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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục, có đạo hàm trên $\mathbb{R}$, $f\left( 2 \right)=16$ và $\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=4$. Tích phân $\int\limits_{0}^{4}{x{f}'\left( \dfrac{x}{2} \right)dx}$ bằng
A. $112$.
B. $144$.
C. $56$.
D. $12$.
A. $112$.
B. $144$.
C. $56$.
D. $12$.
Xét tích phân $I=\int\limits_{0}^{4}{x{f}'\left( \dfrac{x}{2} \right)\text{d}x}$
Đặt: $t=\dfrac{x}{2}\Rightarrow dt=\dfrac{1}{2}dx$
Đổi cận: $x=0\Rightarrow t=0;x=4\Rightarrow t=2$.
Khi đó: $I=\int\limits_{0}^{4}{x{f}'\left( \dfrac{x}{2} \right)\text{d}x}=\int\limits_{0}^{2}{4t{f}'\left( t \right)\text{d}t}=\int\limits_{0}^{2}{4x{f}'\left( x \right)\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=4x \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& u=4\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó: $I=\int\limits_{0}^{2}{4x{f}'\left( x \right)\text{d}x}=\left. 4xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{4f\left( x \right)\text{d}x}$ $=8f\left( 2 \right)-4.4$ $=8.16-16=112$.
Đặt: $t=\dfrac{x}{2}\Rightarrow dt=\dfrac{1}{2}dx$
Đổi cận: $x=0\Rightarrow t=0;x=4\Rightarrow t=2$.
Khi đó: $I=\int\limits_{0}^{4}{x{f}'\left( \dfrac{x}{2} \right)\text{d}x}=\int\limits_{0}^{2}{4t{f}'\left( t \right)\text{d}t}=\int\limits_{0}^{2}{4x{f}'\left( x \right)\text{d}x}$.
Đặt: $\left\{ \begin{aligned}
& u=4x \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& u=4\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Khi đó: $I=\int\limits_{0}^{2}{4x{f}'\left( x \right)\text{d}x}=\left. 4xf\left( x \right) \right|_{0}^{2}-\int\limits_{0}^{2}{4f\left( x \right)\text{d}x}$ $=8f\left( 2 \right)-4.4$ $=8.16-16=112$.
Đáp án A.