do not understand the integration result

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cemile basgul 2020-4-21

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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 2020-4-29

编辑：darova 2020-4-29

在 MATLAB Online 中打开

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 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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Answer 1

Nishant Gupta 2020-5-27

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https://ww2.mathworks.cn/matlabcentral/answers/519677-do-not-understand-the-integration-result#answer_440018

在 MATLAB Online 中打开

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