{"id":34336,"date":"2019-08-19T15:34:00","date_gmt":"2019-08-19T14:34:00","guid":{"rendered":"https:\/\/groupe-reussite.fr\/?page_id=34336"},"modified":"2021-09-17T17:23:02","modified_gmt":"2021-09-17T16:23:02","slug":"cours-en-ligne-corriges-integration-ecs1","status":"publish","type":"page","link":"https:\/\/groupe-reussite.fr\/ressources\/cours-en-ligne-corriges-integration-ecs1\/","title":{"rendered":"Int\u00e9gration en ECG1 – Corrig\u00e9s"},"content":{"rendered":"
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Cours en ligne Maths en ECG1<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en ECG1<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspace Probabilis\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tRaisonnement et Vocabulaire Ensembliste<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tEspaces Vectoriels et Applications Lin\u00e9aires<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Cours en ligne Maths en ECG1<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en ECG1<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tMatrices<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFonctions R\u00e9elles \u00e0 Variables R\u00e9elles<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tProbabilit\u00e9s (univers infini)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFormules de Taylor et D\u00e9veloppements Limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tExtrema et convexit\u00e9<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tS\u00e9ries Num\u00e9riques<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tInt\u00e9gration<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tVariables Al\u00e9atoires Finies<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tConvergence et Approximations<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCONTACTEZ-NOUS<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t\n<\/div>\n
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Corrig\u00e9s : Int\u00e9gration en ECG1<\/h1>\n<\/div>\n

R\u00e9sum\u00e9 de cours<\/a> Exercices<\/a> Corrig\u00e9s<\/a><\/p>\n

Cours en ligne de Maths en ECG1<\/a><\/p>\n

Corrig\u00e9s – Int\u00e9gration<\/h2>\n

Exercice 1 :\u00a0<\/b><\/h3>\n

1)<\/strong> L’expression \"x \mapsto \frac{\ln \left( x \right)}{x}\" (de la forme \"u' u\") se primitive en \"x \mapsto \dfrac12 \ln^2 \left( x \right)\" ainsi<\/p>\n

  <\/span>   <\/span>\"\[\int_1^2 \dfrac{\ln \left( x \right)}{x} dx = \left[ \dfrac12 \ln^2 \left( x \right) \right]_1^2 = \dfrac12 \ln^2 \left( 2 \right).\]\"<\/p>\n<\/p>\n

2)<\/strong> Commen\u00e7ons par lin\u00e9ariser \"\cos^4 \left( t \right).\" On utilise la formule de Moivre-Euler<\/p>\n

\"\cos^4 \left( t \right)\" \"= \left( \dfrac{e^{it} - e^{-it}}{2} \right)^4\"<\/p>\n

\"= \dfrac{e^{4it} + 4 e^{3it } + 6 + 4 e^{- 3it} + e^{- 4it} }{16}\"<\/p>\n

\"= \dfrac18 \cos \left( 4 t \right) + \dfrac12 \cos \left( 3 t \right) + \dfrac38\".<\/p>\n

D’o\u00f9<\/p>\n

\"\int_{0}^{\dfrac{\pi}{2}}\" \"= \left[ \dfrac{\sin \left( 4 t \right)}{32} + \dfrac{\sin \left( 3 t \right)}{6} + \dfrac38 t \right]_0^{\dfrac{\pi}{2}} .\"<\/p>\n

