{"id":11682,"date":"2025-08-26T13:25:31","date_gmt":"2025-08-26T18:25:31","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=11682"},"modified":"2026-04-05T20:24:08","modified_gmt":"2026-04-06T01:24:08","slug":"s2eva2025paoi_t1-centroide-integracion-numerica","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2025paoi_t1-centroide-integracion-numerica\/","title":{"rendered":"s2Eva2025PAOI_T1 coordenadas centroide por integraci\u00f3n num\u00e9rica"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>:\u00a0<a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-2eva30\/2eva2025paoi_t1-centroide-integracion-numerica\/\" data-type=\"post\" data-id=\"11670\">2Eva2025PAOI_T1 coordenadas centroide por integraci\u00f3n num\u00e9rica<\/a><\/p>\n\n\n\n<p>Para el integral con <strong>Qy <\/strong>use f\u00f3rmulas de<strong> Simpson<\/strong> con al menos 3 tramos,<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">Q_y = \\int x dA = \\int_0^5 x^3 dx<\/span>\n\n\n\n<p>a. Para el intervalo de integraci\u00f3n [0,5], al considerar 3 tramos, h = 5\/3 que es mayor que 1. Se prefiere h\u22641 de tal forma que al calcular los errores, la cota del error disminuya por estar en el orden O(h<sup>k<\/sup>), por lo que tramos ser\u00e1 al menos 5 obteniendo h=1.<\/p>\n\n\n\n<p class=\"has-text-align-center\">xi = [ 0, 1, 2, 3, 4, 5]<\/p>\n\n\n\n<p>c. Con 5 tramos, 6 muestras, se usar\u00e1n Simpson de 3\/8 y luego Simpson de 1\/3.<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{3\/8} = \\frac{3}{8} (1) \\left( 0^3+3(1^3)+3(2^3)+3^3\\right) = 20.25<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{1\/3} = \\frac{1}{3} (1) \\left( 3^3+4(4^3)+5^3\\right) = 136 <\/span>\n\n\n\n<p>d. <span class=\"wp-katex-eq katex-display\" data-display=\"true\">Q_y =I_{3\/8} + I_{1\/3} = 20.25+136 =156.25 <\/span><\/p>\n\n\n\n<p>cota de error = O(h<sup>5<\/sup>)+O(h<sup>5<\/sup>) = 2[ O(h<sup>5<\/sup>)] = 2(1<sup>5<\/sup>) = 2<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>F\u00f3rmulas compuestas, tramos: 5\nm\u00e9todos 3:Simpson3\/8, 2:Simpson1\/3, 1:Trapecio, 0:usado\ntramos iguales en xi: &#091;3 0 0 2 0 0]\n&#091;'S38', 20.25]\n&#091;'S13', 136.0]\ntramos iguales en xi: &#091;3 0 0 2 0 0]\n<strong>Integral: 156.25<\/strong><\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>Para el integral con <strong>Qx <\/strong>use <strong>Cuadratura de Gauss<\/strong> de dos puntos<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">Q_x = \\int ydA = \\int_0^5 \\frac{x^4}{2} dx<\/span>\n\n\n\n<p>b. Para cuadratura de Gauss, se tomar\u00edan al menos 3 tramos, dado que la cota de error por truncamiento se aproxima con la&nbsp;\u2245 f<sup>(4)<\/sup>(x)<\/p>\n\n\n\n<p>xi = [ 0, 5\/3, 10\/3, 5]<\/p>\n\n\n\n<p>c. intervalo [0, 5\/3]<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> x_a = \\frac{5\/3+0}{2} - \\left( \\frac{1}{\\sqrt{3}} \\right) \\frac{5\/3-0}{2} =0.352208<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> x_b = \\frac{5\/3+0}{2} + \\left( \\frac{1}{\\sqrt{3}}\\right) \\frac{5\/3-0}{2} =1.314458<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_0 = \\frac{5\/3-0}{2} \\left( f(x_a) +f(1.314458) \\right) <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> = \\frac{5\/3-0}{2} \\left( \\frac{(0.352208)^4}{2} +\\frac{(1.314458) ^4}{2} \\right) = 1.2502 <\/span>\n\n\n\n<p>d. usando el algoritmo se puede obtener el resultado de:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>&#091;xa,xb,f(xa),f(xb)],area\n&#091;0.352208, 1.314458, 