Câu hỏi :

\(\int\limits_0^1 {{e^{3x + 1}}dx} \) bằng

A. \({e^3} - e.\)

B. \(\frac{1}{3}\left( {{e^4} + e} \right).\)

C. \({e^4} - e.\)

D. \(\frac{1}{3}\left( {{e^4} - e} \right).\)

* Đáp án

D

* Hướng dẫn giải

\(\int\limits_0^1 {{e^{3x + 1}}dx} = \frac{1}{3}\int\limits_0^1 {{e^{3x + 1}}d\left( {3x + 1} \right)} = \frac{1}{3}{e^{3x + 1}}\left| \begin{array}{l} 1\\ 0 \end{array} \right. = \frac{1}{3}\left( {{e^4} - e} \right).\)

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