how to find integral

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bhanu kiran vandrangi
bhanu kiran vandrangi on 3 Aug 2021
Answered: Walter Roberson on 4 Aug 2021
74513/(10000*sign(0.99966444607707671821117401123047*e + 15567.00356006622314453125*d*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 99966.4446048736572265625*e*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 143208.73040008544921875*d*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 919819.01861572265625*e*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 6144.93734991550445556640625*d*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 79516.50818634033203125*e*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 1)*(0.99966444607707671821117401123047*e + 15567.00356006622314453125*d*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 99966.4446048736572265625*e*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 143208.73040008544921875*d*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 919819.01861572265625*e*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 6144.93734991550445556640625*d*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 79516.50818634033203125*e*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 1))
my equation looks big but it is in fractional form , how to integrate (from 0 to 2000)this w.r.t 't' alone and by keeping 'd' and 'e' as constant
  2 Comments
Anna Case
Anna Case on 3 Aug 2021
If 'd' and 'e' are defined, you can use numerical integration.
bhanu kiran vandrangi
bhanu kiran vandrangi on 4 Aug 2021
d and e are variables which vary from [0,1] , i need the equation that comes after integration as a function of d,e so that i can optimize the equation to find the suitable values of d,e in that range[0.1]

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Answers (1)

Walter Roberson
Walter Roberson on 4 Aug 2021
First let us enter the equation in without losing precision
S = '74513/(10000*sign(0.99966444607707671821117401123047*e + 15567.00356006622314453125*d*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 99966.4446048736572265625*e*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 143208.73040008544921875*d*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 919819.01861572265625*e*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 6144.93734991550445556640625*d*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 79516.50818634033203125*e*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 1)*(0.99966444607707671821117401123047*e + 15567.00356006622314453125*d*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 99966.4446048736572265625*e*(0.000000006195369979837203611161555727449*exp(-0.031830333683444479966526650949188*t) + 0.000010273947474104261345928534865379*exp(-9.113261252370440103050611146962*t) - 0.00000028014284413763190784152357082348*exp(-0.21190841394611541708927664531825*t)) - 143208.73040008544921875*d*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 919819.01861572265625*e*(0.00000019463729288710096554382289468776*exp(-0.031830333683444479966526650949188*t) + 0.0000011273623338192706455629377160221*exp(-9.113261252370440103050611146962*t) - 0.0000013219996267133105050106678390875*exp(-0.21190841394611541708927664531825*t)) + 6144.93734991550445556640625*d*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 79516.50818634033203125*e*(0.0000061148367096475197968175052665174*exp(-0.031830333683444479966526650949188*t) + 0.0000001237056968493577269185834666132*exp(-9.113261252370440103050611146962*t) - 0.0000062385424065558581219193001743406*exp(-0.21190841394611541708927664531825*t)) - 1))';
SS = str2sym(S)
SS = 
Now let us get an idea of how the equation looks over a range of d, e, t values, to see if integration is likely to work
syms d e t
N = 40;
tvec = linspace(0,2000,N);
[D, E, T] = meshgrid(linspace(0,1,N),linspace(0,1,N),tvec);
Is = double(subs(SS,{d,e,t},{D,E,T}));
minis = min(Is(:)); maxis = max(Is(:));
[minis, maxis]
ans = 1×2
1.0e+04 * 0.0007 2.2206
levels = linspace(minis, maxis, 5);
for K = 1 : length(levels)
isosurface(D, E, T, Is, levels(K));
end
xlabel('d'); ylabel('e'); zlabel('t');
legend(string(levels));
colorbar()
Unfortunately at this angle, the yellow is hidden, but we know it exists somewhere.
That back wall... let's get an idea of what the function looks like there
T1 = linspace(0,2000,5000);
I11 = double(subs(SS, {d,e,t}, {1, 1, T1}));
plot(T1, I11)
Looks like it settles in near 22000 for each value, so the integral would be roughly 22000 * 1700 or so, on the order of 4e7
We can do a numeric integration over t, over a range of d and e values
IntI = trapz(tvec, Is, 3);
surf(D(:,:,1), E(:,:,1), IntI); xlabel('d'); ylabel('e'); zlabel('integral')
4E7 is pretty much what we estimated just a moment ago.
This tells us that to maximize the integral, we should concentrate on e near 1; it is not obvious at the moment whether d has much effect.
I would suggest redoing a plot using e = 1 (constant) and using a range of d and using a denser range of t values, and using trapz() to do numeric integration; this will give you better information about which range is worth looking more closely at.

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