Integration Exercise 2

Imports



In [1]:

    
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns
from scipy import integrate

Indefinite integrals

Here is a table of definite integrals. Many of these integrals has a number of parameters $a$, $b$, etc.

Find five of these integrals and perform the following steps:

Typeset the integral using LateX in a Markdown cell.
Define an integrand function that computes the value of the integrand.
Define an integral_approx funciton that uses scipy.integrate.quad to peform the integral.
Define an integral_exact function that computes the exact value of the integral.
Call and print the return value of integral_approx and integral_exact for one set of parameters.

Here is an example to show what your solutions should look like:

Example

Here is the integral I am performing:

$$ I_1 = \int_0^\infty \frac{dx}{x^2 + a^2} = \frac{\pi}{2a} $$



In [2]:

    
def integrand(x, a):
    return 1.0/(x**2 + a**2)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return 0.5*np.pi/a

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  1.5707963267948966
Exact    :  1.5707963267948966



In [ ]:

    
assert True # leave this cell to grade the above integral

Integral 1

\begin{equation*} \int_{0}^{a}{\sqrt{a^2 - x^2}} dx=\frac{\pi a^2}{4} \end{equation*}



In [30]:

    
# YOUR CODE HERE
def integrand(x, a):
    return (np.sqrt(a**2 - x**2))

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, a, args=(a,))
    return I

def integral_exact(a):
    return (0.25*np.pi*a**2)

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  0.7853981633974481
Exact    :  0.7853981633974483



In [24]:

    
assert True # leave this cell to grade the above integral

Integral 2

\begin{equation*} \int_{0}^{\infty} e^{-ax^2} dx =\frac{1}{2}\sqrt{\frac{\pi}{a}} \end{equation*}



In [34]:

    
# YOUR CODE HERE
def integrand(x, a):
    return np.exp(-a*x**2)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return 0.5*np.sqrt(np.pi/a)

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  0.8862269254527579
Exact    :  0.886226925453



In [32]:

    
assert True # leave this cell to grade the above integral

Integral 3

\begin{equation*} \int_{0}^{\infty} \frac{x}{e^x-1} dx =\frac{\pi^2}{6} \end{equation*}



In [36]:

    
# YOUR CODE HERE
def integrand(x, a):
    return x/(np.exp(x)-1)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return (1/6.0)*np.pi**2

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  1.6449340668482264
Exact    :  1.6449340668482262



In [ ]:

    
assert True # leave this cell to grade the above integral

Integral 4

\begin{equation*} \int_{0}^{\infty} \frac{x}{e^x+1} dx =\frac{\pi^2}{12} \end{equation*}



In [37]:

    
# YOUR CODE HERE
def integrand(x, a):
    return x/(np.exp(x)+1)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return (1/12.0)*np.pi**2

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  0.822467033424113
Exact    :  0.8224670334241131



In [ ]:

    
assert True # leave this cell to grade the above integral

Integral 5

\begin{equation*} \int_{0}^{1} \frac{ln x}{1-x} dx =-\frac{\pi^2}{6} \end{equation*}



In [43]:

    
# YOUR CODE HERE
def integrand(x, a):
    return np.log(x)/(1-x)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, 1, args=(a,))
    return I

def integral_exact(a):
    return (-1.0/6.0)*np.pi**2

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))









    



Numerical:  -1.6449340668482242
Exact    :  -1.6449340668482262



In [ ]:

    
assert True # leave this cell to grade the above integral