Solve indefinite integral with unknown lower limit

The equation I am trying to solve is attached below,

and my Matlab code is

int(11.6 + 2*10^-3*T - 0.67*10^5*(1/T^2),T,T,1650)

The output I am getting is just "-inf", but it should not be like this, could anyone help me figure this out? Thanks

[David Hill]

Why don't you just solve it by hand, not a hard integral.

syms T t
I=int(11.6 + 2*10^-3*T - 0.67*10^5*(1/T^2),T,t,1650);

[Jessica Wan]

It is a lot easier to do it by hand, but it's for an assignment so I have to use Matlab unfortunately

And thank you for your help, the output I got is shown below

piecewise(0 < t, 1445605/66 - 67000/t - (t*(t + 11600))/1000, t <= 0, -Inf)

Do I only consider the part where 0 < t? But if I want to sub this answer somewhere else, how should I specify this?

[Walter Roberson]

If T were 0 then 1/T^2 would be 1/0 which is a problem. If T were negative then T would have to cross 0 on its way to the positive bound, and you would have infinity again.

So... you should consider putting an assumption of positive on your variable. That would allow int() to generate a plain formula.

[Jessica Wan]

@Walter Roberson Thank you for your suggestion. I tried to put a statement before it

t = abs(t);

And now the final answer is

1445605/66 - 67000/abs(t) - abs(t)^2/1000 - (58*abs(t))/5

I assume we can just treat this abs(t) as normal t? Or do you have any better suggestions? Thanks.

Answer 1

[Walter Roberson]

syms t positive

Answer 2

[Matt J]

Looks like there are 2 solutions.

format long g
T0=1650;

rhs=1e-3*T0^2 + 11.66*T0 +  0.67e5/T0;

p=[1e-3,11.66, -rhs, 0.67e5];

r=roots(p);

T=r(r>0 & imag(r)==0)'

T = 1×2

1.0e+00 * 1650 3.05009448006937