{"id":1249,"date":"2015-03-20T17:45:15","date_gmt":"2015-03-20T16:45:15","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1249"},"modified":"2019-05-04T08:48:10","modified_gmt":"2019-05-04T07:48:10","slug":"dm-guide-integrales-n1-calcul-daire","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1249","title":{"rendered":"DM guid\u00e9 Int\u00e9grales n\u00b01 (calcul d’aire)"},"content":{"rendered":"

Sujet :\u00a0Cliquer ici<\/a><\/strong>
\nCorrection en fin de page<\/p>\n

Partie A<\/p>\n

    \n
  1. Courbe de la d\u00e9riv\u00e9e
    \n[peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]Les variations de f nous donne le signe de la d\u00e9riv\u00e9e[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]Situation 1 (le signe de f’ change \u00e0 0,5 environ puisque f change de variation pour cette valeur de x)[\/peekaboo_content]<\/li>\n
  2. Equation de la tangente en A
    \n[peekaboo_link name=\u00a0\u00bbquestion3″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3″] y = x + 2 [\/peekaboo_content]<\/li>\n
  3. D\u00e9terminer\u00a0les valeurs de a et b
    \n[peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb]Utiliser les coordonn\u00e9es des points A et B, sachant que A appartient \u00e0 la courbe de f et que B appartient \u00e0 la courbe de f'[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion4″]Aide 2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]D’apr\u00e8s le point A, on a f (0) = 2 et d’apr\u00e8s le point B, on a f’ (0) = 1[\/peekaboo_content]<\/li>\n
  4. Expression de f(x)
    \n[peekaboo_link name=\u00a0\u00bbquestion4b\u00a0\u00bb]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4b\u00a0\u00bb] f(x)=e^{-x}+2x+1 [\/peekaboo_content]
    \nVariation de f
    \n[peekaboo_link name=\u00a0\u00bbquestion4c\u00a0\u00bb]R\u00e9ponse partielle\u00a0[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4c\u00a0\u00bb]f'(x) change de signe en – ln(2)
    \nRemarque : on connait la courbe de f, il faut donc penser \u00e0 v\u00e9rifier que nos variations sont bien celles donn\u00e9es par le graphique ! [\/peekaboo_content]<\/li>\n<\/ol>\n

    Partie B<\/strong><\/p>\n

      \n
    1. a.\u00a0Variation de la fonction g
      \n[peekaboo_link name=\u00a0\u00bbquestion5″]R\u00e9ponse partielle\u00a0[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]g’ change de signe en 0[\/peekaboo_content]
      \nb.\u00a0Position de la courbe par rapport \u00e0 la droite
      \n[peekaboo_link name=\u00a0\u00bbquestion6″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″]Pour d\u00e9terminer la position de la courbe de f par rapport \u00e0 la droite d’\u00e9quation y = x + 2, on cherche le signe de f(x) – (x+2) c’est-\u00e0-dire le signe de g(x).
      \nRemarque : On peut conjecturer la position en tra\u00e7ant sur le graphique la droite d’\u00e9quation y = x + 2
      \n[\/peekaboo_content]<\/li>\n
    2. Calcul de l’aire\u00a0color\u00e9e en gris
      \n[peekaboo_link name=\u00a0\u00bbquestion12″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion12″] L’aire se calcule par diff\u00e9rence de deux aires. (Aire sous la courbe de f – Aire sous la droite)[\/peekaboo_content]
      \n<\/span> [peekaboo_link name=\u00a0\u00bbquestion13″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion13″]Environ 3,25 unit\u00e9s d’aire, ce qui semble \u00eatre coh\u00e9rent avec le graphique. [\/peekaboo_content]<\/li>\n<\/ol>\n

      Correction : cliquer ici<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

      Sujet :\u00a0Cliquer ici Correction en fin de page Partie A Courbe de la d\u00e9riv\u00e9e [peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]Les variations de f nous donne le signe de la d\u00e9riv\u00e9e[\/peekaboo_content] [peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]Situation 1 (le signe de f’ change \u00e0 0,5 environ puisque f change de variation pour cette valeur de x)[\/peekaboo_content] Equation de la tangente en A […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1249"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1249"}],"version-history":[{"count":8,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1249\/revisions"}],"predecessor-version":[{"id":3418,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1249\/revisions\/3418"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1249"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1249"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1249"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}