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Câu hỏi: Nếu $\int\limits_{-1}^{4}{f\left( x \right)dx}=2$ và $\int\limits_{-1}^{0}{f\left( x \right)dx}=3$ thì $\int\limits_{0}^{4}{\left[ 4{{e}^{2x}}+2f\left( x \right) \right]dx}$ bằng
A. $2{{e}^{8}}-4.$
B. $2{{e}^{8}}-2.$
C. $2{{e}^{8}}+2.$
D. $2{{e}^{8}}+1.$
A. $2{{e}^{8}}-4.$
B. $2{{e}^{8}}-2.$
C. $2{{e}^{8}}+2.$
D. $2{{e}^{8}}+1.$
Ta có $\int\limits_{-1}^{4}{f\left( x \right)dx}=\int\limits_{-1}^{0}{f\left( x \right)dx}+\int\limits_{0}^{4}{f\left( x \right)dx}\Leftrightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=\int\limits_{-1}^{4}{f\left( x \right)dx}-\int\limits_{-1}^{0}{f\left( x \right)dx}=2-3=-1.$
Khi đó $\int\limits_{0}^{4}{\left[ 4{{e}^{2x}}+2f\left( x \right) \right]dx}=4\int\limits_{0}^{4}{{{e}^{2x}}dx}+2\int\limits_{0}^{4}{f\left( x \right)dx}$
$=2{{e}^{2x}}\left| \begin{aligned}
& 4 \\
& 0 \\
\end{aligned} \right.+2.\left( -1 \right)=2{{e}^{8}}-2{{e}^{0}}-2=2{{e}^{8}}-4.$
Khi đó $\int\limits_{0}^{4}{\left[ 4{{e}^{2x}}+2f\left( x \right) \right]dx}=4\int\limits_{0}^{4}{{{e}^{2x}}dx}+2\int\limits_{0}^{4}{f\left( x \right)dx}$
$=2{{e}^{2x}}\left| \begin{aligned}
& 4 \\
& 0 \\
\end{aligned} \right.+2.\left( -1 \right)=2{{e}^{8}}-2{{e}^{0}}-2=2{{e}^{8}}-4.$
Đáp án A.