# my equation has 3 variables but i want to integrate with respect to only one variable ,so that i can optimize the ramaining two from the resulting equation

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my equation(looks big) is as follows , which has three variables (d,e,t) I want to integrate with t from 0 to 2000 (by keeping d,e constant ) so that i can optimize those two using any optimization method . i used int(err,0,2000) command but gettting no result .so how to get past this error?
the eqaution at its basic term is in form 1483/(k-1) , where K=function(d,e,t)
err =
1483/(200*(0.99966444607127868948737159371376*d - 99966.44460712373256683349609375*d*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) + 0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) - 0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + 15567.003559999167919158935546875*e*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) + 0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) - 0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + 919819.0186288356781005859375*d*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) + 0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) - 0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) - 143208.7303999960422515869140625*e*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) + 0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) - 0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) - 79516.5081846714019775390625*d*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) + 0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) - 0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) + 6144.9373500002548098564147949219*e*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) + 0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) - 0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) - 1))

Alan Weiss on 27 Jul 2021
This seems like a solution. Of course, I don't know what your real bounds are on d and e, so I just guessed.
Notice that I used ./ in the first line of the function myfun. That is because the integral function requires vectorized inputs.
lb = [1/2 1/2];
ub = [3/4 3/4];
[sol,fval,eflag,output] = fmincon(@minfn,[1/2 1/2],[],[],[],[],lb,ub)
Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
sol = 1×2
0.7500 0.7500
fval = -5.8613e+04
eflag = 1
output = struct with fields:
iterations: 4 funcCount: 15 constrviolation: 0 stepsize: 2.2605e-04 algorithm: 'interior-point' firstorderopt: 0.0757 cgiterations: 0 message: '↵Local minimum found that satisfies the constraints.↵↵Optimization completed because the objective function is non-decreasing in ↵feasible directions, to within the value of the optimality tolerance,↵and constraints are satisfied to within the value of the constraint tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 3.264963e-07,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.↵↵' bestfeasible: [1×1 struct]
function minit = minfn(x)
d = x(1);
e = x(2);
minit = intval(d,e);
end
function r = intval(d,e)
r = integral(@(t)myfun(d,e,t),0,2000);
end
function val = myfun(d,e,t)
val = 1483./(200*(0.99966444607127868948737159371376*d - ...
99966.44460712373256683349609375*d*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) +...
0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) -...
0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) + ...
15567.003559999167919158935546875*e*(0.0000000061953699799058132798891540549135*exp(-0.031830333683444479980079178105257*t) +...
0.000010273947474151098879779908656928*exp(-9.1132612523704400742108333588476*t) -...
0.00000028014284413259036773946597520535*exp(-0.21190841394611541767881157660725*t)) +...
919819.0186288356781005859375*d*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) +...
0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) -...
0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) -...
143208.7303999960422515869140625*e*(0.00000019463729288918805472585749072323*exp(-0.031830333683444479980079178105257*t) +...
0.0000011273623338164517199144754044937*exp(-9.1132612523704400742108333588476*t) -...
0.0000013219996267057210898032693080495*exp(-0.21190841394611541767881157660725*t)) -...
79516.5081846714019775390625*d*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) +...
0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) -...
0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) +...
6144.9373500002548098564147949219*e*(0.0000061148367096900205219789370403305*exp(-0.031830333683444479980079178105257*t) +...
0.00000012370569685174297169805157636802*exp(-9.1132612523704400742108333588476*t) -...
0.0000062385424065411129723734973140381*exp(-0.21190841394611541767881157660725*t)) - 1));
end
Alan Weiss
MATLAB mathematical toolbox documentation
bhanu kiran vandrangi on 27 Jul 2021
thank you for the solution
i want the equation after integration so that i can use that as a cost function in particleswarm command(PSO) with 0,1 as bounds for bothd,e

R2020b

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