3)<\/strong> On \u00e9crit<\/p>\n

\"\int_0^{\dfrac12} \dfrac{x}{ \sqrt{1 - x^2}} dx\" \"= - \dfrac12 \int_0^{\dfrac12} \dfrac{- 2 x}{\sqrt{ 1 - x^2} } dx.\"<\/div>\n
L’expression \"x \to \dfrac{- 2 x}{\sqrt{ 1 - x^2} }\" (de la forme \"u' u^{- \dfrac12}\") se primitive en \"x \mapsto 2 \sqrt{1 - x^2},\" ainsi<\/div>\n
<\/div>\n
\"- \dfrac12 \int_0^{\dfrac12} \dfrac{- 2 x}{\sqrt{ 1 - x^2} } dx\" \"= -\dfrac12 \left[ 2 \sqrt{1 - x^2} \right]_0^{\dfrac12}\" \"= - \sqrt{1 - \left( \dfrac12 \right)^2 } + 1.\"<\/div>\n
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4)<\/strong> On fait une int\u00e9gration par parties donne, en posant \"u : t \mapsto \dfrac{t^3}{3}\" et \"v : t \mapsto \ln \left( t \right).\" Les fonctions<\/a> \"u\" et \"v\" sont \"C^1\" sur l’intervalle \"\left[ 1 , e \right]\" et :<\/span><\/div>\n
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\"\int_1^e t^2 \ln \left( t \right) dt\" \"= \left[ \dfrac{t^3}{3} \ln \left( t \right) \right]_1^e - \int_1^e \dfrac{t^2}{3} dt\"
\n\"= \dfrac{e^3}{3} - \left[ \dfrac{t^3}{9} \right]_1^e\"
\n\"= \dfrac{e^3}{3} - \dfrac{e^3}{9} + \dfrac19\"
\n\"= \dfrac{2 e^3}{9} + \dfrac19\".<\/div>\n
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5)<\/strong> On fait une int\u00e9gration par parties en posant \"u : t \mapsto \mathrm{Arctan} \left( t \right)\" et \"v : t \mapsto \dfrac{t^2}{2}.\" Les fonctions \"u\" et \"v\" sont \"C^1\" sur \"\left[ 0 , 1 \right]\" et :<\/span><\/div>\n
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\"\int_0^1 t \mathrm{Arctan} \left( t \right) dt\" \"= \left[ \dfrac12 \mathrm{Arctan} \left( t \right) t^2 \right]_0^1 - \dfrac12 \int_0^1 \dfrac{t^2}{1 + t^2} dt\"<\/p>\n

\"= \dfrac12 \mathrm{Arctan} \left( 1 \right) - \int_0^1 \dfrac{t^2 + 1 - 1}{1 + t^2} dt\"
\n\"= \dfrac12 \dfrac{\pi}{4} - \int_0^1 1 dt + \int_0^1 \dfrac{1}{1 + t^2} dt\"
\n\"= \dfrac{\pi}{8} - 1 + \mathrm{Arctan} \left( 1 \right)\"
\n\"= \dfrac{\pi}{8} + \dfrac{\pi}{4} - 1\"
\n\"= \dfrac{3 \pi}{ 8}- 1\".<\/div>\n<\/div>\n<\/div>\n

Exercice 2 :\u00a0<\/b><\/h3>\n

1)<\/strong> Si l’on pose \"y = x^2,\" on commence par remplacer \"x\" par \"y,\" on a donc :<\/p>\n

\n

  <\/span>   <\/span>\"\[\int \dfrac{\sqrt{y}}{ \left(y + 1 \right)^2 }.\]\"<\/p>\n

On exprime \"dx\" en fonction de \"dy.\" Comme \"y = x^2,\" on a \"dy = 2 x dx = 2 \sqrt{y} dx\" soit \"dx = \dfrac{1}{2 \sqrt{y}} dy,\" ainsi<\/div>\n

<\/div>\n
\n

  <\/span>   <\/span>\"\[\dfrac12 \int \dfrac{1}{ \left( y + 1 \right)^2} dy .\]\"<\/p>\n

Il nous reste \u00e0 trouver les bonne bornes : lorsque \"x = 1, y = 1^2 = 1\" et lorsque \"x= 10, y=10^2 = 100,\" d’o\u00f9 finalement :<\/div>\n

<\/div>\n
\n

  <\/span>   <\/span>\"\[\dfrac12 \int_1^{100} \dfrac{1}{ \left( y + 1 \right)^2} dy.\]\"<\/p>\n

Cette derni\u00e8re est plus facile \u00e0 calculer car \"y \to \dfrac{1}{ \left( y + 1 \right)^2}\" se primitive en \"y \to - \dfrac{1}{y + 1},\" d’o\u00f9 :<\/div>\n

