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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 $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{19}{12}$. Tích phân $\int\limits_{-1}^{0}{x.{f}'\left( 2x \right)\text{d}x}$ bằng
A. $-\dfrac{607}{348}$
B. $-\dfrac{20}{3}$
C. $-\dfrac{5}{3}$
D. $-\dfrac{5}{6}$
A. $-\dfrac{607}{348}$
B. $-\dfrac{20}{3}$
C. $-\dfrac{5}{3}$
D. $-\dfrac{5}{6}$
Ta có: $\left\{ \begin{aligned}
& A\left( 1;3 \right)\in g\left( x \right)=ax+b \\
& B\left( -2;-3 \right)\in g\left( x \right)=ax+b \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a+b=3 \\
& -2a+b=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=1 \\
\end{aligned} \right.\Rightarrow g\left( x \right)=2x+1.$
Mà $S=\dfrac{37}{12}\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]\text{d}x}+\int\limits_{0}^{1}{\left[ \left( 2x+1 \right)-f\left( x \right) \right]\text{d}x}=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{\left( 2x+1 \right)\text{d}x}-\int\limits_{-2}^{0}{\left( 2x+1 \right)\text{d}x}=\dfrac{37}{12}\Rightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}=\dfrac{2}{3}\cdot $
Khi đó $\int\limits_{-1}^{0}{x.{f}'\left( 2x \right)\text{d}x}\xrightarrow[\left\{ \begin{smallmatrix}
x=-1\to t=-2 \\
x=0\to t=0
\end{smallmatrix} \right.]{t=2x\to \text{d}t=2\text{d}x}\dfrac{1}{4}\int\limits_{-2}^{0}{t.{f}'\left( t \right)\text{d}t}\xrightarrow[\text{d}v={f}'(t)\text{d}t\to v=f(t)]{u=t\to \text{d}u=\text{d}t}\dfrac{1}{4}\left[ \left. t.f\left( t \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( t \right)\text{d}t} \right]$
$=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2.\left( -3 \right)-\dfrac{2}{3} \right]=-\dfrac{5}{3}$.
& A\left( 1;3 \right)\in g\left( x \right)=ax+b \\
& B\left( -2;-3 \right)\in g\left( x \right)=ax+b \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a+b=3 \\
& -2a+b=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=1 \\
\end{aligned} \right.\Rightarrow g\left( x \right)=2x+1.$
Mà $S=\dfrac{37}{12}\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]\text{d}x}+\int\limits_{0}^{1}{\left[ \left( 2x+1 \right)-f\left( x \right) \right]\text{d}x}=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{\left( 2x+1 \right)\text{d}x}-\int\limits_{-2}^{0}{\left( 2x+1 \right)\text{d}x}=\dfrac{37}{12}\Rightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}=\dfrac{2}{3}\cdot $
Khi đó $\int\limits_{-1}^{0}{x.{f}'\left( 2x \right)\text{d}x}\xrightarrow[\left\{ \begin{smallmatrix}
x=-1\to t=-2 \\
x=0\to t=0
\end{smallmatrix} \right.]{t=2x\to \text{d}t=2\text{d}x}\dfrac{1}{4}\int\limits_{-2}^{0}{t.{f}'\left( t \right)\text{d}t}\xrightarrow[\text{d}v={f}'(t)\text{d}t\to v=f(t)]{u=t\to \text{d}u=\text{d}t}\dfrac{1}{4}\left[ \left. t.f\left( t \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( t \right)\text{d}t} \right]$
$=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2.\left( -3 \right)-\dfrac{2}{3} \right]=-\dfrac{5}{3}$.
Đáp án C.