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Câu hỏi: $\int\limits_{1}^{2}{{{e}^{3x-1}}\text{d}x}$ bằng:
A. $\dfrac{1}{3}\left( {{e}^{5}}-{{e}^{2}} \right)$.
B. $\dfrac{1}{3}{{e}^{5}}-{{e}^{2}}$.
C. ${{e}^{5}}-{{e}^{2}}$.
D. $\dfrac{1}{3}\left( {{e}^{5}}+{{e}^{2}} \right)$.
A. $\dfrac{1}{3}\left( {{e}^{5}}-{{e}^{2}} \right)$.
B. $\dfrac{1}{3}{{e}^{5}}-{{e}^{2}}$.
C. ${{e}^{5}}-{{e}^{2}}$.
D. $\dfrac{1}{3}\left( {{e}^{5}}+{{e}^{2}} \right)$.
Ta có: $\int\limits_{1}^{2}{{{e}^{3x-1}}\text{d}x}$ $=\dfrac{1}{3}\left. {{e}^{3x-1}} \right|_{1}^{2}$ $=\dfrac{1}{3}\left( {{e}^{5}}-{{e}^{2}} \right)$.
Đáp án A.