{"id":2743,"date":"2020-06-19T07:53:38","date_gmt":"2020-06-19T05:53:38","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?page_id=2743"},"modified":"2026-04-20T15:06:53","modified_gmt":"2026-04-20T13:06:53","slug":"methode-point-median","status":"publish","type":"page","link":"https:\/\/www.mathweb.fr\/euclide\/methode-point-median\/","title":{"rendered":"M\u00e9thode du point m\u00e9dian"},"content":{"rendered":"\n<p>La <a rel=\"noreferrer noopener\" href=\"https:\/\/www.youtube.com\/watch?v=FvQg8LgbW18\" target=\"_blank\">m\u00e9thode des rectangles<\/a> (cliquez pour voir une vid\u00e9o explicative), aussi expliqu\u00e9e sur ce site <a href=\"https:\/\/www.mathweb.fr\/euclide\/methode-des-rectangles\/\" target=\"_blank\" aria-label=\"undefined (s\u2019ouvre dans un nouvel onglet)\" rel=\"noreferrer noopener\">ici<\/a>) est une m\u00e9thode algorithmique (que l&#8217;on peut impl\u00e9menter en Python par exemple) qui permet d&#8217;obtenir un encadrement d&#8217;une int\u00e9grale. Je rappelle que pour une fonction positive sur un intervalle [<em>a<\/em>,<em>b<\/em>], l&#8217;int\u00e9grale sur cet intervalle est l&#8217;aire de la partie &#8220;coinc\u00e9e&#8221; entre la courbe repr\u00e9sentative de <em>f<\/em> et l&#8217;axe des abscisse, que l&#8217;on note \\(\\displaystyle\\int_a^b f(x)\\text{d}x.\\)<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"511\" height=\"459\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale01.png\" alt=\"int\u00e9grale d'une fonction positive\" class=\"wp-image-2744\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale01.png 511w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale01-300x269.png 300w\" sizes=\"auto, (max-width: 511px) 100vw, 511px\" \/><figcaption class=\"wp-element-caption\">Int\u00e9grale d&#8217;une fonction positive sur [1 ; 8]<\/figcaption><\/figure>\n<\/div>\n\n\n<h2 class=\"wp-block-heading\">Le principe de la variante<\/h2>\n\n\n\n<p>On subdivise l&#8217;intervalle [<em>a<\/em>;<em>b<\/em>] en <em>n<\/em> intervalles de m\u00eame largeur, \u00e9gale \u00e0 \\(\\frac{b-a}{n}\\). <\/p>\n\n\n\n<p>Ensuite, on note ces intervalles \\(I_0,\\ I_1,\\ &#8230;,\\ I_{n-1}\\) et on note \\(M_k(x_k;f(x_k))\\) le milieu de \\(I_k\\) (le &#8220;point m\u00e9dian&#8221;). <\/p>\n\n\n\n<p>Enfin, on construit alors les rectangles de bases \\(I_k\\) et de hauteur \\(f(x_k)\\).<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"565\" height=\"555\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale02.png\" alt=\"variante de la m\u00e9thode des rectangles en Python\" class=\"wp-image-2745\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale02.png 565w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/06\/integrale02-300x295.png 300w\" sizes=\"auto, (max-width: 565px) 100vw, 565px\" \/><figcaption class=\"wp-element-caption\">Variante de la m\u00e9thode des rectangles<\/figcaption><\/figure>\n<\/div>\n\n\n<ul class=\"wp-block-list\">\n<li>Le premier point \\(M_0\\) a pour abscisse \\(a + \\frac{1}{2}\\times\\frac{b-a}{n}\\), soit \\(a+\\frac{b-a}{2n}\\), et donc pour ordonn\u00e9e \\(f\\left(a+\\frac{b-a}{2n}\\right)\\).<\/li>\n\n\n\n<li>Le deuxi\u00e8me point \\(M_1\\) a pour abscisse \\(a + 3\\times\\frac{1}{2}\\times\\frac{b-a}{n}\\), soit \\(a+\\frac{3(b-a)}{2n}\\), et donc pour ordonn\u00e9e \\(f\\left(a+\\frac{3(b-a)}{2n}\\right)\\).<\/li>\n\n\n\n<li>Etc.<\/li>\n<\/ul>\n\n\n\n<p>Plus on fait grandir <em>n<\/em>, plus les rectangles ont une base petite et plus la somme des aires des rectangles se rapproche de l&#8217;int\u00e9grale cherch\u00e9e.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Impl\u00e9mentation en Python de la variante de la m\u00e9thode des rectangles<\/h2>\n\n\n\n<p>La fonction que j&#8217;ai prise en exemple a pour expression:$$f(x)=\\frac{1}{8}x^3-\\frac{3}{2}x^2+5x-1.