Cho hàm số f(x) có đạo hàm liên tục trên R thỏa mãn f(0) =3 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabgUcaRiaadAgadaqadaqaaiaa % ikdacqGHsislcaWG4baacaGLOaGaayzkaaGaeyypa0JaamiEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG4bGaey4kaSIaaGOm % aiaaykW7caaMc8UaaGPaVlabgcGiIiaadIhacqGHiiIZcqWIDesOaa % a!4FFD! f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\,\,\,\forall x \in \). Tích phân \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaaa!40D2! \int\limits_0^2 {xf'\left( x \right)dx} \) bằng:

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaI0aaabaGaaG4maaaaaaa!386C! - \frac{4}{3}\)

B. \(\frac{2}{3}\)

C. \(\frac{5}{3}\)

D. \(\frac{-10}{3}\)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9maapehabaGaamiEaiaadsgadaqadaqaaiaadA % gadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH % 9aqpcaWG4bGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaamaaee % aaeaqabeaadaahaaWcbeqaaiaaikdaaaaakeaadaWgaaWcbaGaaGim % aaqabaaaaOGaay5bSdGaeyOeI0Yaa8qCaeaacaWGMbWaaeWaaeaaca % WG4baacaGLOaGaayzkaaGaamizaiaadIhacqGH9aqpcaaIYaGaamOz % amaabmaabaGaaGOmaaGaayjkaiaawMcaaiabgkHiTmaapehabaGaam % OzamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4baaleaa % caaIWaaabaGaaGOmaaqdcqGHRiI8aaWcbaGaaGimaaqaaiaaikdaa0 % Gaey4kIipaaSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaaleaacaaI % WaaabaGaaGOmaaqdcqGHRiI8aaaa!6EC1! \int\limits_0^2 {xf'\left( x \right)dx = \int\limits_0^2 {xd\left( {f\left( x \right)} \right) = xf\left( x \right)\left| \begin{array}{l} ^2\\ _0 \end{array} \right. - \int\limits_0^2 {f\left( x \right)dx = 2f\left( 2 \right) - \int\limits_0^2 {f\left( x \right)dx} } } } \)

