Giá trị của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % qadaqaaiaadIhacqGHRaWkcaaIXaaacaGLOaGaayzkaaGaaeyzamaa % CaaaleqabaGaamiEaaaakiaabsgacaWG4baaleaacaaIWaaabaGaaG % ymaaqdcqGHRiI8aaaa!41F4! \int\limits_0^1 {\left( {x + 1} \right){{\rm{e}}^x}{\rm{d}}x} \)
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4bGaey4kaSIaaGymaaqaaiaabsgacaWG % 2bGaeyypa0JaaeyzamaaCaaaleqabaGaamiEaaaakiaabsgacaWG4b % aaaiaawUhaaiabgkDiEpaaceaaeaqabeaacaqGKbGaamyDaiabg2da % 9iaabsgacaWG4baabaGaamODaiabg2da9iaabwgadaahaaWcbeqaai % aadIhaaaaaaOGaay5Eaaaaaa!4EFC! \left\{ \begin{array}{l} u = x + 1\\ {\rm{d}}v = {{\rm{e}}^x}{\rm{d}}x \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = {\rm{d}}x\\ v = {{\rm{e}}^x} \end{array} \right.\)
Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % qadaqaaiaadIhacqGHRaWkcaaIXaaacaGLOaGaayzkaaGaaeyzamaa % CaaaleqabaGaamiEaaaakiaabsgacaWG4baaleaacaaIWaaabaGaaG % ymaaqdcqGHRiI8aOGaeyypa0ZaaqGaaeaadaqadaqaaiaadIhacqGH % RaWkcaaIXaaacaGLOaGaayzkaaGaaeyzamaaCaaaleqabaGaamiEaa % aaaOGaayjcSdWaa0baaSqaaiaaicdaaeaacaaIXaaaaOGaeyOeI0Ya % a8qCaeaacaqGLbWaaWbaaSqabeaacaWG4baaaOGaaeizaiaadIhaaS % qaaiaaicdaaeaacaaIXaaaniabgUIiYdGccqGH9aqpdaqadaqaaiaa % ikdacaqGLbGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabgkHiTmaaei % aabaGaaeyzamaaCaaaleqabaGaamiEaaaaaOGaayjcSdWaa0baaSqa % aiaaicdaaeaacaaIXaaaaOGaeyypa0JaaGOmaiaabwgacqGHsislca % aIXaGaeyOeI0IaaeyzaiabgUcaRiaaigdacqGH9aqpcaqGLbaaaa!6B32! \int\limits_0^1 {\left( {x + 1} \right){{\rm{e}}^x}{\rm{d}}x} = \left. {\left( {x + 1} \right){{\rm{e}}^x}} \right|_0^1 - \int\limits_0^1 {{{\rm{e}}^x}{\rm{d}}x} = \left( {2{\rm{e}} - 1} \right) - \left. {{{\rm{e}}^x}} \right|_0^1 = 2{\rm{e}} - 1 - {\rm{e}} + 1 = {\rm{e}}\)