hello how do I write this function without getting an error saying "variable K has an incorrect value".

my code is: And so far only the (K) is wrong.

N = 1000;
a= 0;
b= pi;
h = (b-a)/N;
x=a:h:b;
f = x.^2.*sin(x);
F = sum(f*h );

a= 1;
b= 2;
h = (b-a)/N;
x=a:h:b;
g =(x.^3.2 + 2.*x.^2.22 +x + 21)  ;
G = sum(g*h) ;

a=-3 ;
b= 3;
h = (b-a)/N;
x=a:h:b;
k =abs(x).*exp(-x).^2 ;
K = sum(k*h);

Answer 1

[KSSV]

a=-3 ;
b= 3;
h = (b-a)/N;
x=a:h:b;
k =abs(x).*exp(-x.^2) ;
K = sum(k*h);

[Reza Ghasemi]

it says code ran without an out put.

i dont know why it doesnt work.

[Reza Ghasemi]

hey KSSV it didn't work do you have any other solutions please??

thank you

[KSSV]

Change the line:

k =abs(x).*exp(-x).^2 ;

to

k =abs(x).*exp(-x.^2) ;

I got K value as 0.99987.

[Reza Ghasemi]

a=-3;
b=3;
h =(b-a)/N;
x=a:h:b;
k = abs(x).*exp(-x.^2) ;
K =sum(k*h);

[Reza Ghasemi]

this is what i did and it says code ran without output.

[KSSV]

What do you mean by ran without output?

[Reza Ghasemi]

[Reza Ghasemi]

the output i meant

[Reza Ghasemi]

[Reza Ghasemi]

these are the errors i get

Answer 2

[Image Analyst]

What is "the above technique"?

Try using trapz() instead of sum.

[Reza Ghasemi]

[Reza Ghasemi]

this is the technique we have to use, the (trapz) didn't work.

[Walter Roberson]

trapz is not going to agree with the required summation; the endpoints are handled differently.

trapz is effectively

[Reza Ghasemi]

so if i want to write the third equation, (K), wouldnt it be

k = abs(x).*exp(-x.^2) ;
K =sum(k*h);

[Walter Roberson]

No, K = which is not the same as

[Reza Ghasemi]

how do i write that? i don't know why KSSV gave me that answer.

[Reza Ghasemi]

uh my bad i gave different question sorry KSSV