Tích phân $I=\int\limits_{0}^{1}{{{x}^{2}}{{e}^{-2x}}dx}$ bằng A. $\frac{1}{4}(1+\frac{5}{{{e}^{2}}}).$ B. $\frac{1}{4}({{e}^{2}}-1).$ C. $\frac{3}{4}(\frac{-5}{{{e}^{2}}}+1).$ D. $\frac{1}{4}(1-\frac{5}{{{e}^{2}}}).$
Đáp án đúng: D Đặt $\left\{ \begin{array}{l}u={{x}^{2}}\\dv={{e}^{-2x}}dx\end{array} \right.=>\left\{ \begin{array}{l}du=2xdx\\v=-\frac{{{e}^{-2x}}}{2}\end{array} \right..$