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Câu hỏi: Biết rằng $\int\limits_{1}^{2}{x{{\left( x-1 \right)}^{n}}dx}=\dfrac{27}{182},$ với $n\in {{\mathbb{N}}^{*}}.$ Mệnh đề nào dưới đây là đúng?
A. $n\le 4.$
B. $n>11.$
C. $4<n\le 8.$
D. $8<n\le 11.$
A. $n\le 4.$
B. $n>11.$
C. $4<n\le 8.$
D. $8<n\le 11.$
Xét $I=\int\limits_{1}^{2}{x{{\left( x-1 \right)}^{n}}dx}$. Đặt $x=1=t\Rightarrow I=\int\limits_{0}^{1}{\left( t+1 \right){{t}^{n}}d\left( t+1 \right)}=\int\limits_{0}^{1}{\left( {{t}^{n+1}}+{{t}^{n}} \right)dt}$
$\Rightarrow I=\left. \left( \dfrac{{{t}^{n+2}}}{n+2}+\dfrac{{{t}^{n+1}}}{n+1} \right) \right|_{0}^{1}=\dfrac{1}{n+2}+\dfrac{1}{n+1}=\dfrac{27}{181}\Rightarrow n=12$.
$\Rightarrow I=\left. \left( \dfrac{{{t}^{n+2}}}{n+2}+\dfrac{{{t}^{n+1}}}{n+1} \right) \right|_{0}^{1}=\dfrac{1}{n+2}+\dfrac{1}{n+1}=\dfrac{27}{181}\Rightarrow n=12$.
Đáp án B.