如何使用循环语句求积分上限未知数。

目的：通过以上画圈方程，求 I*与b的关系。

f(x)与g(x)，分别服从N~（5，0.25），N~（4，0.16）的正态分布。

已将其余的参数赋值，因此自己编写的代码为（代码中I*由t代表）：

syms b double

r=1:1:6;

result=zeros(size(r));

for i=1:length(r)

t=r(i);

fun_1= @(x) 0.5.*(x-b).*normpdf(x,5,0.5);

fun_2= @(x) x.*normpdf(x,5,0.5);

fun_3= @(y) y.*normpdf(y,4,0.4);

fun_4= @(x) (b+0.5.*(x-b)).*normpdf(x,5,0.5);

fun_5= @(y) (b+0.4.*(y-b)).*normpdf(y,4,0.4);

fun_6= @(y) 0.6.*(y-b).*normpdf(y,4,0.4);

fun_7=integral(fun_1,b,6.5).*100.*(8-t)./8.*(1-(integral(fun_2,0,b)-integral(fun_3,0,b)+integral(fun_4,b,6.5)-integral(fun_5,b,5.2))./0.5)...

+integral(fun_6,b,5.2).*(100.*(4.5.*(integral(fun_2,0,b)+integral(fun_4,b,6.5))-5.*(integral(fun_3,0,b)+integral(fun_5,b,5.2)))./2.25)...

+12.5.*t.*(1-(integral(fun_2,0,b)-integral(fun_3,0,b)+integral(fun_4,b,6.5)-integral(fun_5,b,5.2))./0.5)-400;

[b,fval] = fzero(fun_7,0.1);

double(b)

end

运行结果：Error using integral (line 85)

A and B must be floating-point scalars.

错误与问题较多，请教高手前辈，十分感谢！

Answer 1

[vubaen]

syms r0 double

r=1:1:6;

result=zeros(size(r));

for i=1:length(r)

t=r(i);

fun_1= @(x) x.*normpdf(x,5,0.5);

fun_2= @(x) normpdf(x,5,0.5);

fun_3= @(y) y.*normpdf(y,4,0.4);

fun_4= @(y) normpdf(y,4,0.4);

pfa = @(r0) arrayfun(@(r0) 0.5.*(integral(fun_1,r0,6.5)-r0.*integral(fun_2,r0,6.5)).*100.*(8-t)./8.*(1-(integral(fun_1,0,r0)-integral(fun_3,0,r0)+0.5.*(integral(fun_1,r0,6.5)-r0.*integral(fun_2,r0,6.5))-(0.4.*integral(fun_3,r0,5.2)+0.6.*r0.*integral(fun_4,r0,5.2)))./0.5)...

+(0.6.*integral(fun_3,r0,5.2)-0.6.*r0.*integral(fun_4,r0,5.2)).*(100.*(4.5.*(integral(fun_1,0,r0)+0.5.*integral(fun_1,r0,6.5)+r0.*0.5.*integral(fun_2,r0,6.5))-5.*(integral(fun_3,0,r0)+0.4.*integral(fun_3,r0,5.2)+r0.*0.6.*integral(fun_4,r0,5.2)))./2.25)...

+12.5.*t.*(1-(integral(fun_1,0,r0)-integral(fun_3,0,r0)+0.5.*(integral(fun_1,r0,6.5)-r0.*integral(fun_2,r0,6.5))-(0.4.*integral(fun_3,r0,5.2)+0.6.*r0.*integral(fun_4,r0,5.2)))./0.5)-400, r0);

r0 = fzero(pfa,0);

result(i)=vpa(r0);

end

result(i)