{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Regla de Simpson 1/3 para la integral $$\\int_{0}^{3}x^{2}e^{x}$$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "=======================================================================\n",
      "Calculo numérico de la regla del trapecio múltiple\n",
      "=======================================================================\n",
      "número de segmentos = 4\n",
      "valor del límite inferior = 0\n",
      "valor del límite superior = 3\n",
      "-----------------------------------------------------------------------------------------------\n",
      "         n            integral                  et\n",
      "-----------------------------------------------------------------------------------------------\n",
      "         1       180.769832309        83.657665440\n",
      "         2       110.552516971        12.318614871\n",
      "         3       122.993435332        24.958279316\n",
      "         4        99.456833462         1.045675666\n",
      "----------------------------------------------------------------------------------------------\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "def f(x):\n",
    "    z=(x**2)*(math.exp(x))\n",
    "    return z\n",
    " \n",
    "print(\"=======================================================================\")\n",
    "print(\"Calculo numérico de la regla del trapecio múltiple\")\n",
    "print(\"=======================================================================\")\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "n=int(input(\"número de segmentos = \"))\n",
    "a=float(input(\"valor del límite inferior = \"))\n",
    "b=float(input(\"valor del límite superior = \"))\n",
    "\n",
    "print(\"-----------------------------------------------------------------------------------------------\")\n",
    "print(\"{0:>10s}{1:>20s}{2:>20s}\".format(\"n\", \"integral\", \"et\"))\n",
    "print(\"-----------------------------------------------------------------------------------------------\")\n",
    "\n",
    "vv= 98.4276\n",
    "\n",
    "h = (b-a)/n\n",
    " \n",
    "\n",
    "for k in range (n): \n",
    "    sum = f(a)\n",
    "    h = (b-a)/(k+1)\n",
    "    for i in range (1,k,2) :\n",
    "        xi=a + (i*h)\n",
    "        sum = sum + (4 * f(xi)) + (2 * f(xi+h))\n",
    "        \n",
    "        \n",
    "    sum = sum + (4 * f(b-h)) + f(b)\n",
    "    sum = h * sum/3\n",
    "    et=abs((vv-sum)/vv)*100\n",
    "    print(\"{0:10d}{1:20.9f}{2:20.9f}\".format(k+1, sum , et))\n",
    "    \n",
    "\n",
    "\n",
    "print(\"----------------------------------------------------------------------------------------------\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.6.7"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}