'integral' with piecewise expressions

Question

anto 2023-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/1893680-integral-with-piecewise-expressions

评论： anto 2023-1-15

I have the following expression for the angle Beta:

L = 1;
syms y 
Beta = piecewise(0<=y<L/2, pi/2, ...
    L/2 <=y <L*(3/4),pi/2+0.3491, ...
    (3/4)*L <=y <=L,pi/2-0.3491);

i want to compute this integral:

    a = pi/4;
    fun_f=@(y) (1./(50*(1+y)*(cos(a)*cos(Beta)+sin(a)*sin(Beta))+470));
     
    int= integral(fun_f,0,L)
Error using integralCalc/finalInputChecks
Input function must return 'double' or 'single' values. Found 'sym'.

Error in integralCalc/iterateScalarValued (line 315)
                finalInputChecks(x,fx);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 87)
Q = integralCalc(fun,a,b,opstruct);

Does anybody know how to calculate the integral with using integral, but with Beta defined as a piecewise function?

I looked for documentation for piecewise but I don't really know if it's needed to use it. I'd prefer not to.

If you need further information i'll be happy to provide it in order to solve my problem

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Answer 1

Torsten 2023-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/1893680-integral-with-piecewise-expressions#answer_1148210

编辑：Torsten 2023-1-13

在 MATLAB Online 中打开

L = 1;

Beta = @(y,L) pi/2.*((0<=y) & (y<L/2)) + (pi/2+0.3491).*((L/2<=y) & (y<3/4*L)) + (pi/2-0.3491).*((3/4*L<=y) & (y<=L));

figure(1)

plot((0:0.01:L),Beta(0:0.01:L,L))

a = pi/4;

fun_f=@(y,L) (1./(50*(1+y).*(cos(a)*cos(Beta(y,L))+sin(a)*sin(Beta(y,L)))+470));

figure(2)

plot((0:0.01:L),fun_f(0:0.01:L,L))

format long

int_value= integral(@(y)fun_f(y,L),0,L)

int_value =

0.001918690636037

int_value_improved = integral(@(y)fun_f(y,L),0,L/2) + integral(@(y)fun_f(y,L),L/2,3*L/4) + integral(@(y)fun_f(y,L),3/4*L,L)

int_value_improved =

0.001918689980576

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anto 2023-1-14

在 MATLAB Online 中打开

I tried implementing it on my code: but it gives me the error : 'Incorrect number or types of inputs or outputs for function 'cos'.'

LATO = 1;
intersezione_34LATO= (3/4)*LATO;
i = 1;
p = [0.125000000000000	0];
L_AB= 0.625000000000000;
alpha_1 = pi/2;
        Beta = @(l) (pi/2+0.3491).*(LATO/2<=(p(i,1)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i)) <intersezione_34LATO)) + (pi/2-0.3491).*(intersezione_34LATO<=(p(i,2)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i))<= LATO));
        fun_f=@(l) (1./(cos(Beta(l))));
        tau_f(i) = integral(@(l)fun_f(l),0,L_AB(i),'AbsTol',0,'RelTol',1e-20,'ArrayValued',false); 
Incorrect number or types of inputs or outputs for function 'cos'.

Error in solution (line 9)
        fun_f=@(l) (1./(cos(Beta(l))));

Error in solution (line 11)
        tau_f(i) = integral(@(l)fun_f(l),0,L_AB(i),'AbsTol',0,'RelTol',1e-20,'ArrayValued',false); 

Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 87)
Q = integralCalc(fun,a,b,opstruct);

I even tried with the syntax for Beta you suggested:

LATO = 1;
intersezione_34LATO= (3/4)*LATO;
i = 1;
p = [0.125000000000000	0];
L_AB= 0.625000000000000;
alpha_1 = pi/2;
   
        
        Beta = @(l,p,alpha_1,LATO) (pi/2+0.3491).*(LATO/2<=(p(i,1)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i)) <intersezione_34LATO)) + (pi/2-0.3491).*(intersezione_34LATO<=(p(i,2)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i))<= LATO));
                fun_f=@(l) (1./(cos(Beta(l,p,alpha_1,LATO))));
                
                  tau_f(i) = integral(@(l)fun_f(l),0,L_AB(i),'AbsTol',0,'RelTol',1e-20,'ArrayValued',false);

this is part of a for cycle. but the syntax is as you suggested, but for some reasons it still gives:

Incorrect number or types of inputs or outputs for function 'cos'.

