# How can I plot integral function for fun3 ?

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laila elatrash on 4 Nov 2019
Commented: laila elatrash on 4 Nov 2019
fun3 = fun1 - fun 2 , where;
fun1= @(x1) (i*w*x1./U)*(-1./r.^3)+(3*(x1*x1)./(r.^5));
fun2=fun2= @(x1) exp(i*w*x1./U).*(((k2.^2)*x1*x1./r.^3)*exp(-k2*r+k*x1)-(k2*k*x1*1./(r.^2))*exp(-k2*r+k*x1)-(k2*1./(r.^2))*exp(-k2*r+k*x1)...
+(2*k2*x1*x1./(r.^4))*exp(-k2*r+k*x1)-(k2*k*x1*1./(r.^2))*exp(-k2*r+k*x1)+((k.^2)*1./r)*exp(-k2*r+k*x1)...
-(k*1*x1./(r.^3))*exp(-k2*r+k*x1)-(k2*x1*x1./(r.^4))*exp(-k2*r+k*x1)+(k*x1*1./(r.^3))*exp(-k2*r+k*x1)...
+(1./(r.^3))*exp(-k2*r+k*x1)-(3*x1*x1./(r.^5))*exp(-k2*r+k*x1));
where: x2=0; ru=1; mu=1; w=0.01; U=1;
r = sqrt((x1).^2 + (x2).^2 + (x3).^2); k2 =sqrt((k.^2)+(h.^2)); h =sqrt(-i*ru*w./mu);

Nadir Altinbas on 4 Nov 2019
at the first stage you should define a boundry
laila elatrash on 4 Nov 2019
you means interval values integral?
[a,b] = [-inf , x1]
is it what you means?

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