why I am getting 'array indices must be positive intergers or logical value' error?

a=1.4492;

b=-3.4497;

n=100;

h=(b-a)/n;

f(a)=integ_trape(a);

f(b)=integ_trape(b);

r=0.5*(f(a)+f(b));

for i=1:1:n-1

r=r+f(a+i*h);

break;

I=h*r;

end

Answer 1

[Star Strider]

Some of the posted code makes no sense, especially the break call in the for loop, since that will result in only the first value of ‘r’ being calculated, and nothing else.

See if this is closer to what you want:

a=1.4492;
b=-3.4497;
n=100;
h=(b-a)/n;
f = @(q) integ_trape(q);
r=0.5*(f(a)+f(b));
for i=1:1:n-1
    r = r+f(a+i*h);
%     break;
    I(i) = h*r;
end

I

I = 1×99

-0.0117 -0.0348 -0.0691 -0.1143 -0.1702 -0.2366 -0.3132 -0.3998 -0.4961 -0.6020 -0.7172 -0.8414 -0.9744 -1.1160 -1.2660 -1.4240 -1.5900 -1.7635 -1.9445 -2.1327 -2.3278 -2.5296 -2.7378 -2.9523 -3.1728 -3.3990 -3.6308 -3.8679 -4.1100 -4.3569

function fun3=integ_trape(x)
fun3=-x^2-2*x+5;
end

The ‘f’ anonymous function is also not necessary (it just calls ‘integ_trape’) however I kept it in for convenience.

.

[Raikamal Mukherjee]

thank you!I am beginner in MATLAB,stil can't understand why

f(a)=integ_trape(a);

did'nt work as I called function in bisection method and it did work.

a=1;
b=2;
n=100;
tol=0.0001;
fa=func_rootfind_1(a);
fb=func_rootfind_1(b);
if(fa.*fb < 0)
    for i=1:1:n;
        c=(a+b)./2;
        fa=func_rootfind_1(a);
        fb=func_rootfind_1(b);
        fc=func_rootfind_1(c);
        if(abs(fc) < tol || abs(a-b) < tol)
            root=c;
            %i=n+1;
            break;
        else
            if(fa.*fc < 0)
                b=c;
            else
                a=c;
            end
            %i=i+1;
        end
    end
else
    root=100000;
end
root         


function fun1=func_rootfind_1(x)
z=x^3-x-1;
fun1=z

[Star Strider]

As always, my pleasure!