62. Cho \(\int\limits_0^1 {x{e^{2022x}}{\rm{d}}x}  = a{e^{2022}} + b\),\(a,b \in \mathbb{Q}\). Tính \(\frac{a}{b}\).

Câu hỏi:

62. Cho \(\int\limits_0^1 {x{e^{2022x}}{\rm{d}}x}  = a{e^{2022}} + b\),\(a,b \in \mathbb{Q}\). Tính \(\frac{a}{b}\).

A. \(\frac{1}{{2021}}\).

B. \(\frac{1}{{2022}}\).

C. \(2022\).

D. \(2021\).

Lời giải

Đặt \(\left\{ \begin{array}{l}u = x\\{\rm{d}}v = {e^{2022x}}{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = {\rm{d}}x\\v = \frac{1}{{2022}}{e^{2022x}}\end{array} \right.\). 

Suy ra \(\int\limits_0^1 {x{e^{2022x}}{\rm{d}}x}  = \)\(\left. {\frac{{x{e^{2022x}}}}{{2022}}} \right|_0^1 – \frac{1}{{2022}}\int\limits_0^1 {{e^{2022x}}{\rm{d}}x} \)\( = \frac{{{e^{2022}}}}{{2022}} – \left. {\frac{1}{{{{2022}^2}}}{e^{2022x}}} \right|_0^1\)\( = \frac{{{e^{2022}}}}{{2022}} – \frac{1}{{{{2022}^2}}}\left( {{e^{2022}} – 1} \right)\)\( = \frac{{2021}}{{{{2022}^2}}}{e^{2022}} + \frac{1}{{{{2022}^2}}}\). 

Vậy \(a = \frac{{2021}}{{{{2022}^2}}},{\rm{ }}b = \frac{1}{{{{2022}^2}}} \Rightarrow \frac{a}{b} = 2021\).

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Thuộc chủ đề: Trắc nghiệm Tích phân