Câu 22 Mã đề 101 Thi THPTQG 2018
\(\int\limits^1_0e^{3x-1}\text{d}x\) bằng
\(\frac{1}{3}\left(e^5-e^2\right)\).\(\frac{1}{3}e^5-e^2\).\(e^5-e^2\).\(\frac{1}{3}\left(e^5+e^2\right)\).Hướng dẫn giải:\(\int e^{3x-1}\text{d}x=\frac{1}{3}\int e^{3x-1}\text{d}\left(3x-1\right)=\frac{1}{3}e^{3x-1}+C\). Do đó \(\int\limits^2_1e^{3x-1}\text{d}x=\frac{1}{3}\left(e^5-e^2\right)\).