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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và có $f\left( -2 \right)=2;f\left( 0 \right)=1.$ Tính $I=\int\limits_{-2}^{0}{\dfrac{{f}'\left( x \right)-f\left( x \right)}{{{e}^{x}}}dx}.$
A. $I=1-2{{e}^{2}}$.
B. $I=1-2{{e}^{-2}}$.
C. $I=1+2{{e}^{2}}$.
D. $I=1+2{{e}^{-2}}$.
A. $I=1-2{{e}^{2}}$.
B. $I=1-2{{e}^{-2}}$.
C. $I=1+2{{e}^{2}}$.
D. $I=1+2{{e}^{-2}}$.
Ta có:
$I=\int\limits_{-2}^{0}{\dfrac{{f}'\left( x \right)-f\left( x \right)}{{{e}^{x}}}dx}=\int\limits_{-2}^{0}{\dfrac{{f}'\left( x \right).{{e}^{x}}-f\left( x \right){{\left( {{e}^{x}} \right)}^{\prime }}}{{{e}^{2x}}}dx}=\int\limits_{-2}^{0}{{{\left( \dfrac{f\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}dx}=\left. \dfrac{f\left( x \right)}{{{e}^{x}}} \right|_{-2}^{0}=1-2{{e}^{2}}$.
$I=\int\limits_{-2}^{0}{\dfrac{{f}'\left( x \right)-f\left( x \right)}{{{e}^{x}}}dx}=\int\limits_{-2}^{0}{\dfrac{{f}'\left( x \right).{{e}^{x}}-f\left( x \right){{\left( {{e}^{x}} \right)}^{\prime }}}{{{e}^{2x}}}dx}=\int\limits_{-2}^{0}{{{\left( \dfrac{f\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}dx}=\left. \dfrac{f\left( x \right)}{{{e}^{x}}} \right|_{-2}^{0}=1-2{{e}^{2}}$.
Đáp án A.