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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và đường thẳng $\left( d \right):g\left( x \right)=ax+b$ có đồ thị như hình vẽ.
Biết diện tích miền tô đậm bằng $\dfrac{37}{12}$ và $\int_{0}^{1} f(x) \mathrm{d} x=\dfrac{19}{12}$. Tích phân $\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x$ bằng
A. $-\dfrac{5}{3}$.
B. $-\dfrac{607}{348}$.
C. $-\dfrac{5}{6}$.
D. $-\dfrac{20}{3}$.
A. $-\dfrac{5}{3}$.
B. $-\dfrac{607}{348}$.
C. $-\dfrac{5}{6}$.
D. $-\dfrac{20}{3}$.
Xét $\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x$.
Đặt $t=2x\Rightarrow dt=2dx\Rightarrow dx=\dfrac{dt}{2}$ ; $x=\dfrac{t}{2}$.
Đôi cận : $x=-1\Rightarrow t=-2 ; x=0\Rightarrow t=0$.
Suy ra : $I=\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x=\int\limits_{-2}^{0}{\dfrac{t}{2}}{f}'\left( t \right)\dfrac{\text{d}t}{2}=\dfrac{1}{4}\int\limits_{-2}^{0}{t.{f}'\left( t \right)dt}=\dfrac{1}{4}\int\limits_{-2}^{0}{x.{f}'\left( x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Suy ra $I=\dfrac{1}{4}\left[ \left. xf\left( x \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]$.
Xét hàm số $g\left( x \right)=ax+b$. Đồ thị hàm số đi qua điểm $\left( -2 ; -3 \right)$ và $\left( 1 ; 3 \right)$ nên ta có hệ phương trình $\left\{ \begin{aligned}
& -2a+b=-3 \\
& a+b=3 \\
\end{aligned} \right.\Leftrightarrow a=2; b=1\Rightarrow g\left( x \right)=2x+1$
Diện tích miềm tô đậm :
$S=\int\limits_{-2}^{0}{\left[ f\left( x \right)-g\left( x \right) \right]}dx+\int\limits_{0}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]}dx=\dfrac{37}{12}$.
$S=\int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]} \text{d}x+\int\limits_{0}^{1}{\left[ 2x+1-f\left( x \right) \right]} \text{d}x=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right) }\text{d}x-\int\limits_{-2}^{0}{\left( 2x+1 \right)} \text{d}x+\int\limits_{0}^{1}{\left( 2x+1 \right)}\text{d}x-\int\limits_{0}^{1}{f\left( x \right) }\text{d}x=\dfrac{37}{12}\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right) }\text{d}x=\dfrac{2}{3}$.
Suy ra $I=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]=-\dfrac{5}{3}$.
Đặt $t=2x\Rightarrow dt=2dx\Rightarrow dx=\dfrac{dt}{2}$ ; $x=\dfrac{t}{2}$.
Đôi cận : $x=-1\Rightarrow t=-2 ; x=0\Rightarrow t=0$.
Suy ra : $I=\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x=\int\limits_{-2}^{0}{\dfrac{t}{2}}{f}'\left( t \right)\dfrac{\text{d}t}{2}=\dfrac{1}{4}\int\limits_{-2}^{0}{t.{f}'\left( t \right)dt}=\dfrac{1}{4}\int\limits_{-2}^{0}{x.{f}'\left( x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
Suy ra $I=\dfrac{1}{4}\left[ \left. xf\left( x \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]$.
Xét hàm số $g\left( x \right)=ax+b$. Đồ thị hàm số đi qua điểm $\left( -2 ; -3 \right)$ và $\left( 1 ; 3 \right)$ nên ta có hệ phương trình $\left\{ \begin{aligned}
& -2a+b=-3 \\
& a+b=3 \\
\end{aligned} \right.\Leftrightarrow a=2; b=1\Rightarrow g\left( x \right)=2x+1$
Diện tích miềm tô đậm :
$S=\int\limits_{-2}^{0}{\left[ f\left( x \right)-g\left( x \right) \right]}dx+\int\limits_{0}^{1}{\left[ g\left( x \right)-f\left( x \right) \right]}dx=\dfrac{37}{12}$.
$S=\int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]} \text{d}x+\int\limits_{0}^{1}{\left[ 2x+1-f\left( x \right) \right]} \text{d}x=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right) }\text{d}x-\int\limits_{-2}^{0}{\left( 2x+1 \right)} \text{d}x+\int\limits_{0}^{1}{\left( 2x+1 \right)}\text{d}x-\int\limits_{0}^{1}{f\left( x \right) }\text{d}x=\dfrac{37}{12}\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right) }\text{d}x=\dfrac{2}{3}$.
Suy ra $I=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)dx} \right]=-\dfrac{5}{3}$.
Đáp án A.