<\/div>\n
\"\dfrac12 \int_1^{100} \dfrac{1}{ \left( y + 1 \right)^2} dy\" \"= \dfrac12 \left[ - \dfrac{1}{y + 1} \right]_1^{100}\" \"= \dfrac12 \left( - \dfrac{1}{101} + \dfrac12 \right).\"<\/div>\n
<\/div>\n
2)<\/strong> On va un peu plus vite : l’int\u00e9grale, apr\u00e8s le changement de variable, est\u00a0<\/span><\/p>\n

  <\/span>   <\/span>\"\[\int_{0}^{\dfrac{\pi}{2}} \cos^2 \left( \theta \right) d \theta.\]\"<\/p>\n

Pour calculer cette int\u00e9grale, il faut lin\u00e9ariser \"\cos^2 \left( \theta \right).\" On utilise les formules de Moivre-Euler :\"\cos^2 \left( \theta \right)\" \"= \left( \dfrac{e^{i\theta} + e^{- i \theta} }{2} \right)^2\"
\n\"= \dfrac{e^{2i \theta} + e^{- 2i \theta } + 2}{4}\"
\n\"= \dfrac{\cos \left( 2 \theta \right) + 1}{2}\".<\/p>\n

Ainsi<\/p>\n

\n

  <\/span>   <\/span>\"\[\int_{0}^{\dfrac{\pi}{2}} \cos^2 \left( \theta \right) d \theta = \int_0^{\dfrac{\pi}{2}} \dfrac{\cos \left( 2 \theta \right) + 1}{2} d \theta = \dfrac{\pi}{4}.\]\"<\/p>\n<\/p>\n

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PRENDRE DES COURS PARTICULIERS EN ECG1<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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C\u2019est reprendre le contr\u00f4le<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers en pr\u00e9pa ECG<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
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Exercice 3 :\u00a0<\/b><\/h3>\n

1)<\/strong> On a<\/p>\n

\"I_{0 ,q} = \int_a^b \left( b - t \right)^q dt\"<\/p>\n

\"= \left[ - \dfrac{ \left( b - t \right)^{q + 1} }{q + 1} \right]_a^b\"
\n\"= \dfrac{ \left( b- a \right)^{q + 1} }{q + 1}\".<\/div>\n
<\/div>\n
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2)<\/strong> On fait une int\u00e9gration par parties en posant \"u : t \mapsto \left( t- a \right)^p\" et \"v : t \mapsto - \dfrac{ \left( b - t \right)^{q + 1}}{q + 1}.\" Les fonctions \"u\" et \"v\" sont \"C^1\" sur \"\left[ a , b \right]\" et :<\/span><\/p>\n

\"I_{p , q}\" \"= \left[ - \dfrac{\left( t - a \right)^p \left( b - t \right)^{q + 1} }{q + 1} \right]_a^b\" \"+ \dfrac{p}{q + 1} \int_a^b \left( t -a \right)^{p - 1} \left( b - t \right)^{q + 1} dt\".<\/div>\n
Comme \"\left[ - \dfrac{\left( t - a \right)^p \left( b - t \right)^{q + 1} }{q + 1} \right]_a^b = 0,\" on a \"I_{p , q} = \dfrac{p}{q + 1} I_{p - 1 , q +1}.\"<\/div>\n
<\/div>\n
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3)<\/strong> Si l’on applique \"k\" fois (avec \"0 \le k \le p\") la relation pr\u00e9c\u00e9dente, on a :<\/span><\/p>\n

\"I_{p , q}\" \"= \dfrac{p \left( p - 1 \right) \times \cdots \times \left( p - \left( k - 1 \right) \right)}{\left( q + 1 \right)\times \left( q + 2 \right) \times \cdots \times \left( q + k \right)} I_{p - k , q + k}.\"
\nEn prenant \"k = p,\" on a<\/div>\n
<\/div>\n
\n

  <\/span>   <\/span>\"\[I_{p , q} = \dfrac{p! q!}{\left( q + p \right)!} I_{0 , p + q}.\]\"<\/p>\n

Comme \"I_{0 , p + q} = \dfrac{\left( b - a \right)^{p + q + 1}}{p + q + 1}\" d’o\u00f9 finalement<\/div>\n