Theo bài ra ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaey4kaSIaamOzamaabmaa % baGaaGOmaiabgkHiTiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWG4b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadIhacqGHRaWk % caaIYaGaeyiaIiIaamiEaiabgIGiolabl2riHkabgkDiElaadAgada % qadaqaaiaaicdaaiaawIcacaGLPaaacqGHRaWkcaWGMbWaaeWaaeaa % caaIYaaacaGLOaGaayzkaaGaeyypa0JaaGOmaiabgkDiElaadAgada % qadaqaaiaaikdaaiaawIcacaGLPaaacqGH9aqpcaaIYaGaeyOeI0Ia % amOzamaabmaabaGaaGimaaGaayjkaiaawMcaaiabg2da9iabgkHiTi % aaigdaaeaacqGHshI3daWdXbqaaiaadIhacaWGMbGaai4jamaabmaa % baGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0JaeyOeI0 % IaaGOmaiabgkHiTmaapehabaGaamOzamaabmaabaGaamiEaaGaayjk % aiaawMcaaiaadsgacaWG4bGaeyypa0JaeyOeI0IaaGOmaiabgkHiTm % aapehabaGaamOzamaabmaabaGaamiDaaGaayjkaiaawMcaaiaadsga % caWG0baaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aaWcbaGaaGimaa % qaaiaaikdaa0Gaey4kIipaaSqaaiaaicdaaeaacaaIYaaaniabgUIi % Ydaaaaa!8BB1! \begin{array}{l} f\left( x \right) + f\left( {2 - x} \right) = {x^2} - 2x + 2\forall x \in R \Rightarrow f\left( 0 \right) + f\left( 2 \right) = 2 \Rightarrow f\left( 2 \right) = 2 - f\left( 0 \right) = - 1\\ \Rightarrow \int\limits_0^2 {xf'\left( x \right)dx = - 2 - \int\limits_0^2 {f\left( x \right)dx = - 2 - \int\limits_0^2 {f\left( t \right)dt} } } \end{array}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaikdacqGHsislcaWG4bGaeyO0H4TaamizaiaadshacqGH9aqp % cqGHsislcaWGKbGaamiEaaaa!42B1! t = 2 - x \Rightarrow dt = - dx\). Đổi cận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIWaGaeyO0H4TaamiDaiabg2da9iaaikda % aeaacaWG4bGaeyypa0JaaGOmaiabgkDiElaadshacqGH9aqpcaaIWa % aaaiaawUhaaaaa!46BF! \left\{ \begin{array}{l} x = 0 \Rightarrow t = 2\\ x = 2 \Rightarrow t = 0 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHsh % I3daWdXbqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9maapehabaGaamOzamaabmaabaGaaGOmaiabgk % HiTiaadIhaaiaawIcacaGLPaaaaSqaaiaaicdaaeaacaaIYaaaniab % gUIiYdaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aOGaamizaiaadI % haaeaacqGHshI3caaIYaWaa8qCaeaacaWGMbWaaeWaaeaacaWG4baa % caGLOaGaayzkaaGaamizaiaadIhacqGH9aqpdaWdXbqaaiaadAgada % qadaqaaiaadIhaaiaawIcacaGLPaaacaWGKbGaamiEaiabgUcaRmaa % pehabaGaamOzamaabmaabaGaaGOmaiabgkHiTiaadIhaaiaawIcaca % GLPaaacaWGKbGaamiEaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipa % aSqaaiaaicdaaeaacaaIYaaaniabgUIiYdaaleaacaaIWaaabaGaaG % OmaaqdcqGHRiI8aaGcbaGaeyO0H4TaaGOmamaapehabaGaamOzamaa % bmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0Zaa8 % qCaeaadaWadaqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaa % cqGHRaWkcaWGMbWaaeWaaeaacaaIYaGaeyOeI0IaamiEaaGaayjkai % aawMcaaaGaay5waiaaw2faaiaadsgacaWG4baaleaacaaIWaaabaGa % aGOmaaqdcqGHRiI8aaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaO % qaaiabgkDiElaaikdadaWdXbqaaiaadAgadaqadaqaaiaadIhaaiaa % wIcacaGLPaaacaWGKbGaamiEaiabg2da9maapehabaWaaeWaaeaaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaadIhacqGH % RaWkcaaIYaaacaGLOaGaayzkaaGaamizaiaadIhaaSqaaiaaicdaae % aacaaIYaaaniabgUIiYdaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8 % aaGcbaGaeyO0H4TaaGOmamaapehabaGaamOzamaabmaabaGaamiEaa % GaayjkaiaawMcaaiaadsgacaWG4bGaeyypa0ZaaeWaaeaadaWcaaqa % aiaadIhadaahaaWcbeqaaiaaiodaaaaakeaacaaIZaaaaiabgkHiTi % aadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamiEaaGa % ayjkaiaawMcaamaaeeaaeaqabeaadaahaaWcbeqaaiaaikdaaaaake % aadaWgaaWcbaGaaGimaaqabaaaaOGaay5bSdaaleaacaaIWaaabaGa % aGOmaaqdcqGHRiI8aOGaeyypa0ZaaSaaaeaacaaI4aaabaGaaG4maa % aaaeaacqGHshI3daWdXbqaaiaadAgadaqadaqaaiaadIhaaiaawIca % caGLPaaacaWGKbGaamiEaiabg2da9maalaaabaGaaGinaaqaaiaaio % daaaaaleaacaaIWaaabaGaaGOmaaqdcqGHRiI8aaaaaa!D48F! \begin{array}{l} \Rightarrow \int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {f\left( {2 - x} \right)} } dx\\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {f\left( x \right)dx + \int\limits_0^2 {f\left( {2 - x} \right)dx} } } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {\left[ {f\left( x \right) + f\left( {2 - x} \right)} \right]dx} } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \int\limits_0^2 {\left( {{x^2} - 2x + 2} \right)dx} } \\ \Rightarrow 2\int\limits_0^2 {f\left( x \right)dx = \left( {\frac{{{x^3}}}{3} - {x^2} + 2x} \right)\left| \begin{array}{l} ^2\\ _0 \end{array} \right.} = \frac{8}{3}\\ \Rightarrow \int\limits_0^2 {f\left( x \right)dx = \frac{4}{3}} \end{array}\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaamOzaiaacEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWG % KbGaamiEaiabg2da9iabgkHiTiaaikdacqGHsisldaWcaaqaaiaais % daaeaacaaIZaaaaiabg2da9iabgkHiTmaalaaabaGaaGymaiaaicda % aeaacaaIZaaaaaWcbaGaaGimaaqaaiaaikdaa0Gaey4kIipaaaa!4A2E! \int\limits_0^2 {xf'\left( x \right)dx = - 2 - \frac{4}{3} = - \frac{{10}}{3}} \)

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