Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaGaamiEaiaadwgadaahaaWcbeqaaiaaikdacaWG4baa % aOGaamizaiaadIhaaSqaaiaaicdaaeaacaaIYaGaaGimaiaaigdaca % aI3aaaniabgUIiYdaaaa!4398! I = \int\limits_0^{2017} {x{e^{2x}}dx} \)
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4baabaGaamyzamaaCaaaleqabaGaaGOm % aiaadIhaaaGccaWGKbGaamiEaiabg2da9iaadsgacaWG2baaaiaawU % haaiabgkDiEpaaceaaeaqabeaacaWGKbGaamyDaiabg2da9iaadsga % caWG4baabaGaamODaiabg2da9maalaaabaGaamyzamaaCaaaleqaba % GaaGOmaiaadIhaaaaakeaacaaIYaaaaaaacaGL7baaaaa!4FB0! \left\{ \begin{array}{l} u = x\\ {e^{2x}}dx = dv \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = dx\\ v = \frac{{{e^{2x}}}}{2} \end{array} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaGaamiEaiaadwgadaahaaWcbeqaaiaaikdacaWG4baa % aOGaamizaiaadIhaaSqaaiaaicdaaeaacaaIYaGaaGimaiaaigdaca % aI3aaaniabgUIiYdGccqGH9aqpdaabcaqaaiaadIhacaGGUaWaaSaa % aeaacaWGLbWaaWbaaSqabeaacaaIYaGaamiEaaaaaOqaaiaaikdaaa % aacaGLiWoadaqhaaWcbaGaaGimaaqaaiaaikdacaaIWaGaaGymaiaa % iEdaaaGccqGHsisldaWdXbqaamaalaaabaGaamyzamaaCaaaleqaba % GaaGOmaiaadIhaaaaakeaacaaIYaaaaiaadsgacaWG4baaleaacaaI % WaaabaGaaGOmaiaaicdacaaIXaGaaG4naaqdcqGHRiI8aOGaeyypa0 % ZaaqGaaeaacaWG4bGaaiOlamaalaaabaGaamyzamaaCaaaleqabaGa % aGOmaiaadIhaaaaakeaacaaIYaaaaaGaayjcSdWaa0baaSqaaiaaic % daaeaacaaIYaGaaGimaiaaigdacaaI3aaaaOGaeyOeI0YaaqGaaeaa % daWcaaqaaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaaGcbaGaaG % inaaaaaiaawIa7amaaDaaaleaacaaIWaaabaGaaGOmaiaaicdacaaI % XaGaaG4naaaakiabg2da9iaaikdacaaIWaGaaGymaiaaiEdacaGGUa % WaaSaaaeaacaWGLbWaaWbaaSqabeaacaaI0aGaaGimaiaaiodacaaI % 0aaaaaGcbaGaaGOmaaaacqGHsisldaWcaaqaaiaadwgadaahaaWcbe % qaaiaaisdacaaIWaGaaG4maiaaisdaaaaakeaacaaI0aaaaiabgUca % RmaalaaabaGaaGymaaqaaiaaisdaaaGaeyypa0ZaaSaaaeaacaaI0a % GaaGimaiaaiodacaaIZaGaamyzamaaCaaaleqabaGaaGinaiaaicda % caaIZaGaaGinaaaakiabgUcaRiaaigdaaeaacaaI0aaaaaaa!8E2A! I = \int\limits_0^{2017} {x{e^{2x}}dx} = \left. {x.\frac{{{e^{2x}}}}{2}} \right|_0^{2017} - \int\limits_0^{2017} {\frac{{{e^{2x}}}}{2}dx} = \left. {x.\frac{{{e^{2x}}}}{2}} \right|_0^{2017} - \left. {\frac{{{e^{2x}}}}{4}} \right|_0^{2017} = 2017.\frac{{{e^{4034}}}}{2} - \frac{{{e^{4034}}}}{4} + \frac{1}{4} = \frac{{4033{e^{4034}} + 1}}{4}\)