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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f\left( 4 \right)=1$ và $\int\limits_{-2}^{2}{xf\left( x+2 \right)d\text{x}}=5$ khi đó $\int\limits_{0}^{4}{\left[ {{x}^{2}}{f}'\left( x \right)+4f\left( x \right) \right]d\text{x}}$ bằng
A. $-6$
B. 4
C. $-10$
D. 6
A. $-6$
B. 4
C. $-10$
D. 6
Đặt $t=x+2$ ta có: $\int\limits_{02}^{2}{xf\left( x+2 \right)d\text{x}}=5\Leftrightarrow \int\limits_{0}^{4}{\left( t-2 \right)f\left( t \right)dt}=5\Rightarrow \int\limits_{0}^{4}{\left( x-2 \right)f\left( x \right)d\text{x}}=5$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2\text{xdx} \\
& v=f\left( x \right) \\
\end{aligned} \right. $ suy ra $ \int\limits_{0}^{4}{{{x}^{2}}{f}'\left( x \right)d\text{x}}=\left. {{x}^{2}}f\left( x \right) \right|_{0}^{4}-\int\limits_{0}^{4}{2\text{x}f\left( x \right)d\text{x}}$
$=16f\left( 4 \right)-2\int\limits_{0}^{4}{xf\left( x \right)d\text{x}}$
Do đó $\int\limits_{0}^{4}{\left[ {{x}^{2}}{f}'\left( x \right)+4f\left( x \right) \right]d\text{x}}=16-2\int\limits_{0}^{2}{\left[ xf\left( x \right)-2f\left( x \right) \right]d\text{x}}=16-2\int\limits_{0}^{4}{\left( x-2 \right)f\left( x \right)d\text{x}}$
$=16-2.5=6$.
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={f}'\left( x \right)d\text{x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2\text{xdx} \\
& v=f\left( x \right) \\
\end{aligned} \right. $ suy ra $ \int\limits_{0}^{4}{{{x}^{2}}{f}'\left( x \right)d\text{x}}=\left. {{x}^{2}}f\left( x \right) \right|_{0}^{4}-\int\limits_{0}^{4}{2\text{x}f\left( x \right)d\text{x}}$
$=16f\left( 4 \right)-2\int\limits_{0}^{4}{xf\left( x \right)d\text{x}}$
Do đó $\int\limits_{0}^{4}{\left[ {{x}^{2}}{f}'\left( x \right)+4f\left( x \right) \right]d\text{x}}=16-2\int\limits_{0}^{2}{\left[ xf\left( x \right)-2f\left( x \right) \right]d\text{x}}=16-2\int\limits_{0}^{4}{\left( x-2 \right)f\left( x \right)d\text{x}}$
$=16-2.5=6$.
Đáp án D.