Here is the homework assignment I'm trying to solve:

A further improvement of the approximate integration method from the last question is to divide the area under the f(x) curve into n equally-spaced trapezoids.

Based on this idea, the following formula can be derived for approximating the integral:

!(https://www.dropbox.com/s/q84mx8r5ml1q7n1/Screenshot%202017-10-01%2016.09.32.png?dl=0)!

where h is the width of the trapezoids, h=(b−a)/n, and xi=a+ih,i∈0,...,n, are the coordinates of the sides of the trapezoids. The figure above visualizes the idea of the trapezoidal rule.

Implement this formula in a Python function trapezint( f,a,b,n ). You may need to check and see if b > a, otherwise you may need to swap the variables.

For instance, the result of trapezint( math.sin,0,0.5*math.pi,10 ) should be 0.9979 (with some numerical error). The result of trapezint( abs,-1,1,10 ) should be 2.0

This is my code but It doesn't seem to return the right values.
For print ((trapezint( math.sin,0,0.5*math.pi,10)))
I get 0.012286334153465965, when I am suppose to get 0.9979
For print (trapezint(abs, -1, 1, 10))
I get 0.18000000000000002, when I am suppose to get 1.0.

import math
def trapezint(f,a,b,n):g = 0if b>a:h = (b-a)/float(n)for i in range (0,n):k = 0.5*h*(f(a+i*h) + f(a + (i+1)*h))g = g + kreturn gelse:a,b=b,ah = (b-a)/float(n)for i in range(0,n):k = 0.5*h*(f(a + i*h) + f(a + (i + 1)*h))g = g + kreturn gprint ((trapezint( math.sin,0,0.5*math.pi,10)))
print (trapezint(abs, -1, 1, 10))

Essentially, your return g statement was indented, when it should not have been.

Also, I removed your duplicated code, so it would adhere to "DRY" "Don't Repeat Yourself" principle, which prevents errors, and keeps code simplified and more readable.

import math
def trapezint(f, a, b, n):g = 0if b > a:h = (b-a)/float(n)else:h = (a-b)/float(n)for i in range (0, n):k = 0.5 * h * ( f(a + i*h) + f(a + (i+1)*h) )g = g + kreturn gprint ( trapezint( math.sin, 0, 0.5*math.pi, 10) )
print ( trapezint(abs, -1, 1, 10) )

0.9979429863543573
1.0000000000000002