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Câu hỏi: Khẳng định nào sau đây đúng về kết quả $\int\limits_{1}^{e}{{{x}^{3}}\ln xdx}=\dfrac{3{{e}^{a}}+1}{b}$ ?
A. $a-b=12$.
B. $a-b=4$.
C. $a.b=64$.
D. $ab=46$.
A. $a-b=12$.
B. $a-b=4$.
C. $a.b=64$.
D. $ab=46$.
Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{4}}}{4} \\
\end{aligned} \right.$.
Khi đó $\int\limits_{1}^{e}{{{x}^{3}}\ln xdx}=\left. \dfrac{{{x}^{4}}}{4}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{{{x}^{4}}}{4}\dfrac{1}{x}dx}=\dfrac{{{e}^{4}}}{4}-\dfrac{1}{4}\int\limits_{1}^{e}{{{x}^{3}}dx}=\dfrac{{{e}^{4}}}{4}-\left. \dfrac{{{x}^{4}}}{16} \right|_{1}^{e}=\dfrac{3{{e}^{4}}+1}{16}\Rightarrow a=4,b=16$.
Ta có: $a-b=-12,a.b=64$.
Vậy $a.b=64$.
& u=\ln x \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{4}}}{4} \\
\end{aligned} \right.$.
Khi đó $\int\limits_{1}^{e}{{{x}^{3}}\ln xdx}=\left. \dfrac{{{x}^{4}}}{4}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{{{x}^{4}}}{4}\dfrac{1}{x}dx}=\dfrac{{{e}^{4}}}{4}-\dfrac{1}{4}\int\limits_{1}^{e}{{{x}^{3}}dx}=\dfrac{{{e}^{4}}}{4}-\left. \dfrac{{{x}^{4}}}{16} \right|_{1}^{e}=\dfrac{3{{e}^{4}}+1}{16}\Rightarrow a=4,b=16$.
Ta có: $a-b=-12,a.b=64$.
Vậy $a.b=64$.
Đáp án C.