Simpson's rule in Python

Question

For a numerical methods class, I need to write a program to evaluate a definite integral with Simpson\'s composite rule. I already got this far (see below), but my answer is

Answer 1

There is my code (i think that is the most easy method). I done this in jupyter notebook. The easiest and most accurate code for Simpson method is 1/3.

Explanation

For standard method (a=0, h=4, b=12) and f=100-(x^2)/2

We got: n= 3.0, y0 = 100.0, y1 = 92.0, y2 = 68.0, y3 = 28.0,

So simpson method = h/3*(y0+4*y1+2*y2+y3) = 842,7 (this is not true). Using 1/3 rule we got:

h = h/2= 4/2= 2 and then:

n= 3.0, y0 = 100.0, y1 = 98.0, y2 = 92.0, y3 = 82.0, y4 = 68.0, y5 = 50.0, y6 = 28.0,

Now we calculate the integral for each step (n=3 = 3 steps):

Integral of the first step: h/3*(y0+4*y1+y2) = 389.3333333333333

Integral of the second step: h/3*(y2+4*y3+y4) = 325.3333333333333

Integral of the third step: h/3*(y4+4*y5+y6) = 197.33333333333331

Sum all, and we get 912.0 AND THIS IS TRUE

x=0
b=12
h=4
x=float(x)
h=float(h)
b=float(b)
a=float(x)
def fun(x): 
    return 100-(x**2)/2
h=h/2
l=0  #just numeration
print('n=',(b-x)/(h*2))
n=int((b-a)/h+1)
y = []   #tablica/lista wszystkich y / list of all "y"
yf = []
for i in range(n):
    f=fun(x)
    print('y%s' %(l),'=',f)
    y.append(f)
    l+=1
    x+=h
print(y,'\n')
n1=int(((b-a)/h)/2)  
l=1
for i in range(n1):
    nf=(h/3*(y[+0]+4*y[+1]+y[+2]))
    y=y[2:]  # with every step, we deleting 2 first "y" and we move 2 spaces to the right, i.e. y2+4*y3+y4
    print('Całka dla kroku/Integral for a step',l,'=',nf)
    yf.append(nf)
    l+=1
print('\nWynik całki/Result of the integral =', sum(yf) )

At the beginning I added simple data entry:

d=None
while(True):
    print("Enter your own data or enter the word "test" for ready data.\n")
    x=input ('Enter the beginning of the interval (a): ') 
    if x == 'test':
        x=0
        h=4  #krok (Δx)
        b=12 #granica np. 0>12  
        #w=(20*x)-(x**2)  lub   (1+x**3)**(1/2)
        break
    h=input ('Enter the size of the integration step (h): ')
    if h == 'test':
        x=0
        h=4 
        b=12 
        break
    b=input ('Enter the end of the range (b): ')
    if b == 'test':
        x=0
        h=4  
        b=12 
        break 
    d=input ('Give the function pattern: ')
    if d == 'test':
        x=0
        h=4  
        b=12
        break
    elif d != -9999.9:
        break

x=float(x)
h=float(h)
b=float(b)
a=float(x)

if d == None or d == 'test':
    def fun(x): 
        return 100-(x**2)/2 #(20*x)-(x**2)
else:
    def fun(x): 
        w = eval(d)
        return  w
h=h/2
l=0  #just numeration
print('n=',(b-x)/(h*2))
n=int((b-a)/h+1)
y = []   #tablica/lista wszystkich y / list of all "y"
yf = []
for i in range(n):
    f=fun(x)
    print('y%s' %(l),'=',f)
    y.append(f)
    l+=1
    x+=h
print(y,'\n')
n1=int(((b-a)/h)/2)  
l=1
for i in range(n1):
    nf=(h/3*(y[+0]+4*y[+1]+y[+2]))
    y=y[2:]
    print('Całka dla kroku/Integral for a step',l,'=',nf)
    yf.append(nf)
    l+=1
print('\nWynik całki/Result of the integral =', sum(yf) )