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Câu hỏi: Cho hàm số $f\left( x \right)$ có ${f}'\left( x \right)$ và ${f}''\left( x \right)$ liên tục trên đoạn $\left[ 1;3 \right]$. Biết $f\left( 1 \right)=1,$ $f\left( 3 \right)=81,$ ${f}'\left( 1 \right)=4,$ ${f}'\left( 3 \right)=108$. Giá trị của $\int\limits_{1}^{3}{\left( 4-2x \right){f}''\left( x \right)dx}$ bằng
A. $48.$
B. $-64.$
C. $-48.$
D. $64.$
Xét: $I=\int\limits_{1}^{3}{\left( 4-2x \right){f}''\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=4-2x \\
& dv={f}''\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-2dx \\
& v={f}'\left( x \right) \\
\end{aligned} \right.$
Khi đó: $I=\left. \left( 4-2x \right){f}'\left( x \right) \right|_{1}^{3}+2\int\limits_{1}^{3}{{f}'\left( x \right)dx}=\left. \left( 4-2x \right){f}'\left( x \right) \right|_{1}^{3}+\left. 2f\left( x \right) \right|_{1}^{3}$.
$=-2{f}'\left( 3 \right)-2{f}'\left( 1 \right)+2f\left( 3 \right)-2f\left( 1 \right)=-2.108-2.4+2.81-2.1=-64$.
A. $48.$
B. $-64.$
C. $-48.$
D. $64.$
Xét: $I=\int\limits_{1}^{3}{\left( 4-2x \right){f}''\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=4-2x \\
& dv={f}''\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-2dx \\
& v={f}'\left( x \right) \\
\end{aligned} \right.$
Khi đó: $I=\left. \left( 4-2x \right){f}'\left( x \right) \right|_{1}^{3}+2\int\limits_{1}^{3}{{f}'\left( x \right)dx}=\left. \left( 4-2x \right){f}'\left( x \right) \right|_{1}^{3}+\left. 2f\left( x \right) \right|_{1}^{3}$.
$=-2{f}'\left( 3 \right)-2{f}'\left( 1 \right)+2f\left( 3 \right)-2f\left( 1 \right)=-2.108-2.4+2.81-2.1=-64$.
Đáp án B.