{"id":2918,"date":"2017-11-11T06:00:21","date_gmt":"2017-11-11T11:00:21","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=2918"},"modified":"2025-12-23T08:14:05","modified_gmt":"2025-12-23T13:14:05","slug":"s1eva2011ti_t1-encontrar-%ce%b1-en-integral","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva20\/s1eva2011ti_t1-encontrar-%ce%b1-en-integral\/","title":{"rendered":"s1Eva2011TI_T1 Encontrar \u03b1 en integral"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2011TI_T1 Encontrar \u03b1 en integral<\/a><\/p>\n\n\n\n

Desarrollo Anal\u00edtico<\/h2>\n\n\n\n

Se iguala la ecuaci\u00f3n al valor buscado = 10, y se resuelve<\/p>\n\n\n \\int_{\\alpha}^{2\\alpha} x e^{x}dx = 10<\/span>\n\n\n\n

siendo: \u03bc = x , \u03b4v = ex<\/sup>, \u03b4u = \u03b4x , v = ex<\/sup><\/p>\n\n\n \\int u dv = uv - \\int v \\delta u <\/span>\n\n\n xe^x \\Big|_{\\alpha}^{2 \\alpha} - \\int_{\\alpha}^{2\\alpha} e^{x}dx - 10 = 0<\/span>\n\n\n 2\\alpha e^{2 \\alpha} -\\alpha e^{\\alpha} - (e^{2\\alpha} - e^{\\alpha}) - 10 = 0<\/span>\n\n\n (2\\alpha-1)e^{2 \\alpha}+ (1-\\alpha) e^{\\alpha} - 10 = 0<\/span>\n\n\n\n

la funci\u00f3n a usar en el m\u00e9todo es<\/p>\n\n\n f(\\alpha) = (2\\alpha-1)e^{2 \\alpha}+ (1-\\alpha)e^{\\alpha} -10<\/span>\n\n\n\n

Se obtiene la derivada para el m\u00e9todo de Newton Raphson<\/p>\n\n\n f'(\\alpha) = 2e^{2 \\alpha} + 2(2\\alpha-1)e^{2 \\alpha} - e^{\\alpha} + (1-\\alpha) e^{\\alpha}<\/span>\n\n\n f'(\\alpha) = (2 + 2(2\\alpha-1))e^{2 \\alpha} +(-1 + (1-\\alpha)) e^{\\alpha}<\/span>\n\n\n f'(\\alpha) = 4\\alpha e^{2 \\alpha} -\\alpha e^{\\alpha}<\/span>\n\n\n\n

la f\u00f3rmula para el m\u00e9todo de Newton-Raphson<\/p>\n\n\n \\alpha_{i+1} = \\alpha_i - \\frac{f(\\alpha)}{f'(\\alpha)} <\/span>\n\n\n\n

se prueba con \u03b10<\/sub> = 1, se evita el valor de cero por la indeterminaci\u00f3n que se da por f'(0) = 0<\/p>\n\n\n\n

iteraci\u00f3n 1:<\/p>\n\n\n f(1) = (2(1)-1)e^{2(1)}+ (1-(1))e^{(1)} -10<\/span>\n\n\n f'(1) = 4(1) e^{2 (1)} -(1) e^{(1)}<\/span>\n\n\n \\alpha_{1} = 1- \\frac{f(1)}{f'(1)} <\/span>\n\n\n\n

\\alpha_{1} = 1.0973 <\/span>
error = 0.0973<\/p>\n\n\n\n

iteraci\u00f3n 2:
f(2) = (2(1.0973)-1)e^{2(1.0973)}+ (1-(1.0973))e^{(1.0973)} -10<\/span><\/p>\n\n\n f'(2) = 4(1.0973) e^{2 (1.0973)} -(1.0973) e^{(1.0973)}<\/span>\n\n\n \\alpha_{2} = 1.0973 - \\frac{f(1.0973)}{f'(1.0973)} <\/span>\n\n\n\n

\\alpha_{2} = 1.0853 <\/span>
error = 0.011941<\/p>\n\n\n\n

iteraci\u00f3n 3:
f(3) = (2(1.0853)-1)e^{2(1.0853)}+ (1-(1.0853))e^{(1.0853)} -10<\/span><\/p>\n\n\n f'(3) = 4(1.0853) e^{2 (1.0853)} -(1.0853) e^{(1.0853)}<\/span>\n\n\n \\alpha_{3} = 1.0853- \\frac{f(1.0853)}{f'(1.0853)} <\/span>\n\n\n\n

\\alpha_{3} = 1.0851 <\/span>
error = 0.00021951<\/p>\n\n\n\n

[  xi,          xnuevo,      f(xi),      f'(xi),       tramo ]\n[  1.           1.0973      -2.6109      26.8379       0.0973]\n[  1.0973e+00   1.0853e+00   4.3118e-01   3.6110e+01   1.1941e-02]\n[  1.0853e+00   1.0851e+00   7.6468e-03   3.4836e+01   2.1951e-04]\n[  1.0851e+00   1.0851e+00   2.5287e-06   3.4813e+01   7.2637e-08]\nraiz:  1.08512526549<\/code><\/pre>\n\n\n\n
se obtiene el valor de la ra\u00edz con 4 iteraciones, con error de aproximaci\u00f3n de 7.2637e-08<\/p>\n\n\n\n
Desarrollo con Python<\/h2>\n\n\n\n
\"Newton-Raphson<\/figure>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2011TI_T1 Encontrar \u03b1 en integral Desarrollo Anal\u00edtico Se iguala la ecuaci\u00f3n al valor buscado = 10, y se resuelve siendo: \u03bc = x , \u03b4v = ex, \u03b4u = \u03b4x , v = ex la funci\u00f3n a usar en el m\u00e9todo es Se obtiene la derivada para el m\u00e9todo de Newton Raphson la f\u00f3rmula […]<\/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":[45],"tags":[58,54],"class_list":["post-2918","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2918","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=2918"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2918\/revisions"}],"predecessor-version":[{"id":18729,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2918\/revisions\/18729"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=2918"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=2918"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=2918"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}