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Câu hỏi: Cho $\int\limits_{0}^{1}{(x-1){{e}^{2x}}}dx=a+b{{e}^{2}}$, với $a;b\in \mathbb{Q}, a,b$ là các phân số tối giản. Tổng $a+b$ bằng
A. $-3$.
B. $\dfrac{1}{2}$.
C. $1$.
D. $5$.
A. $-3$.
B. $\dfrac{1}{2}$.
C. $1$.
D. $5$.
Đặt $u=x-1\Rightarrow \text{d}u=\text{d}x$ ; $\text{d}v={{e}^{2x}}\text{d}x,$ chọn $v=\dfrac{1}{2}{{e}^{2x}}$.
$\int\limits_{0}^{1}{(x-1){{e}^{2x}}}dx=\left. \dfrac{1}{2}(x-1){{e}^{2x}} \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{1}{2}{{e}^{2x}}\text{d}x}$ $=\dfrac{1}{2}-\left. \dfrac{1}{4}{{e}^{2x}} \right|_{0}^{1}=\dfrac{1}{2}-\left( \dfrac{1}{4}{{e}^{2}}-\dfrac{1}{4} \right)=\dfrac{3}{4}-\dfrac{1}{4}{{e}^{2}}$.
$\Rightarrow a=\dfrac{3}{4};b=-\dfrac{1}{4}$.
Vậy $a+b=\dfrac{1}{2}$
$\int\limits_{0}^{1}{(x-1){{e}^{2x}}}dx=\left. \dfrac{1}{2}(x-1){{e}^{2x}} \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{1}{2}{{e}^{2x}}\text{d}x}$ $=\dfrac{1}{2}-\left. \dfrac{1}{4}{{e}^{2x}} \right|_{0}^{1}=\dfrac{1}{2}-\left( \dfrac{1}{4}{{e}^{2}}-\dfrac{1}{4} \right)=\dfrac{3}{4}-\dfrac{1}{4}{{e}^{2}}$.
$\Rightarrow a=\dfrac{3}{4};b=-\dfrac{1}{4}$.
Vậy $a+b=\dfrac{1}{2}$
Đáp án B.