{"id":235119,"date":"2022-06-22T17:17:49","date_gmt":"2022-06-22T15:17:49","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=235119"},"modified":"2025-09-29T11:25:35","modified_gmt":"2025-09-29T09:25:35","slug":"methode-de-calculs-de-primitives","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/methode-de-calculs-de-primitives\/","title":{"rendered":"M\u00e9thode de calculs de primitives"},"content":{"rendered":"\n

Vous travaillez actuellement sur des calculs de primitives<\/strong> ? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9 \u00e0 la m\u00e9thode de calculs de primitives<\/strong>, ces notions n’auront plus aucun secret pour vous gr\u00e2ce \u00e0 des m\u00e9thodologies de pointe !<\/p>\n

Pour aller plus loin et vraiment ma\u00eetriser ces techniques, d\u00e9couvre comment les appliquer efficacement avec nos prof particulier de math\u00e9matiques<\/strong><\/a>; c’est l’assurance de r\u00e9sultats \u00e0 la hauteur de tes ambitions en pr\u00e9pa. \ud83c\udf93<\/p>\n

\u00a0<\/p>\n\n\n\n

M\u00e9thodes de calculs des primitives <\/h2>\n\n\n\n

Proposition : Int\u00e9gration par parties<\/h3>\n\n\n\n

<\/p>\nSoient \"u\" et \"v\" deux fonctions de classe \"\mathcal{C}^1\" sur un segment \"[a,b]\", alors :\n

  <\/span>   <\/span>\"\[\int_a^b u'(t)v(t)\mathrm{d}t=\big[u(t)v(t)\big]_a^b-\int_a^b u(t)v'(t)\mathrm{d}t.\]\"<\/p>\n\n\n\n

D\u00e9monstration<\/h3>\n\n\n\n

<\/p>\nIl suffit d’int\u00e9grer la relation \"(uv)'=u'v+uv'\" entre \"a\" et \"b\".\n\n\n\n

Exemple<\/h3>\n\n\n\n

<\/p>\nSoit \"x\in\mathbb{R}\". Calculons \"\displaystyle I=\int_0^x t e^t\mathrm{d}t\" par int\u00e9gration par parties :\non pose \"u'(t)=e^t\" de sorte que \"u(t)=e^t\" et \"v(t)=t\". Les fonctions \"u\" et \"v\" sont de classe \"\mathcal{C}^1\" sur \"[0,x]\" si \"x\" est positif (ou \"[x,0]\" si \"x\" est n\u00e9gatif), par int\u00e9gration par parties :\n

  <\/span>   <\/span>\"\[I=\big[te^t\big]_0^x- \int_0^x e^t\mathrm{d}t=xe^x-e^x+1=(x-1)e^x+1.\]\"<\/p>\nLes primitives de \"x\mapsto xe^x\" sont donc les fonctions \"x\mapsto (x-1)e^x+C\" avec \"C\in\mathbb{R}\".\n\n\n\n

Proposition : Changement de variable<\/h3>\n\n\n\n

<\/p>\nSoit \"f\" une fonction continue sur un intervalle \"I\" et \"\varphi\" une fonction de classe \"\mathcal{C}^1\" sur \"[a,b ]\" \u00e0 valeurs dans \"I\". Alors :\n

  <\/span>   <\/span>\"\[\int_a^b f(\varphi(x))\varphi'(x)\mathrm{d}x=\int_{\varphi(a)}^{\varphi(b)} f(y)\mathrm{d}y.\]\"<\/p>\n\n\n\n

D\u00e9monstration<\/h3>\n\n\n\n

<\/p>\nEn utilisant les formules de d\u00e9rivations (notamment la d\u00e9riv\u00e9e d’une compos\u00e9e), on peut montrer que les fonctions \"\displaystyle{x\mapsto \int_a^x f(\varphi(t))\varphi'(t)\mathrm{d}t}\" et \"\displaystyle{x\mapsto\int_{\varphi(a)}^{\varphi(x)} f(t)\mathrm{d}t}\" sont des primitives de \"x\mapsto f(\varphi(x))\varphi'(x)\" sur \"[a,b]\" qui s’annulent en \"a\". On en d\u00e9duit que ces fonctions sont \u00e9gales.\n\n\n\n

Exemple<\/h3>\n\n\n\n

<\/p>\nCalculons \"\displaystyle I=\int_0^1 \sqrt{1-t^2}\mathrm{d}t\" par changement de variable : on pose \"t=\cos(u)\", alors :\n

  • \"\sqrt{1-t^2}=\sqrt{1-\cos(u)^2}=|\sin(u)|\" ; <\/li>\n
  • on d\u00e9termine les nouvelles bornes : lorsque \"t=0\", \"u=\dfrac{\pi}{2}\" et lorsque \"t=1\", \"u=0\"; <\/li>\n
  • on calcule l’\u00e9l\u00e9ment diff\u00e9rentiel : \"\mathrm{d}t=-\sin(u)\mathrm{d}u\". <\/li> \nPar changement de variable, on a :\n

      <\/span>   <\/span>\"\[I=\int_{\pi/2}^0 -|\sin(u)|\sin(u)\mathrm{d}u=\int_0^{\pi/2} \sin(u)^2\mathrm{d}u=\int_0^{\pi/2} \dfrac{1+\cos(2u)}{2}\mathrm{d}u=\dfrac{\pi}{4}.\]\"<\/p>\n\n\n\n

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    Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/em>(\u00e9ditions Vuibert, juin 2021) <\/em>\u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n<\/div>\n<\/div>\n\n\n

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    \n 3.7\/5 - (3 votes) <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

    Vous travaillez actuellement sur des calculs de primitives ? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9 \u00e0 la m\u00e9thode de (…)<\/p>\n","protected":false},"author":158,"featured_media":161830,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-235119","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235119","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=235119"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/235119\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/161830"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=235119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=235119"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=235119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}