do not understand the integration result

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cemile basgul
cemile basgul 2020-4-21
评论： cemile basgul 2020-6-4
when I integrate this via: int(1/4624*exp(12028*(2367-t/643*t+2199060)),t,0,360)
I get this result: 1933501^(1/2)*pi^(1/2)*erf((720*1933501^(1/2))/643)*exp(26478763956))/55617472
What does this number mean? why am I getting a value like this?
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cemile basgul
cemile basgul 2020-4-29
编辑：darova 2020-4-29
Not to be mistaken by the paranthesis, I wrote variable numbers as below. However, I am still getting those long numbers such as
This is for F(1)
(538948486624482741*exp(-4579610173792908711849542041234325/912795614384412005898158547664896))/39076167421235083376000 - (1851567729550599*exp(-28376783941740538497326829612455/3135926614906861015272882438144))/156304669684940333504 - (6612410559153235*exp(33062052795766175/1693660223635456)*ei(-52899284473225880/1851567729550599))/19538083710617541688 + (6612410559153235*exp(33062052795766175/1693660223635456)*ei(-13224821118306470000/538948486624482741))/19538083710617541688
A=4624;
E=10^5;
R=8.314; %Universal gas constant
Tref=616.15; %Melting point of PEEK in K (343C)
T=T0; %Temperature of constant points
syms t
for m=1:101
a(m)=sum(Temperature{m}>=485) ; %constant degree of healing points
b(m)=sum(Temperature{m}<485); %non-isothermal degree of healing points
timea(m)=a(m).*0.003;
timeb(m)=b(m).*0.003;
c(m)=(Temperature{m}(a(m)+1)-Temperature{m}(it(m)))/(time(m)-timea(m)); %slope of the curve
d(m)=Temperature{m}(a(m)+1)-(c(m)*timea(m)); %constant for the linear line equation
F(m)=int(1/(A*exp((E/R)*((1/(c(m)*t+d(m)))-(1/Tref)))),t,timea(m),time(m));
end
cemile basgul
cemile basgul 2020-4-29
If I use Integral instead of Int, it works, I mean gives a number value such as 3.4059e-09
A=4624;
E=10^5;
R=8.314; %Universal gas constant
Tref=616.15; %Melting point of PEEK in K (343C)
T=T0; %Temperature of constant points
syms t
p=NaN(1,101);
for m=1:101
a(m)=sum(Temperature{m}>=485) ; %constant degree of healing points
b(m)=sum(Temperature{m}<485); %non-isothermal degree of healing points
timea(m)=a(m).*0.003;
timeb(m)=b(m).*0.003;
c(m)=(Temperature{m}(a(m)+1)-Temperature{m}(it(m)))/(time(m)-timea(m)); %slope of the curve
d(m)=Temperature{m}(a(m)+1)-(c(m)*timea(m)); %constant for the linear line equation
F=integral(@(t)(1./(A.*exp((E./R)*((1./(c(m).*t+d(m)))-(1./Tref))))),timea(m),time(m));
p(m)=F^(1/4);
end

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回答（1 个）

Nishant Gupta
Nishant Gupta 2020-5-27
If you use vpa function for evaluation, the result will be inf since its a very large number.
syms t;
>> expr = 1/4624*exp(12028*(2367-t/643*t+2199060));
>> F = int(expr,[0 360])
F =
(1933501^(1/2)*pi^(1/2)*erf((720*1933501^(1/2))/643)*exp(26478763956))/55617472
>> vpa(F)
ans =
Inf
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cemile basgul
cemile basgul 2020-6-4
do you know why int and integral gives different results for this? I am not sure which one to use.

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