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Câu hỏi: Giá trị tích phân $\int\limits_{0}^{100}{x.{{e}^{2x}}dx}$ bằng
A. $\dfrac{1}{4}\left( 199{{e}^{200}}-1 \right)$.
B. $\dfrac{1}{2}\left( 199{{e}^{200}}-1 \right)$.
C. $\dfrac{1}{4}\left( 199{{e}^{200}}+1 \right)$.
D. $\dfrac{1}{2}\left( 199{{e}^{200}}+1 \right)$.
A. $\dfrac{1}{4}\left( 199{{e}^{200}}-1 \right)$.
B. $\dfrac{1}{2}\left( 199{{e}^{200}}-1 \right)$.
C. $\dfrac{1}{4}\left( 199{{e}^{200}}+1 \right)$.
D. $\dfrac{1}{2}\left( 199{{e}^{200}}+1 \right)$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{100}{x.{{e}^{2x}}dx}=\dfrac{1}{2}x{{e}^{2x}}\left| \begin{aligned}
& 100 \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{100}{{{e}^{2x}}dx}=50{{e}^{200}}-\dfrac{1}{4}{{e}^{2x}}\left| \begin{aligned}
& 100 \\
& 0 \\
\end{aligned} \right.=50{{e}^{200}}-\dfrac{1}{4}{{e}^{200}}+\dfrac{1}{4}=\dfrac{1}{4}\left( 199{{e}^{200}}+1 \right)$
& u=x \\
& dv={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{100}{x.{{e}^{2x}}dx}=\dfrac{1}{2}x{{e}^{2x}}\left| \begin{aligned}
& 100 \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{100}{{{e}^{2x}}dx}=50{{e}^{200}}-\dfrac{1}{4}{{e}^{2x}}\left| \begin{aligned}
& 100 \\
& 0 \\
\end{aligned} \right.=50{{e}^{200}}-\dfrac{1}{4}{{e}^{200}}+\dfrac{1}{4}=\dfrac{1}{4}\left( 199{{e}^{200}}+1 \right)$
Đáp án C.