Do you know why? Thanks

Torsten 2023-1-14

编辑：Torsten 2023-1-14

在 MATLAB Online 中打开

I doubt that Beta is what you want since the result is of type "logical".

What function do you want to use (in a mathematical notation) ?

But if you think everything is as wanted - here is the result:

LATO = 1;

intersezione_34LATO= (3/4)*LATO;

i = 1;

p = [0.125000000000000 0];

L_AB= 0.625000000000000;

alpha_1 = pi/2;

Beta = @(l,p,alpha_1,LATO) (pi/2+0.3491).*(LATO/2<=(p(i,1)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i)) <intersezione_34LATO)) + (pi/2-0.3491).*(intersezione_34LATO<=(p(i,2)+l.*sin(alpha_1(i)))) & ((p(i,2)+l.*sin(alpha_1(i))<= LATO));

fun_f=@(l) (1./(cos(double(Beta(l,p,alpha_1,LATO)))));

l=0:0.01:L_AB;

plot(l,fun_f(l))

tau_f(i) = integral(@(l)fun_f(l),0,L_AB(i))%,'AbsTol',0,'RelTol',1e-20,'ArrayValued',false)

tau_f = 0.8377

anto 2023-1-15

OK, thanks a lot. With double it works, have a good day

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Answer 2

Walter Roberson 2023-1-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1893680-integral-with-piecewise-expressions#answer_1148505

Use matlabFunction with the piecewise expression, giving the 'file' option and 'optimize' false. And when you integral specify 'arrayvalued' true

matlabFunction can convert piecewise to if/else but only when writing to file, and the result cannot accept vectors

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Walter Roberson 2023-1-14

在 MATLAB Online 中打开

format long g

L = 1;

syms y

Pi = sym(pi);

Beta = piecewise(0<=y<L/2, Pi/2, ...

L/2 <= y < L*(3/4), Pi/2 + sym(3491)/10^4, ...

(3/4)*L <= y <=L, Pi/2 - sym(3491)/10^4);

a = Pi/4;

fun_f = (1./(50*(1+y)*(cos(a)*cos(Beta)+sin(a)*sin(Beta))+470))

fun_f =

result_symbolic = int(fun_f, y, 0, L)

result_symbolic =

result_vpa = vpa(result_symbolic, 16)

result_vpa =

0.001918689980575797

fun_f_h = matlabFunction(fun_f, 'vars', y, 'File', 'fun_f.m', 'optimize', false)

fun_f_h = function_handle with value:

@fun_f

result_numeric = integral(fun_f_h, 0, L, 'arrayvalued', true)

result_numeric =

0.00191869063603731

dbtype fun_f.m

1 function fun_f = fun_f(y) 2 %FUN_F 3 % FUN_F = FUN_F(Y) 4 5 % This function was generated by the Symbolic Math Toolbox version 9.2. 6 % 14-Jan-2023 22:18:56 7 8 if ((y < 1.0./2.0) & (0.0 <= y)) 9 fun_f = 1.0./((sqrt(2.0).*(y.*5.0e+1+5.0e+1))./2.0+4.7e+2); 10 elseif ((y < 3.0./4.0) & (1.0./2.0 <= y)) 11 fun_f = 1.0./((y.*5.0e+1+5.0e+1).*((sqrt(2.0).*cos(pi./2.0+3.491e-1))./2.0+(sqrt(2.0).*sin(pi./2.0+3.491e-1))./2.0)+4.7e+2); 12 elseif ((y <= 1.0) & (3.0./4.0 <= y)) 13 fun_f = 1.0./((y.*5.0e+1+5.0e+1).*((sqrt(2.0).*cos(pi./2.0-3.491e-1))./2.0+(sqrt(2.0).*sin(pi./2.0-3.491e-1))./2.0)+4.7e+2); 14 else 15 fun_f = NaN; 16 end

anto 2023-1-15

Thanks a lot

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