integrate with the upper limit is a function

Hello!

How to integrate a function on Matlab with the upper limit is a function, for exemple the following integral:

i tried this code:

U= vpa(int(f,s,0,t^2))

but it returns int(f,s,0,t^2)!!

is there any method to do it?

[Star Strider]

The Symbolic Math Toolbox appropriately considers an otherwise undefined function.

Define it and you get an actual result —

syms s t 
f(s) = 1 / (s^2 + 2*s + 1)

f(s) = 

U = vpa(int(f,s,0,t^2))

U = 

.

[L.GREEN]

there is the code i used:

syms s t

f=sin(0.008414709848078965066525023216303*exp(-2*s^(1/2)) + 0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303) + 1

U= vpa(int(f,s,0,t^2))

the code returns

numeric::int(sin((4850749969438627*exp(-2*s^(1/2)))/576460752303423488 + (3104479980440721*cos(s)^2)/36893488147419103232 + 4850749969438627/576460752303423488) + 1, s == 0..t^2)

Why i dont have the result of the integral?

[Torsten]

Why i dont have the result of the integral?

MATLAB cannot compute an analytical antiderivative of f. And such an antiderivative would be necessary because the result is not a numerical value, but a function of the symbolic variable t.

[L.GREEN]

Is there any other alternative to do it on Matlab? or to obtain the result of the integral as a function of the symbolic variable t?

[Torsten]

Is there any other alternative to do it on Matlab?

If you want to keep t symbolic: no.

If you give t numerical values, you can use "integral" for evaluation.

f = @(s) sin(0.008414709848078965066525023216303*exp(-2*s.^(1/2)) + 0.00008414709848078965066525023216303*cos(s).^2 + 0.008414709848078965066525023216303) + 1;
F = @(t)integral(f,0,t^2);
tnum = 0:0.1:5;
Fnum = arrayfun(@(t)F(t),tnum);
plot(tnum,Fnum)
grid on

[Star Strider]

The int function is not doing the integration (or the subsequent substitution). Not all functions have analytic integrals, and yours is apparently one of them.

syms s t
f(s) = sin(0.008414709848078965066525023216303*exp(-2*s^(1/2)) + 0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303) + 1

f(s) = 

U1 = int(f,s)

U1(s) = 

% U = vpa(int(f,s,0,t^2))
% U = simplify(U, 100)

Consider instead integrating this numerically if you want a numeric result, using the integral function.

.

[L.GREEN]

The inconvient of the intergal function is that we need to use '.^ 'and/or './'

and I obtaint the function f after several calculus on Matlab the result I obtient is without '.^ 'and/or './'

[Walter Roberson]

If you have working with a mix of scalar constants and scalar symbolic variables, then you can use ^ and / instead of .^ and ./ . You only need .^ and ./ if you are working with non-scalar values.

syms s t
f1 = sin(0.008414709848078965066525023216303*exp(-2*s^(1/2)) + 0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303) + 1

f1 = 

f2 = sin(0.008414709848078965066525023216303*exp(-2*s.^(1/2)) + 0.00008414709848078965066525023216303*cos(s).^2 + 0.008414709848078965066525023216303) + 1

f2 = 

isAlways(f1 == f2)

ans = logical
   1

[Torsten]

The inconvient of the intergal function is that we need to use '.^ 'and/or './'

If you have a symbolic function and want to convert it to a function that can be used together with "integral", use "matlabFunction". Here, you will also get .^, ./ and .* operators.

Another way is to use the 'ArrayValues',true option for "integral" that passes one value for s to the function at a time. The disadvantage is that this option might slow down the integration:

f = @(s) sin(0.008414709848078965066525023216303*exp(-2*s^(1/2)) + 0.00008414709848078965066525023216303*cos(s)^2 + 0.008414709848078965066525023216303) + 1;
F = @(t) integral(f,0,t^2,'ArrayValued',1);
tnum = 0:0.1:5;
Fnum = arrayfun(@(t)F(t),tnum);
plot(tnum,Fnum)
grid on