Integration Exercise 2

Imports



In [20]:

    
%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 [21]:

    
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 [22]:

    
assert True # leave this cell to grade the above integral

Integral 1

YOUR ANSWER HERE: $$ \LARGE \int_0^\infty e^{-ax} cos(bx) dx = \frac{a}{a^2+b^2} $$



In [23]:

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

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

def integral_exact(a, b):
    return a/(a**2 + b**2)

print("Numerical: ", integral_approx(1.0, 3.0))
print("Exact    : ", integral_exact(1.0,3.0))









    



Numerical:  0.10000000000001068
Exact    :  0.1



In [24]:

    
assert True # leave this cell to grade the above integral

Integral 2

YOUR ANSWER HERE: $$ \LARGE{ \int_0^\infty \frac{dx}{\sqrt{a^2 - x^2}} = \frac{\pi}{2} }$$



In [41]:

    
# YOUR CODE HERE
def integrand(x, a):
    v = np.sqrt(a**2 - x**2)
    return v**-1

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 0.5*np.pi

print("Numerical: ", integral_approx(3.0))
print("Exact    : ", integral_exact(3.0))









    



Numerical:  1.57079632680332
Exact    :  1.5707963267948966



In [26]:

    
assert True # leave this cell to grade the above integral

Integral 3

YOUR ANSWER HERE: $$ \LARGE{ \int_0^a \sqrt{a^2 - x^2}dx = \frac{\pi a^2}{4} }$$



In [29]:

    
# 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 [28]:

    
assert True # leave this cell to grade the above integral

Integral 4

YOUR ANSWER HERE: $$ \LARGE \int_0^\infty e^{-ax} sin(bx) dx = \frac{b}{a^2+b^2} $$



In [37]:

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

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

def integral_exact(a, b):
    return b/(a**2 + b**2)

print("Numerical: ", integral_approx(1.0, 3.0))
print("Exact    : ", integral_exact(1.0,3.0))









    



Numerical:  0.3000000000000034
Exact    :  0.3



In [ ]:

    
assert True # leave this cell to grade the above integral

Integral 5

YOUR ANSWER HERE: $$ \LARGE \int_0^\infty e^{-ax^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} $$



In [38]:

    
# 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 [ ]:

    
assert True # leave this cell to grade the above integral