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