Python实现数值积分方式

原理：

利用复化梯形公式，复化Simpson公式，计算积分。

步骤：

import math

"""测试函数"""
def f(x,i):
 if i == 1:
   return (4 - (math.sin(x)) ** 2) ** 0.5
 if i == 2:
   if x == 0:
     return 1
   else:
     return math.sin(x) / x
 if i == 3:
   return (math.exp(x)) / (4 + x ** 2)
 if i == 4:
   return math.log(1+x,math.e) / (1 + x ** 2)

"""打印显示函数"""  
def p(i,n):
 return "第" + str(i) + "题,n=" + str(n) + "时的积分值为："

"""复化Simpson函数"""
def Simpson(a, b, n, i):
 h = (b - a) / (2 * n)
 F0 = f(a,i) + f(b,i)
 F1 = 0
 F2 = 0
 for j in range(1,2 * n):
   x = a + (j * h)
   if j ％ 2 == 0:
     F2 = F2 + f(x,i)
   else:
     F1 = F1 + f(x,i)
 SN = (h * (F0 + 2 * F2 + 4 * F1)) / 3
 print("复化Simpson函数" + p(i,n) + str("％-10.7f"％(SN)))
 return SN

def T(a, b, n, i):
 h = (b - a) / n
 F0 = f(a,i) + f(b,i)
 F = 0
 for j in range(1,n):
   x = a + (j * h)
   F = F + f(x,i)
 SN = (h * (F0 + 2 * F)) / 2
 print("复化梯形函数" + p(i,n) + str("％-10.7f"％(SN)))
 return SN

def SimpsonTimes(x):
 n = 1
 y = Simpson(0, math.pi/4, n, 1)
 while(abs(y - 1.5343916) > x):
   n = n + 1
   y = Simpson(0, math.pi/4, n, 1)
 else:
   return n

def Times(x):
 n = 1
 y = T(0, math.pi/4, n, 1)
 while(abs(y - 1.5343916) > x):
   n = n + 1
   y = T(0, math.pi/4, n, 1)
 else:
   return n

"""
 测试部分  
"""
Simpson(0, math.pi/4, 10, 1)
Simpson(0, 1, 10, 2)
Simpson(0, 1, 10, 3)
Simpson(0, 1, 10, 4)
Simpson(0, math.pi/4, 20, 1)
Simpson(0, 1, 20, 2)
Simpson(0, 1, 20, 3)
Simpson(0, 1, 20, 4)

T(0, math.pi/4, 10, 1)
T(0, 1, 10, 2)
T(0, 1, 10, 3)
T(0, 1, 10, 4)
T(0, math.pi/4, 20, 1)
T(0, 1, 20, 2)
T(0, 1, 20, 3)
T(0, 1, 20, 4)

print("复化梯形函数求解第一问，精度为0.00001时需要" + str(Times(0.00001)) + "个步数")
print("复化Simpson函数求解第一问，精度为0.00001时需要" + str(SimpsonTimes(0.00001)) + "个步数")
print("复化梯形函数求解第一问，精度为0.000001时需要" + str(Times(0.000001)) + "个步数")
print("复化Simpson函数求解第一问，精度为0.000001时需要" + str(SimpsonTimes(0.000001)) + "个步数")

来源：https://blog.csdn.net/MachineRandy/article/details/82634757