In [67]:
    
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns
from scipy import integrate
    
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:
integrand function that computes the value of the integrand.integral_approx funciton that uses scipy.integrate.quad to peform the integral.integral_exact function that computes the exact value of the integral.integral_approx and integral_exact for one set of parameters.Here is an example to show what your solutions should look like:
Here is the integral I am performing:
$$ I_1 = \int_0^\infty \frac{dx}{x^2 + a^2} = \frac{\pi}{2a} $$
In [68]:
    
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))
    
    
In [69]:
    
assert True # leave this cell to grade the above integral
    
In [70]:
    
def integrand(x, a):
    return np.sin(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.pi/2, args=(a,))
    return I
def integral_exact(a):
    return np.pi/4
print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))
    
    
In [71]:
    
assert True # leave this cell to grade the above integral
    
In [127]:
    
def integrand(x, m, a):
    return (x*np.sin((m*x)))/(x**2 + a**2)
def integral_approx(m, a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(m,a,))
    return I
def integral_exact(m, a):
    return ((np.pi)*.5) *(np.exp(-1*m*a))
print("Numerical: ", integral_approx(1.0,1.0))
print("Exact    : ", integral_exact(1.0,1.0))
    
    
In [73]:
    
assert True # leave this cell to grade the above integral
    
In [120]:
    
def integrand(a,x):
    return np.sin((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.pi/2), args=(a))
    return I
def integral_exact(a):
    return .5*(np.sqrt((np.pi)/np.pi*2*np.pi))
print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))
    
    
In [ ]:
    
assert True # leave this cell to grade the above integral
    
In [125]:
    
def integrand(a, x, 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(1.0,1.0))
print("Exact    : ", integral_exact(1.0,1.0))
    
    
In [ ]:
    
assert True # leave this cell to grade the above integral
    
In [101]:
    
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))
    
    
In [ ]:
    
assert True # leave this cell to grade the above integral