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  1. {
  2. "cells": [
  3. {
  4. "cell_type": "markdown",
  5. "metadata": {},
  6. "source": [
  7. "### Resolver la siguiente integral con la Regla del Simpson:"
  8. ]
  9. },
  10. {
  11. "cell_type": "markdown",
  12. "metadata": {},
  13. "source": [
  14. "$$\\int_{0}^{3}x^{2}e^{x}$$"
  15. ]
  16. },
  17. {
  18. "cell_type": "code",
  19. "execution_count": 9,
  20. "metadata": {},
  21. "outputs": [
  22. {
  23. "name": "stdout",
  24. "output_type": "stream",
  25. "text": [
  26. "=================================================\n",
  27. "Cálculo numérico de la regla del simpson\n",
  28. "=================================================\n",
  29. "Dame el número de segmentos : 4\n",
  30. "Dame el valor del límite inferior : 0\n",
  31. "Dame el valor del límite superior : 3\n",
  32. "-----------------------------------------------------------------------------------------------\n",
  33. " n integral et\n",
  34. "-----------------------------------------------------------------------------------------------\n",
  35. " 1 180.769832309 83.657665440\n",
  36. " 2 110.552516971 12.318614871\n",
  37. " 3 122.993435332 24.958279316\n",
  38. " 4 99.456833462 1.045675666\n",
  39. "----------------------------------------------------------------------------------------------\n"
  40. ]
  41. }
  42. ],
  43. "source": [
  44. "import math\n",
  45. "\n",
  46. "def f(x):\n",
  47. " z=(x**2)*(math.exp(x))\n",
  48. " return z\n",
  49. " \n",
  50. "print(\"=================================================\")\n",
  51. "print(\"Cálculo numérico de la regla del simpson\")\n",
  52. "print(\"=================================================\")\n",
  53. "\n",
  54. "\n",
  55. "\n",
  56. "\n",
  57. "n=int(input(\"Dame el número de segmentos : \"))\n",
  58. "a=float(input(\"Dame el valor del límite inferior : \"))\n",
  59. "b=float(input(\"Dame el valor del límite superior : \"))\n",
  60. "\n",
  61. "print(\"-----------------------------------------------------------------------------------------------\")\n",
  62. "print(\"{0:>10s}{1:>20s}{2:>20s}\".format(\"n\", \"integral\", \"et\"))\n",
  63. "print(\"-----------------------------------------------------------------------------------------------\")\n",
  64. "\n",
  65. "vv= 98.4276\n",
  66. "\n",
  67. "h = (b-a)/n\n",
  68. " \n",
  69. "\n",
  70. "for j in range (n): \n",
  71. " sum = f(a)\n",
  72. " h = (b-a)/(j+1)\n",
  73. " for i in range (1,j,2) :\n",
  74. " xi=a + (i*h)\n",
  75. " sum = sum + (4 * f(xi)) + (2 * f(xi+h))\n",
  76. " \n",
  77. " \n",
  78. " sum = sum + (4 * f(b-h)) + f(b)\n",
  79. " sum = h * sum/3\n",
  80. " et=abs((vv-sum)/vv)*100\n",
  81. " print(\"{0:10d}{1:20.9f}{2:20.9f}\".format(j+1, sum , et))\n",
  82. " \n",
  83. "\n",
  84. "\n",
  85. "print(\"----------------------------------------------------------------------------------------------\")"
  86. ]
  87. }
  88. ],
  89. "metadata": {
  90. "kernelspec": {
  91. "display_name": "Python 3",
  92. "language": "python",
  93. "name": "python3"
  94. },
  95. "language_info": {
  96. "codemirror_mode": {
  97. "name": "ipython",
  98. "version": 3
  99. },
  100. "file_extension": ".py",
  101. "mimetype": "text/x-python",
  102. "name": "python",
  103. "nbconvert_exporter": "python",
  104. "pygments_lexer": "ipython3",
  105. "version": "3.7.4"
  106. }
  107. },
  108. "nbformat": 4,
  109. "nbformat_minor": 2
  110. }
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