{"id":4094,"date":"2020-05-06T14:45:11","date_gmt":"2020-05-06T13:45:11","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=4094"},"modified":"2020-05-12T10:43:57","modified_gmt":"2020-05-12T09:43:57","slug":"dm-guide-integrale-et-fonction-ln","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=4094","title":{"rendered":"DM Guid\u00e9 Int\u00e9grale et fonction ln 090520"},"content":{"rendered":"<p><strong>Partie A<\/strong><\/p>\n<ol>\n<li>[peekaboo_link name=\u00a0\u00bbquestionA1&Prime;]Valeurs de u(1) et u(0)<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionA1&Prime;] u(1)=0 \u00a0car le point A(1,0) appartient \u00e0 la courbe de u et u(0)=4 \u00a0car le point B(0,4) appartient \u00e0 la courbe de u<br \/>\n[\/peekaboo_content]<\/li>\n<li>[peekaboo_link name=\u00a0\u00bbquestionA2&Prime;]Limite de u en +\u221e<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionA2&Prime;] La limite est \u00e9gale \u00e0 1 car la droite D qui a pour \u00e9quation y=1 est asymptote \u00e0 la courbe de u en +\u221e<br \/>\n[\/peekaboo_content][peekaboo_link name=\u00a0\u00bbquestionA22&Prime;]Valeur de a[\/peekaboo_link]<br \/>\n[peekaboo_content name=\u00a0\u00bbquestionA22&Prime;] Il faut calculer la limite de u(x) puis dire que cette limite vaut 1, ce qui donne une information sur a.<br \/>\n[\/peekaboo_content]<\/li>\n<li>[peekaboo_link name=\u00a0\u00bbquestionA3&Prime;]Trouver l&rsquo;expression de u(x) c&rsquo;est-\u00e0-dire trouver les valeurs de b et c.<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionA3&Prime;] Traduire u(1)=0 et u(0)=4 \u00e0 partir de u(x) ce qui donne deux \u00e9quations d&rsquo;inconnues b et c et donc \u00a0un syst\u00e8me \u00e0 r\u00e9soudre.<br \/>\nRemarque : b=-5 et c=4[\/peekaboo_content]<\/li>\n<\/ol>\n<p><strong>Partie B<\/strong><\/p>\n<ol>\n<li>[peekaboo_link name=\u00a0\u00bbquestionB1&Prime;]Limite de f en 0<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionB1&Prime;] Pour faire apparaitre xln(x), il faut r\u00e9duire f(x) au m\u00eame d\u00e9nominateur<br \/>\n[\/peekaboo_content]<\/li>\n<li>[peekaboo_link name=\u00a0\u00bbquestionB2&Prime;]Limite de f en +\u221e<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionB2&Prime;] On tombe sur une F.I. (mettre x en facteur ce qui fait apparaitre <img src='http:\/\/s0.wp.com\/latex.php?latex=%5Cdfrac%7B%5Cln+x%7D%7Bx%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\dfrac{\\ln x}{x}' title='\\dfrac{\\ln x}{x}' class='latex' \/>)<br \/>\n[\/peekaboo_content]<\/li>\n<li>[peekaboo_link name=\u00a0\u00bbquestionB3&Prime;]Signe de f'(x)<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionB3&Prime;] Remarque : si f'(x) = u(x) alors le signe de f'(x) est celui de u(x) donc d&rsquo;apr\u00e8s la courbe de u, f'(x) est positif si 0&lt;x&lt;1 et si x&gt;4 et f'(x)&lt;0 si 1&lt;x&lt;4.<br \/>\nLe signe de f'(x) doit \u00eatre trouv\u00e9 par le calcul mais le graphique vous permet de v\u00e9rifier vos r\u00e9sultats[\/peekaboo_content]<\/li>\n<\/ol>\n<p><strong>Partie C<\/strong><\/p>\n<ol>\n<li>[peekaboo_link name=\u00a0\u00bbquestionC1&Prime;]Aire n\u00b01<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionC1&Prime;] Attention, la courbe est situ\u00e9e sous l&rsquo;axe des abscisses donc l&rsquo;aire n&rsquo;est pas \u00e9gale \u00e0 l&rsquo;int\u00e9grale mais \u00e0 l&rsquo;oppos\u00e9 de l&rsquo;int\u00e9grale.