{"id":2270,"date":"2018-08-29T10:55:47","date_gmt":"2018-08-29T15:55:47","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=2270"},"modified":"2026-04-05T20:14:15","modified_gmt":"2026-04-06T01:14:15","slug":"s2eva2018ti_t2-deducir-simpson-13","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2018ti_t2-deducir-simpson-13\/","title":{"rendered":"s2Eva2018TI_T2 Deducir Simpson 1\/3"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 2Eva2018TI_T2 Deducir Simpson 1\/3<\/a><\/p>\n\n\n\n

Para el problema, se usan los puntos:  [a,f(a)], [b,f(b)] y [c,f(c)]
por donde pasa la curva f(x) aproximada a un polinomio de grado 2, f(x) \\approx p(x) <\/span><\/p>\n\n\n \\int_a^b f(x) dx \\approx \\int_a^b p(x) dx<\/span>\n\n\n p(x) = L_a f(a) + L_c f(c) + L_b f(b) <\/span>\n\n\n \\int_a^b p(x) dx = \\int_a^b \\Big[ L_a f(a) +L_c f(c) + L_b f(b) \\Big] dx =<\/span>\n\n\n = \\int_a^b L_a f(a) dx + \\int_a^b L_c f(c) dx + \\int_a^b L_b f(b) dx <\/span>\n\n\n \\int_a^b p(x) dx = I_1 + I_2 + I_3 <\/span>\n\n\n\n

Como referencia se usa la gr\u00e1fica para relacionar a<\/strong>, b<\/strong>, c<\/strong> y h<\/strong>:<\/p>\n\n\n\n

\"regla<\/figure>\n\n\n\n
\n\n\n\n

Primer Integral<\/strong><\/em><\/p>\n\n\n\n

Para el primer integral I_1= \\int_a^b L_a f(a) dx <\/span> se tiene que:<\/p>\n\n\n L_a = \\frac{(x-b)(x-c)}{(a-b)(a-c)} = \\frac{(x-b)(x-c)}{(-2h)(-h)}<\/span>\n\n\n L_a = \\frac{(x-b)(x-c)}{2h^2} <\/span>\n\n\n\n

se convierte en:<\/p>\n\n\n I_1 = \\int_a^b \\frac{(x-b)(x-c)}{2h^2} f(a) dx = \\frac{f(a)}{2h^2} \\int_a^b (x-b)(x-c)dx <\/span>\n\n\n\n

Para dejar la parte del integral en funci\u00f3n de h<\/strong>, a<\/strong> y b<\/strong>, teniendo que c<\/strong> est\u00e1 en la mitad de [a,b], es decir c<\/strong> = (a+b)\/2 , se usa:<\/p>\n\n\n u = x-a <\/span>\n\n\n\n

por lo que \\frac{du}{dx}=1<\/span> y du = dx<\/span><\/p>\n\n\n x-c = (u+a) - \\frac{a+b}{2} <\/span>\n\n\n = u+ \\frac{a-b}{2} = u - \\frac{b-a}{2} <\/span>\n\n\n x-c = u-h <\/span>\n\n\n x-b = (u+a)-b = u-2\\Big(\\frac{b-a}{2}\\Big) = u-2h <\/span>\n\n\n\n

Se actualiza el integral de x<\/strong> entre [a,b]  usando u = x-a, que se convierte el rango para u<\/strong> en [0, b-a] que es lo mismo que [0,2h]<\/p>\n\n\n \\int_a^b (x-b)(x-c)dx = \\int_0^{2h} (u-2h)(u-h)du =<\/span>\n\n\n = \\int_0^{2h} \\Big( u^2 - 2hu - uh + 2h^2 \\Big) du <\/span>\n\n\n = \\int_0^{2h} \\Big( u^2 - 3hu + 2h^2 \\Big) du <\/span>\n\n\n = \\frac{u^3}{3}- 3h\\frac{u^2}{2}+ 2h^2u \\Big|_0^{2h} <\/span>\n\n\n = \\frac{(2h)^3}{3}- 3h\\frac{(2h)^2}{2} + 2h^2(2h) -(0-0+0) <\/span>\n\n\n = \\frac{8h^3}{3}- 6h^3 + 4h^3 =\\frac{8h^3}{3}- 2h^3 <\/span>\n\n\n = \\frac{2h^3}{3}<\/span>\n\n\n\n

resultado que se usa en I1<\/sub><\/p>\n\n\n I_1= \\frac{f(a)}{2h^2}\\frac{2h^3}{3} =\\frac{h}{3} f(a) <\/span>\n\n\n\n

que es el primer t\u00e9rmino de la f\u00f3rmula general de Simpson 1\/3<\/p>\n\n\n\n

\n\n\n\n

Segundo Integral<\/strong><\/em><\/p>\n\n\n\n

Para el Segundo integral I_2= \\int_a^b L_c f(c) dx <\/span> se tiene que:<\/p>\n\n\n L_c = \\frac{(x-a)(x-b)}{(c-a)(c-b)} = \\frac{(x-a)(x-b)}{(h)(-h)}<\/span>\n\n\n L_c = \\frac{(x-a)(x-b)}{-h^2} <\/span>\n\n\n\n

se convierte en:<\/p>\n\n\n I_2 = \\frac{f(c)}{-h^2} \\int_a^b (x-a)(x-b) dx <\/span>\n\n\n = \\frac{f(c)}{-h^2} \\int_0^{2h} (u)(u-2h) du <\/span>\n\n\n\n

siendo:<\/p>\n\n\n \\int_0^{2h}(u^2-2hu)du=\\Big(\\frac{u^3}{3}-2h\\frac{u^2}{2}\\Big)\\Big|_0^{2h} <\/span>\n\n\n =\\frac{(2h)^3}{3}-h(2h)^2-(0-0)<\/span>\n\n\n =\\frac{8h^3}{3}-4h^3 = -\\frac{4h^3}{3} <\/span>\n\n\n\n

usando en I2<\/sub><\/p>\n\n\n I_2 = \\frac{f(c)}{-h^2}\\Big(-\\frac{4h^3}{3}) = \\frac{h}{3}4f(c) <\/span>\n\n\n\n

\n\n\n\n

Tarea<\/strong>: Continuar las operaciones para y tercer integral para llegar a la f\u00f3rmula general de Simpson 1\/3:<\/p>\n\n\n I = \\frac{h}{3} \\Big( f(a)+4f(c) + f(b) \\Big) <\/span>\n","protected":false},"excerpt":{"rendered":"

Ejercicio: 2Eva2018TI_T2 Deducir Simpson 1\/3 Para el problema, se usan los puntos:  [a,f(a)], [b,f(b)] y [c,f(c)]por donde pasa la curva f(x) aproximada a un polinomio de grado 2, Como referencia se usa la gr\u00e1fica para relacionar a, b, c y h: Primer Integral Para el primer integral se tiene que: se convierte en: Para dejar […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[58,54],"class_list":["post-2270","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2270","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=2270"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2270\/revisions"}],"predecessor-version":[{"id":23863,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2270\/revisions\/23863"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=2270"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=2270"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=2270"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}