$$Je vais donc d&#8217;abord d\u00e9finir cette fonction:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def f(x):\n    return x**3\/8 - 3*x**2\/2 + 5*x - 1<\/pre>\n\n\n\n<p>Ensuite je d\u00e9finis une fonction <em><strong>area(a,b,n)<\/strong><\/em> qui devra retourner la valeur approch\u00e9e de l&#8217;int\u00e9grale cherch\u00e9e sur l&#8217;intervalle [<em>a<\/em>;<em>b<\/em>] avec notre m\u00e9thode, en prenant un <em>n<\/em> &#8220;assez grand&#8221;.<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def area(a,b,n):\n    s = 0 # somme des aires initialement nulle\n    largeur = (b - a) \/ n\n    for k in range(n):\n        xk = a + (k + 0.5) * largeur # abscisse des \"points m\u00e9dians\"\n        s = s + largeur * f(xk)\n        \n    return s<\/pre>\n\n\n\n<p>Afin de voir les diff\u00e9rents r\u00e9sultats pour diff\u00e9rentes valeurs de <em>n<\/em>, on cr\u00e9\u00e9e une boucle sur <em>n<\/em>:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">for n in range(100,2100,100):\n    print(\"Avec n = {}, on obtient une aire de : {}.\".format(n , area(1,8,n)))<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">Avec n = 100, on obtient une aire de : 22.968214062500007.\nAvec n = 200, on obtient une aire de : 22.968616015625.\nAvec n = 300, on obtient une aire de : 22.968690451388888.\nAvec n = 400, on obtient une aire de : 22.968716503906258.\nAvec n = 500, on obtient une aire de : 22.968728562499976.\nAvec n = 600, on obtient une aire de : 22.968735112847224.\nAvec n = 700, on obtient une aire de : 22.968739062499978.\nAvec n = 800, on obtient une aire de : 22.96874162597657.\nAvec n = 900, on obtient une aire de : 22.968743383487663.\nAvec n = 1000, on obtient une aire de : 22.96874464062501.\nAvec n = 1100, on obtient une aire de : 22.96874557076446.\nAvec n = 1200, on obtient une aire de : 22.968746278211807.\nAvec n = 1300, on obtient une aire de : 22.968746828772172.\nAvec n = 1400, on obtient une aire de : 22.968747265624987.\nAvec n = 1500, on obtient une aire de : 22.968747618055556.\nAvec n = 1600, on obtient une aire de : 22.968747906494144.\nAvec n = 1700, on obtient une aire de : 22.96874814554499.\nAvec n = 1800, on obtient une aire de : 22.968748345871923.\nAvec n = 1900, on obtient une aire de : 22.96874851540862.\nAvec n = 2000, on obtient une aire de : 22.968748660156262.<\/pre>\n\n\n\n<p>Pour <em>n<\/em> = 100, on obtient une valeur approch\u00e9e de l&#8217;int\u00e9grale au milli\u00e8me. S<\/p>\n\n\n\n<p>Si l&#8217;on souhaite une meilleur approximation, il faut aller jusqu&#8217;\u00e0 <em>n<\/em>=2000 pour une valeur approch\u00e9e au moins \u00e0 \\(10^{6}\\) pr\u00e8s.<\/p>\n\n\n\n<p>Cette variante est appel\u00e9e <em>m\u00e9thode du point m\u00e9dian<\/em>. <\/p>\n\n\n\n<p>Il existe d&#8217;autres m\u00e9thodes. Par exemple, la <a href=\"https:\/\/www.mathweb.fr\/euclide\/2024\/11\/21\/approximation-dune-aire-methode-des-trapezes\/\">m\u00e9thode des trap\u00e8zes<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>La m\u00e9thode des rectangles (cliquez pour voir une vid\u00e9o explicative), aussi expliqu\u00e9e sur ce site ici) est une m\u00e9thode algorithmique (que l&#8217;on peut impl\u00e9menter en Python par exemple) qui permet d&#8217;obtenir un encadrement d&#8217;une int\u00e9grale. Je rappelle que pour une fonction positive sur un intervalle [a,b], l&#8217;int\u00e9grale sur cet intervalle [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":11623,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-2743","page","type-page","status-publish","has-post-thumbnail","hentry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>M\u00e9thode du point m\u00e9dian - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Je vous pr\u00e9sente ici une variante de la m\u00e9thode des rectangles permettant d&#039;obtenir un encadrement d&#039;une int\u00e9grale.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/methode-point-median\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"M\u00e9thode du point m\u00e9dian - 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