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Câu hỏi: Nếu $\int\limits_{-2}^{0}{\left( 4-{{e}^{-x/2}} \right)}dx=a+2be$ thì giá trị của $a+2b$ là
A. $12$.
B. $9$.
C. $12,5$.
D. $8$.
Ta có: $I=\int\limits_{-2}^{0}{4dx-\int\limits_{-2}^{0}{{{e}^{-x/2}}dx}}$ $=\left. 4x \right|_{-2}^{0}+\left. 2{{e}^{-x/2}} \right|_{-2}^{0}$ $=8+2\left( 1-e \right)$ $=10-2e$.
$\Rightarrow a=10 ; b=-1$ $\Rightarrow a+2b=8$.
A. $12$.
B. $9$.
C. $12,5$.
D. $8$.
Ta có: $I=\int\limits_{-2}^{0}{4dx-\int\limits_{-2}^{0}{{{e}^{-x/2}}dx}}$ $=\left. 4x \right|_{-2}^{0}+\left. 2{{e}^{-x/2}} \right|_{-2}^{0}$ $=8+2\left( 1-e \right)$ $=10-2e$.
$\Rightarrow a=10 ; b=-1$ $\Rightarrow a+2b=8$.
Đáp án D.