Solve indefinite integral with unknown lower limit

2022 3 5
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更新时间：2022 3 5
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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

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David Hill
David Hill 2022-3-5
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
Jessica Wan 2022-3-5
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
Walter Roberson 2022-3-5
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
Jessica Wan 2022-3-5
@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.

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Matt J
Matt J 2022-3-5
编辑：Walter Roberson 2022-3-5
Ran in:

0 个投票

Ran in:
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

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