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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ và thỏa mãn $f\left( 0 \right)=-2$ và $f\left( x \right)+f\left( 4-x \right)={{x}^{2}}-4x+1,\forall x\in \mathbb{R}$. Tích phân $\int\limits_{0}^{2}{x.{f}'\left( 2x \right)dx}$ bằng
A. $\dfrac{23}{6}$.
B. $\dfrac{23}{24}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{19}{12}$.
A. $\dfrac{23}{6}$.
B. $\dfrac{23}{24}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{19}{12}$.
Đặt $t=2x\Rightarrow dt=2dx$ nên
$I=\int\limits_{0}^{2}{x.{f}'\left( 2x \right)dx}=\int\limits_{0}^{4}{\dfrac{t}{2}{f}'\left( t \right).\dfrac{dt}{2}=\dfrac{1}{4}\int\limits_{0}^{4}{t{f}'\left( t \right)dt}=4\dfrac{1}{4}\int\limits_{0}^{4}{x{f}'\left( x \right)dx}}$
Lại có: $\int\limits_{0}^{4}{x{f}'\left( x \right)dx}=\left. xf\left( x \right) \right|_{0}^{4}-\int\limits_{0}^{4}{f\left( x \right)dx}=f\left( 4 \right)-\int\limits_{0}^{4}{f\left( x \right)dx}$
Mặt khác $f\left( x \right)+f\left( 4-x \right)={{x}^{2}}-4x+1\Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}+\int\limits_{0}^{4}{f\left( 4-x \right)dx}=\dfrac{-20}{3}\left( * \right)$
Do $\int\limits_{0}^{4}{f\left( 4-x \right)dx}=-\int\limits_{0}^{4}{f\left( 4-x \right)d\left( 4-x \right)}=-\int\limits_{4}^{0}{f\left( u \right)du}=\int\limits_{0}^{4}{f\left( u \right)du}$
Suy ra $\left( * \right)\Leftrightarrow 2\int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{-20}{3}\Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{-10}{3}$
Thay $x=0$ vào giả thiết ta được $f\left( 0 \right)+f\left( 4 \right)=1\Rightarrow f\left( 4 \right)=3$ nên $I=\dfrac{23}{6}$.
$I=\int\limits_{0}^{2}{x.{f}'\left( 2x \right)dx}=\int\limits_{0}^{4}{\dfrac{t}{2}{f}'\left( t \right).\dfrac{dt}{2}=\dfrac{1}{4}\int\limits_{0}^{4}{t{f}'\left( t \right)dt}=4\dfrac{1}{4}\int\limits_{0}^{4}{x{f}'\left( x \right)dx}}$
Lại có: $\int\limits_{0}^{4}{x{f}'\left( x \right)dx}=\left. xf\left( x \right) \right|_{0}^{4}-\int\limits_{0}^{4}{f\left( x \right)dx}=f\left( 4 \right)-\int\limits_{0}^{4}{f\left( x \right)dx}$
Mặt khác $f\left( x \right)+f\left( 4-x \right)={{x}^{2}}-4x+1\Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}+\int\limits_{0}^{4}{f\left( 4-x \right)dx}=\dfrac{-20}{3}\left( * \right)$
Do $\int\limits_{0}^{4}{f\left( 4-x \right)dx}=-\int\limits_{0}^{4}{f\left( 4-x \right)d\left( 4-x \right)}=-\int\limits_{4}^{0}{f\left( u \right)du}=\int\limits_{0}^{4}{f\left( u \right)du}$
Suy ra $\left( * \right)\Leftrightarrow 2\int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{-20}{3}\Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{-10}{3}$
Thay $x=0$ vào giả thiết ta được $f\left( 0 \right)+f\left( 4 \right)=1\Rightarrow f\left( 4 \right)=3$ nên $I=\dfrac{23}{6}$.
Đáp án A.