Biết rằng $\int\limits_{0}^{1}{x{{\text{e}}^{2x}}dx}=a{{\text{e}}^{2}}+b$ (với$a,b\in \mathbb{Q}$). Tính$P=a+b$. A. $P=0$. B. $P=\frac{1}{2}$. C. $P=\frac{1}{4}$. D. $P=1$.
Đáp án đúng: B Chọn B Xét tích phân $\int\limits_{0}^{1}{x{{\text{e}}^{2x}}dx}$. Đặt$\left\{ \begin{array}{l}x=u\\{{\text{e}}^{2x}}dx=\text{d}v\end{array} \right.\Rightarrow \left\{ \begin{array}{l}\text{d}x=\text{d}u\\\frac{1}{2}{{\text{e}}^{2x}}=v\end{array} \right.$. Khi đó$\int\limits_{0}^{1}{x{{\text{e}}^{2x}}dx}$$=\left. \frac{1}{2}x{{\text{e}}^{2x}} \right|_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{{{\text{e}}^{2x}}dx}$$=\frac{1}{2}{{\text{e}}^{2}}-\left. \frac{1}{4}{{\text{e}}^{2x}} \right|_{0}^{1}$$=\frac{1}{2}{{\text{e}}^{2}}-\frac{1}{4}{{\text{e}}^{2}}+\frac{1}{4}$$=\frac{1}{4}{{\text{e}}^{2}}+\frac{1}{4}$. $\Rightarrow $$a=\frac{1}{4}$,$b=\frac{1}{4}$. Vậy$P=\frac{1}{2}$.
