Câu hỏi:
13. \(\int\limits_1^2 {x{{\left( {x – 1} \right)}^{2021}}{\rm{d}}x} \) bằng
A. \(\frac{1}{{2021}} + \frac{1}{{2022}}\).
B. \(\frac{1}{{2021}} – \frac{1}{{2022}}\).
C. \(\frac{1}{{2022}} + \frac{1}{{2023}}\).
D. \(\frac{1}{{2022}} – \frac{1}{{2023}}\).
Lời giải
Đặt \(t = x – 1\) \( \Rightarrow {\rm{d}}t = {\rm{d}}x\)
Đổi cận: \(x = 1 \Rightarrow t = 0;\,x = 2 \Rightarrow t = 1\).
\(\int\limits_1^2 {x{{\left( {x – 1} \right)}^{2021}}{\rm{d}}x} = \int\limits_0^1 {\left( {t + 1} \right){t^{2021}}{\rm{d}}t} = \int\limits_0^1 {\left( {{t^{2022}} + {t^{2021}}} \right){\rm{d}}t = \left( {\frac{{{t^{2023}}}}{{2023}} + \frac{{{t^{2022}}}}{{2022}}} \right)\left| {\begin{array}{*{20}{c}}{^1}\\{_0}\end{array}} \right.} = \frac{1}{{2023}} + \frac{1}{{2022}}\).
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Thuộc chủ đề: Trắc nghiệm Tích phân
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