<br \/>\nSinon vous trouveriez une aire n\u00e9gative !<br \/>\n<img src='http:\/\/s0.wp.com\/latex.php?latex=%5Cmathscr%7BA%7D%3D-%5Cint_%7B1%7D%5E4+u%28x%29+dx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathscr{A}=-\\int_{1}^4 u(x) dx' title='\\mathscr{A}=-\\int_{1}^4 u(x) dx' class='latex' \/><br \/>\n[\/peekaboo_content]<\/li>\n<li>[peekaboo_link name=\u00a0\u00bbquestionC2&Prime;]Aire n\u00b02<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionC2&Prime;] On s&rsquo;int\u00e9resse \u00e0 l&rsquo;aire du domaine situ\u00e9 sous la courbe de u \u00e0 partir du point B(x=4) et jusqu&rsquo;au point de coordonn\u00e9es (<img src='http:\/\/s0.wp.com\/latex.php?latex=%5Clambda&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\lambda' title='\\lambda' class='latex' \/> ; 0) et situ\u00e9 au dessus de l&rsquo;axe des abscisses.<br \/>\n[\/peekaboo_content]<br \/>\n[peekaboo_link name=\u00a0\u00bbquestionC22&Prime;]Aide \u00a0pour derni\u00e8re question<br \/>\n[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionC22&Prime;] <img src='http:\/\/s0.wp.com\/latex.php?latex=%5Cmathscr%7BA%7D_%5Clambda+%3D%5Cint_%7B4%7D%5E%7B%5Clambda%7D+u%28x%29+dx%3Df%28%5Clambda%29-f%284%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathscr{A}_\\lambda =\\int_{4}^{\\lambda} u(x) dx=f(\\lambda)-f(4)' title='\\mathscr{A}_\\lambda =\\int_{4}^{\\lambda} u(x) dx=f(\\lambda)-f(4)' class='latex' \/> car f est une primitive de u<br \/>\n<img src='http:\/\/s0.wp.com\/latex.php?latex=%5Cmathscr%7BA%7D+%3D+%5Cmathscr%7BA%7D_%5Clambda&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\mathscr{A} = \\mathscr{A}_\\lambda' title='\\mathscr{A} = \\mathscr{A}_\\lambda' class='latex' \/> ssi <img src='http:\/\/s0.wp.com\/latex.php?latex=f%28%5Clambda%29%3D-3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(\\lambda)=-3' title='f(\\lambda)=-3' class='latex' \/><br \/>\nLe probl\u00e8me revient \u00e0 se poser la question suivante :<br \/>\nd&rsquo;apr\u00e8s le tableau de variation de f, peut-on avoir une image qui prenne la valeur -3 ?<br \/>\n[\/peekaboo_content]<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Partie A [peekaboo_link name=\u00a0\u00bbquestionA1&Prime;]Valeurs de u(1) et u(0) [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionA1&Prime;] u(1)=0 \u00a0car le point A(1,0) appartient \u00e0 la courbe de u et u(0)=4 \u00a0car le point B(0,4) appartient \u00e0 la courbe de u [\/peekaboo_content] [peekaboo_link name=\u00a0\u00bbquestionA2&Prime;]Limite de u en +\u221e [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestionA2&Prime;] La limite est \u00e9gale \u00e0 1 car la droite D qui a pour [&hellip;]<\/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\/4094"}],"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=4094"}],"version-history":[{"count":17,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/4094\/revisions"}],"predecessor-version":[{"id":4160,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/4094\/revisions\/4160"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4094"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4094"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4094"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}