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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${f}'\left( x \right)=\left( x+3 \right){{e}^{x}}\left( \forall x\in \mathbb{R} \right)$ và $f\left( 0 \right)=5$. Tính $I=\int\limits_{0}^{3}{f\left( x \right)dx}$
A. $I=4{{e}^{3}}-10$.
B. $I=4{{e}^{3}}+8$.
C. $I=4{{e}^{3}}+10$.
D. $I=4{{e}^{3}}-8$.
A. $I=4{{e}^{3}}-10$.
B. $I=4{{e}^{3}}+8$.
C. $I=4{{e}^{3}}+10$.
D. $I=4{{e}^{3}}-8$.
Ta có: $f\left( x \right)=f\left( x+2 \right){{e}^{x}}dx$. Đặt $\left\{ \begin{aligned}
& u=x+3 \\
& dv={{e}^{x}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Suy ra $f\left( x \right)=\left( x+3 \right){{e}^{x}}-\int\limits_{{}}^{{}}{{{e}^{x}}dx}+C=\left( x+2 \right){{e}^{x}}+C$
Mặt khác $f\left( 0 \right)=2{{e}^{0}}+C=5\Leftrightarrow C=3\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{x}}+3$
$I=\int\limits_{0}^{3}{f\left( x \right)dx}=\int\limits_{0}^{3}{\left( x+2 \right){{e}^{x}}dx}+\int\limits_{0}^{3}{3dx}=\left( x+1 \right){{e}^{x}}\left| \begin{aligned}
& ^{3} \\
& _{0} \\
\end{aligned} \right.+9=4{{e}^{3}}+8$
& u=x+3 \\
& dv={{e}^{x}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v={{e}^{x}} \\
\end{aligned} \right.$
Suy ra $f\left( x \right)=\left( x+3 \right){{e}^{x}}-\int\limits_{{}}^{{}}{{{e}^{x}}dx}+C=\left( x+2 \right){{e}^{x}}+C$
Mặt khác $f\left( 0 \right)=2{{e}^{0}}+C=5\Leftrightarrow C=3\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{x}}+3$
$I=\int\limits_{0}^{3}{f\left( x \right)dx}=\int\limits_{0}^{3}{\left( x+2 \right){{e}^{x}}dx}+\int\limits_{0}^{3}{3dx}=\left( x+1 \right){{e}^{x}}\left| \begin{aligned}
& ^{3} \\
& _{0} \\
\end{aligned} \right.+9=4{{e}^{3}}+8$
Đáp án B.