Để tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9maapehabaWaaeWaaeaacaWG4bGaey4kaSIaaGymaaGaayjkaiaa % wMcaaiaaikdadaahaaWcbeqaaiaadIhaaaaabaGaaGimaaqaaiaaig % daa0Gaey4kIipaaaa!41A8! M = \int\limits_0^1 {\left( {x + 1} \right){2^x}} \) bằng phương pháp tích phân từng phần ta đặt u = x + 1 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaabaaa % aaaaaapeGaamODaiabg2da9iaaikdadaahaaWcbeqaaiaadIhaaaGc % caWGKbGaamiEaaaa!3CD2! {\rm{d}}v = {2^x}dx\). Tìm du và tính M

A. du = dx và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9maalaaabaGaaG4maaqaaiGacYgacaGGUbGaaGOmaaaacqGHRaWk % daWcaaqaaiaaigdaaeaadaqadaqaaiGacYgacaGGUbGaaGOmaaGaay % jkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!41F8! M = \frac{3}{{\ln 2}} + \frac{1}{{{{\left( {\ln 2} \right)}^2}}}\)

B. du = dx và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9maalaaabaGaaG4maaqaaiGacYgacaGGUbGaaGOmaaaacqGHsisl % daWcaaqaaiaaigdaaeaadaqadaqaaiGacYgacaGGUbGaaGOmaaGaay % jkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!4203! M = \frac{3}{{\ln 2}} - \frac{1}{{{{\left( {\ln 2} \right)}^2}}}\)

C. du = 1 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9iaaiodaciGGSbGaaiOBaiaaikdacqGHsisldaqadaqaaiGacYga % caGGUbGaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaa % a!4128! M = 3\ln 2 - {\left( {\ln 2} \right)^2}\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadw % hacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadIhadaahaaWc % beqaaiaaikdaaaGccqGHRaWkcaWG4baaaa!3E31! {\rm{d}}u = \frac{1}{2}{x^2} + x\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9maalaaabaGaaG4maaqaaiGacYgacaGGUbGaaGOmaaaacqGHsisl % daWcaaqaaiaaigdaaeaadaqadaqaaiGacYgacaGGUbGaaGOmaaGaay % jkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!4203! M = \frac{3}{{\ln 2}} - \frac{1}{{{{\left( {\ln 2} \right)}^2}}}\)

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Môn: Toán Lớp 12

Chủ đề: Nguyên Hàm - Tích Phân Và Ứng Dụng

Bài: Tích phân

Lời giải:

Báo sai

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4bGaey4kaSIaaGymaaqaaiaabsgaqaaa % aaaaaaWdbiaadAhacqGH9aqpcaaIYaWaaWbaaSqabeaacaWG4baaaO % GaaeizaiaadIhaaaWdaiaawUhaaaaa!429A! \left\{ \begin{array}{l} u = x + 1\\ {\rm{d}}v = {2^x}{\rm{d}}x \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aai % qaaqaabeqaaabaaaaaaaaapeGaaeizaiaadwhacqGH9aqpcaqGKbGa % amiEaaqaaiaadAhacqGH9aqpdaWcaaqaaiaaikdadaahaaWcbeqaai % aadIhaaaaakeaaciGGSbGaaiOBaiaaikdaaaaaa8aacaGL7baaaaa!450D! \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = \frac{{{2^x}}}{{\ln 2}} \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 % da9maabmaabaGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPaaadaWc % aaqaaiaaikdadaahaaWcbeqaaiaadIhaaaaakeaaciGGSbGaaiOBai % aaikdaaaWaaqqaaqaabeqaaiaaigdaaeaacaaIWaaaaiaawEa7aiab % gkHiTmaapehabaWaaSaaaeaacaaIYaWaaWbaaSqabeaacaWG4baaaa % GcbaGaciiBaiaac6gacaaIYaaaaiaabsgacaWG4baaleaacaaIWaaa % baGaaGymaaqdcqGHRiI8aaaa!4EED! M = \left( {x + 1} \right)\frac{{{2^x}}}{{\ln 2}}\left| \begin{array}{l} 1\\ 0 \end{array} \right. - \int\limits_0^1 {\frac{{{2^x}}}{{\ln 2}}{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaI0aaabaGaciiBaiaac6gacaaIYaaaaiabgkHiTmaalaaa % baGaaGymaaqaaiGacYgacaGGUbGaaGOmaaaacqGHsisldaWcaaqaai % aaikdadaahaaWcbeqaaiaadIhaaaaakeaadaqadaqaaiGacYgacaGG % UbGaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaGcda % abbaabaeqabaGaaGymaaqaaiaaicdaaaGaay5bSdaaaa!49D8! = \frac{4}{{\ln 2}} - \frac{1}{{\ln 2}} - \frac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}}\left| \begin{array}{l} 1\\ 0 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIZaaabaGaciiBaiaac6gacaaIYaaaaiabgkHiTmaalaaa % baGaaGymaaqaamaabmaabaGaciiBaiaac6gacaaIYaaacaGLOaGaay % zkaaWaaWbaaSqabeaacaaIYaaaaaaaaaa!4130! = \frac{3}{{\ln 2}} - \frac{1}{{{{\left( {\ln 2} \right)}^2}}}\)