The Simpson's 1-3 (h / 3) method in Matlab

The C code that finds the following integral according to the Simpson's 1-3 (h / 3) method is given below. Fill in the blanks on the code appropriately.

I want to solve this question below in Matlab but i didn't do it. This is simple question but i can't do it. If someone will help me, i will be very happy.

C code version:

I tried this code block in Matlab:

% Ask for user input

% Lower bound (a)

a = input('What is your lower bound (a)?')

% Upper bound (b)

b = input('What is your upper bound (b)?')

% Subintervals

N = input('How many subintervals (N)?')

% Defining function

f = @(x,e) (e*x+sin(x))

% Finding h

h=(b-a)/N;

% Finding the values of x for each interval

x=linspace(a,b,N);

% Calculating the integral

for i = 1:N-1

I(i)= (h/3)*(f(x(i))+(4*f((x(i)+x(i+1))/2))+f(x(i+1)));

end

answer1 = sum(I)

disp(I)

% adding f(b) to sum

val2=ff(length(xx));

sum=val1+val2+sum;% set sum

% set result

result=sum*h/3;

Answer 1

[Steven Lord]

% Ask for user input
% Lower bound (a)
a = input('What is your lower bound (a)?')
% Upper bound (b)
b = input('What is your upper bound (b)?')
% Subintervals
N = input('How many subintervals (N)?')
% Defining function
f = @(x,e) (e*x+sin(x))

You've defined your function to accept two inputs but ...

% Finding h
h=(b-a)/N; 
% Finding the values of x for each interval
x=linspace(a,b,N);
% Calculating the integral
for i = 1:N-1
I(i)= (h/3)*(f(x(i))+(4*f((x(i)+x(i+1))/2))+f(x(i+1)));

you call your function with just one. Since the first term in f is supposed to compute the exponential of x, that's not the right way to define it.

f = @(x) exp(x)+sin(x); % Added by Steve L

Now this f matches the problem statement and it can be (must be) called with just one input argument.

end
answer1 = sum(I)

Since this is a homework assignment I would define answer1 to be 0 before the loop and write the body of the loop closer to the way the C code does, just adding to answer1 each time. The mod function will be of use to you if you do. Keep in mind that MATLAB indexing is 1-based unlike C's 0-based indexing.

disp(I)
% adding f(b) to sum
val2=ff(length(xx));

Typo. ff should be f, and you shouldn't call it with length(x) as input. You should call it with the last element of x, x(end).

sum=val1+val2+sum;% set sum
% set result
result=sum*h/3;

You can't use sum both as a variable and as a function in the same script. Change the two instances of sum in the last line to answer1. In addition, you haven't defined a variable val1 in your code.