0.007694, 1.492648] 1.250285\n&#091;xa,xb,f(xa),f(xb)],area\n&#091;2.018874, 2.981125, 8.306298, 39.490340] 39.830532\n&#091;xa,xb,f(xa),f(xb)],area\n&#091;3.685541, 4.647791, 92.251874, 233.322542] 271.312014\nIntegral:  <strong>312.3928<\/strong>3264746234<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<figure class=\"wp-block-image alignright\"><a href=\"http:\/\/blog.espol.edu.ec\/analisisnumerico\/files\/2025\/08\/centroide01.png\"><img decoding=\"async\" src=\"http:\/\/blog.espol.edu.ec\/analisisnumerico\/files\/2025\/08\/centroide01-300x227.png\" alt=\"superficie para centroide\" class=\"wp-image-11671\" \/><\/a><\/figure>\n\n\n\n<p>Para el integral de \u00e1rea A:<\/p>\n\n\n\n<p>se plantea usando Simpson de 3\/8 o cualquier otro m\u00e9todo num\u00e9rico. con h=1<\/p>\n\n\n\n<p>xi = [ 0, 1, 2, 3, 4, 5]<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> A = \\int f(x) dx = \\int_0^5 x^2 dx <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{3\/8} = \\frac{3}{8} (1) \\left( 0^2+3(1^2)+3(2^2)+3^2\\right) = 9<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{1\/3} = \\frac{1}{3} (1) \\left( 3^2+4(4^2)+5^2\\right) = 32.6666 <\/span>\n\n\n\n<p>d. <span class=\"wp-katex-eq katex-display\" data-display=\"true\">Q_y =I_{3\/8} + I_{1\/3} = 9+32.6666 =41.6666 <\/span><\/p>\n\n\n\n<p>cota de error = O(h<sup>5<\/sup>)+O(h<sup>5<\/sup>) = 2[ O(h<sup>5<\/sup>)] = 2(1<sup>5<\/sup>) = 2<\/p>\n\n\n\n<p>usando el algoritmo se obtiene:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>F\u00f3rmulas compuestas, tramos: 5\nm\u00e9todos 3:Simpson3\/8, 2:Simpson1\/3, 1:Trapecio, 0:usado\ntramos iguales en xi: &#091;3 0 0 2 0 0]\n&#091;'S38', 9.0]\n&#091;'S13', 32.666666666666664]\ntramos iguales en xi: &#091;3 0 0 2 0 0]\nIntegral: <strong>41.6666<\/strong>66666666664<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\" \/>\n\n\n\n<p>e. Determine las coordenadas del centroide seg\u00fan las f\u00f3rmulas presentadas.<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\bar{x} = \\frac{Q_y}{A}= \\frac{156.25}{41.6666} = 3.75<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> \\bar{y} = \\frac{Q_x}{A}= \\frac{312.5}{41.6666} = 7.5 <\/span>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"640\" height=\"480\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2025\/08\/centroide02.png\" alt=\"centroide marcada coordenada\" class=\"wp-image-18340\" \/><\/figure>\n\n\n\n<p><strong>literal f<\/strong> se deja como tarea.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ejercicio:\u00a02Eva2025PAOI_T1 coordenadas centroide por integraci\u00f3n num\u00e9rica Para el integral con Qy use f\u00f3rmulas de Simpson con al menos 3 tramos, a. Para el intervalo de integraci\u00f3n [0,5], al considerar 3 tramos, h = 5\/3 que es mayor que 1. Se prefiere h\u22641 de tal forma que al calcular los errores, la cota del error disminuya [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[58,54],"class_list":["post-11682","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11682","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=11682"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11682\/revisions"}],"predecessor-version":[{"id":23883,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11682\/revisions\/23883"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=11682"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=11682"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=11682"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}