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Câu hỏi: $\int\limits_{0}^{1}{{{e}^{3x+1}}dx}$ bằng
A. ${{e}^{3}}-e.$
B. $\dfrac{1}{3}\left( {{e}^{4}}+e \right).$
C. ${{e}^{4}}-e.$
D. $\dfrac{1}{3}\left( {{e}^{4}}-e \right).$
A. ${{e}^{3}}-e.$
B. $\dfrac{1}{3}\left( {{e}^{4}}+e \right).$
C. ${{e}^{4}}-e.$
D. $\dfrac{1}{3}\left( {{e}^{4}}-e \right).$
Ta có $\int\limits_{0}^{1}{{{e}^{3x+1}}dx}=\dfrac{1}{3}\int\limits_{0}^{1}{{{e}^{3x+1}}d\left( 3x+1 \right)}=\dfrac{1}{3}{{e}^{3x+1}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{3}\left( {{e}^{4}}-e \right).$
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{3}\left( {{e}^{4}}-e \right).$
Đáp án D.