<\/div>\n
\n

  <\/span>   <\/span>\"\[I_{p , q} = \dfrac{p! q!}{\left( q + p \right)!} \times \dfrac{\left( b - a \right)^{p + q + 1}}{p + q + 1} = \dfrac{p! q! \left( b - a \right)^{p + q + 1}}{\left( p + q + 1 \right)!}.\]\"<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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Exercice 4 :\u00a0<\/b><\/h3>\n

1)<\/strong> On calcule<\/p>\n

\"I_{n + 1 } - I_n\" \"= \int_0^1 \dfrac{t^{n + 1}}{1 + t^2} dt - \int_0^1 \dfrac{t^n}{1 + t^2} dt\" \"= \int_0^1 t^n \dfrac{t - 1}{ 1 + t^2} dt.\"
\nComme \"t - 1 \le 0\" pour \"0 \le t \le 1\" ainsi \"\int_0^1 t^n \dfrac{t - 1}{ 1 + t^2} dt \le 0\" et \"I_{n + 1} - I_n \le 0\" et donc la suite \"\left( I_n \right)_{n \in \mathbb{N}}\" est d\u00e9croissante.<\/div>\n
<\/div>\n
Pour voir que la suite<\/a> \"\left( I_n \right)_{n \in \mathbb{N}}\" est minor\u00e9e, il suffit de voir que pour tout \"n \in \mathbb{N},\" \"I_n \ge 0.\"<\/div>\n

2)<\/strong> On encadre \"I_n\" par deux suites ayant \"0\" pour limite.<\/p>\n

Pour \"0\le t \le 1,\" on a \"0 \le \dfrac{1}{1 + t^2} \le \dfrac12\" puis \"0 \le \dfrac{t^n}{1 + t^2} \le \dfrac12 t^n\" et \"0 \le I_n \le \dfrac12 \int_0^1 t^n = \dfrac{1}{2 \left( n + 1 \right)}.\" Comme \"\lim_{n \to +\infty} \dfrac{1}{2 \left( n + 1 \right)} = 0,\" on en d\u00e9duit que gr\u00e2ce au th\u00e9or\u00e8me d’encadrement que \"\lim_{n \to +\infty} I_n = 0.\"<\/div>\n
<\/div>\n
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3)<\/strong> On une int\u00e9gration par parties en posant \"u : t \mapsto \dfrac{1}{1 + t^2}\" et \"v : t \mapsto \dfrac{t^{n + 1} }{n + 1}.\" \"u\" et \"v\" sont \"C^1\" sur \"\left[ 0 , 1 \right],\" on a donc<\/span><\/p>\n

\"I_n = \left[ \dfrac{1}{1 + t^2} \times \dfrac{t^{n + 1}}{n + 1} \right]_0^1 + \dfrac{2}{n + 1} \int_0^1 \dfrac{t^{n +2} }{1 + t^2} dt\"
\n\"= \dfrac{1}{2 \left( n + 1 \right)} + \dfrac{2}{n + 1} \int_0^1 \dfrac{t^{n +2} }{1 + t^2} dt\".<\/div>\n
De sorte que<\/div>\n
<\/div>\n
\"n I_n = \dfrac12 \times \dfrac{n}{n + 1} + \dfrac{2n}{n + 1} \int_0^1 \dfrac{t^{n + 2}}{1 + t^2} dt.\"
\nEn proc\u00e9dant comme ci-dessus, on montre que \"\lim_{n \to +\infty} \int_0^1 \dfrac{t^{n + 2}}{1 + t^2} dt = 0.\"<\/div>\n
<\/div>\n
De plus, \"\lim_{n \to +\infty} \dfrac12 \times \dfrac{n}{n + 1} = \dfrac12,\" on obtient :<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[\lim_{n \to +\infty} n I_n = \dfrac12 .\]\"<\/p>\n<\/div>\n

M\u00eame si certains chapitres n’ont pas encore \u00e9t\u00e9 travaill\u00e9s en cours, vous pouvez d’ores et d\u00e9j\u00e0 en prendre connaissance et en d\u00e9couvrir les grandes notions avec les cours en ligne d’ECG1 en maths :<\/div>\n
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