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Câu hỏi: Biết $\int{x{{e}^{2x}}}\text{d}x=ax{{e}^{2x}}+b{{e}^{2x}}+C\left( a,b\in \mathbb{Q},C\in \mathbb{R} \right)$. Tính tích $a.b.$
A. $ab=-\dfrac{1}{8}\cdot $
B. $ab=-\dfrac{1}{4}\cdot $
C. $ab=\dfrac{1}{8}\cdot $
D. $ab=\dfrac{1}{4}\cdot $
A. $ab=-\dfrac{1}{8}\cdot $
B. $ab=-\dfrac{1}{4}\cdot $
C. $ab=\dfrac{1}{8}\cdot $
D. $ab=\dfrac{1}{4}\cdot $
Đặt $u=x\Rightarrow \text{d}u=\text{d}x$
$\text{d}v={{e}^{2x}}\text{d}x\Rightarrow v=\dfrac{1}{2}{{e}^{2x}}$
Khi đó $\int{x{{e}^{2x}}}\text{d}x=\dfrac{1}{2}x{{e}^{2x}}-\int{\dfrac{1}{2}{{e}^{2x}}\text{d}x}=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C.$
Vậy $a=\dfrac{1}{2},b=-\dfrac{1}{4}\Rightarrow a.b=-\dfrac{1}{8}.$
$\text{d}v={{e}^{2x}}\text{d}x\Rightarrow v=\dfrac{1}{2}{{e}^{2x}}$
Khi đó $\int{x{{e}^{2x}}}\text{d}x=\dfrac{1}{2}x{{e}^{2x}}-\int{\dfrac{1}{2}{{e}^{2x}}\text{d}x}=\dfrac{1}{2}x{{e}^{2x}}-\dfrac{1}{4}{{e}^{2x}}+C.$
Vậy $a=\dfrac{1}{2},b=-\dfrac{1}{4}\Rightarrow a.b=-\dfrac{1}{8}.$
Đáp án A.