Quiz — Intégrales dépendant d'un paramètre.pdf

Question 1 sur 10 20:00

[{"id":85745,"question":"Soit la fonction définie par $f(x) = \\int_{0}^{1} e^{tx} dt$. Que vaut $f'(x)$ ?","option_a":"$f'(x) = \\int_{0}^{1} t e^{tx} dt$","option_b":"$f'(x) = \\int_{0}^{1} e^{tx} dt$","option_c":"$f'(x) = x e^{x}$","option_d":"$f'(x) = e^{x} - 1$","option_e":"","option_f":"","bonne_reponse":"a","explication":"D'après le théorème de Leibniz, on peut dériver sous le signe intégral : $f'(x) = \\int_{0}^{1} \\frac{\\partial}{\\partial x} e^{tx} dt = \\int_{0}^{1} t e^{tx} dt$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"$f'(x) = \\\\int_{0}^{1} t e^{tx} dt$\", \"b\": \"$f'(x) = \\\\int_{0}^{1","_debug_options_count":4},{"id":85746,"question":"L'intégrale $\\int_{a}^{b} f(x, \\lambda) dx$ est toujours continue par rapport à $\\lambda$ si $f$ est continue en $(x, \\lambda)$.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est une conséquence du théorème de continuité sous le signe intégral : si $f$ est continue sur un compact, l'intégrale est continue par rapport au paramètre.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":85747,"question":"Soit $F(\\lambda) = \\int_{0}^{\\lambda} \\sin(x + \\lambda) dx$. Que vaut $F'(\\lambda)$ ?","option_a":"$F'(\\lambda) = \\sin(2\\lambda)$","option_b":"$F'(\\lambda) = \\cos(\\lambda) + \\int_{0}^{\\lambda} \\cos(x + \\lambda) dx$","option_c":"$F'(\\lambda) = \\cos(2\\lambda)$","option_d":"$F'(\\lambda) = \\sin(\\lambda)$","option_e":"","option_f":"","bonne_reponse":"b","explication":"En appliquant le théorème de Leibniz : $F'(\\lambda) = \\sin(\\lambda + \\lambda) + \\int_{0}^{\\lambda} \\cos(x + \\lambda) dx = \\cos(\\lambda) + \\int_{0}^{\\lambda} \\cos(x + \\lambda) dx$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"$F'(\\\\lambda) = \\\\sin(2\\\\lambda)$\", \"b\": \"$F'(\\\\lambda) = \\\\cos(\\","_debug_options_count":4},{"id":85748,"question":"L'intégrale $\\int_{0}^{1} \\frac{1}{x + \\lambda} dx$ converge pour tout $\\lambda \u003E 0$.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour $\\lambda \u003E 0$, la fonction $\\frac{1}{x + \\lambda}$ est continue sur $[0, 1]$, donc l'intégrale converge.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":85749,"question":"Soit $G(\\lambda) = \\int_{0}^{1} \\ln(x + \\lambda) dx$. Que vaut $G'(1)$ ?","option_a":"$G'(1) = 0$","option_b":"$G'(1) = \\int_{0}^{1} \\frac{1}{x + 1} dx$","option_c":"$G'(1) = \\ln(2)$","option_d":"$G'(1) = 1$","option_e":"","option_f":"","bonne_reponse":"b","explication":"D'après le théorème de Leibniz : $G'(\\lambda) = \\int_{0}^{1} \\frac{1}{x + \\lambda} dx$. Donc $G'(1) = \\int_{0}^{1} \\frac{1}{x + 1} dx$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"$G'(1) = 0$\", \"b\": \"$G'(1) = \\\\int_{0}^{1} \\\\frac{1}{x + 1} dx$\",","_debug_options_count":4},{"id":85750,"question":"L'intégrale $\\int_{0}^{\\infty} e^{-\\lambda x} dx$ converge pour tout $\\lambda \u003E 0$.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Pour $\\lambda \u003E 0$, l'intégrale $\\int_{0}^{\\infty} e^{-\\lambda x} dx$ converge et vaut $\\frac{1}{\\lambda}$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":85751,"question":"Soit $H(\\lambda) = \\int_{0}^{\\pi} \\sin(x\\lambda) dx$. Que vaut $H(\\lambda)$ ?","option_a":"$H(\\lambda) = \\frac{1 - \\cos(\\pi \\lambda)}{\\lambda}$","option_b":"$H(\\lambda) = \\frac{2}{\\lambda}$","option_c":"$H(\\lambda) = 0$","option_d":"$H(\\lambda) = \\sin(\\pi \\lambda)$","option_e":"","option_f":"","bonne_reponse":"a","explication":"En calculant l'intégrale : $H(\\lambda) = \\left[-\\frac{\\cos(x\\lambda)}{\\lambda}\\right]_{0}^{\\pi} = \\frac{1 - \\cos(\\pi \\lambda)}{\\lambda}$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"$H(\\\\lambda) = \\\\frac{1 - \\\\cos(\\\\pi \\\\lambda)}{\\\\lambda}$\", \"b\":","_debug_options_count":4},{"id":85752,"question":"Si $f(x, \\lambda)$ est dérivable par rapport à $\\lambda$ et que $\\frac{\\partial f}{\\partial \\lambda}$ est continue, alors $\\int_{a}^{b} f(x, \\lambda) dx$ est dérivable par rapport à $\\lambda$.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est une conséquence directe du théorème de dérivation sous le signe intégral (théorème de Leibniz).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":85753,"question":"Soit $I(\\lambda) = \\int_{0}^{1} x^{\\lambda} dx$. Que vaut $I(2)$ ?","option_a":"$I(2) = \\frac{1}{3}$","option_b":"$I(2) = 1$","option_c":"$I(2) = \\frac{1}{2}$","option_d":"$I(2) = 2$","option_e":"","option_f":"","bonne_reponse":"a","explication":"En calculant l'intégrale : $I(\\lambda) = \\left[\\frac{x^{\\lambda + 1}}{\\lambda + 1}\\right]_{0}^{1} = \\frac{1}{\\lambda + 1}$. Donc $I(2) = \\frac{1}{3}$.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"$I(2) = \\\\frac{1}{3}$\", \"b\": \"$I(2) = 1$\", \"c\": \"$I(2) = \\\\frac{1","_debug_options_count":4},{"id":85754,"question":"L'intégrale $\\int_{-1}^{1} \\frac{1}{x + \\lambda} dx$ converge pour tout $\\lambda \\neq 0$.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Pour $\\lambda \\neq 0$, la fonction $\\frac{1}{x + \\lambda}$ est continue sur $[-1, 1]$ si $\\lambda \\notin [-1, 1]$. Sinon, l'intégrale est impropre et peut diverger.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]