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Câu hỏi: Cho hàm số bậc ba $y=f\left( x \right)$ và hàm số bậc nhất $y=g\left( x \right)$ có đồ thị như hình bên dưới.
Biết diện tích phần tô đậm bằng $\dfrac{37}{12}$ và $\int_{0}^{1}{f}\left( x \right)\text{d}x=\dfrac{19}{12}$. Giá trị $\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x$ 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. $-\dfrac{20}{3}$.
$I=\int_{-1}^{0}{x}.{f}'\left( 2x \right)\text{d}x$
Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x$ Suy ra $I=\int\limits_{-2}^{0}{\dfrac{t}{2}{f}'\left( t \right)\dfrac{\text{dt}}{2}}=\dfrac{1}{4}\int\limits_{-2}^{0}{x{f}'\left( x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}x={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
Suy ra: $I=\dfrac{1}{4}\int\limits_{-2}^{0}{x{f}'\left( x \right)\text{d}x}=\dfrac{1}{4}\left[ \left. xf\left( x \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]\left( ** \right)$.
Quan sát đồ thị ta thấy:
$f\left( 1 \right)=3; f\left( -2 \right)=-3$.
Gọi $g\left( x \right)=ax+b\Rightarrow \left\{ \begin{aligned}
& a+b=3 \\
& -2a+b=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=1 \\
\end{aligned} \right.\Rightarrow g\left( x \right)=2x+1$.
$\int\limits_{-2}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x}=\int\limits_{-2}^{0}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x}=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-2x-1 \right]\text{d}x=}\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-2x-1 \right]\text{d}x=}\dfrac{37}{12}$
$\begin{aligned}
& \Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x-}\int\limits_{0}^{1}{f\left( x \right)\text{d}x=}\dfrac{37}{12}-4 \\
& \Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x-}\dfrac{19}{12}=\dfrac{37}{12}-4\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}=\dfrac{2}{3} \\
\end{aligned}$.
$I=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2\left( -3 \right)-\dfrac{2}{3} \right]=-\dfrac{5}{3}$.
Đặt $t=2x\Rightarrow \text{d}t=2\text{d}x$ Suy ra $I=\int\limits_{-2}^{0}{\dfrac{t}{2}{f}'\left( t \right)\dfrac{\text{dt}}{2}}=\dfrac{1}{4}\int\limits_{-2}^{0}{x{f}'\left( x \right)\text{d}x}$.
Đặt $\left\{ \begin{aligned}
& u=x \\
& \text{d}x={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
Suy ra: $I=\dfrac{1}{4}\int\limits_{-2}^{0}{x{f}'\left( x \right)\text{d}x}=\dfrac{1}{4}\left[ \left. xf\left( x \right) \right|_{-2}^{0}-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]\left( ** \right)$.
Quan sát đồ thị ta thấy:
$f\left( 1 \right)=3; f\left( -2 \right)=-3$.
Gọi $g\left( x \right)=ax+b\Rightarrow \left\{ \begin{aligned}
& a+b=3 \\
& -2a+b=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2 \\
& b=1 \\
\end{aligned} \right.\Rightarrow g\left( x \right)=2x+1$.
$\int\limits_{-2}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x}=\int\limits_{-2}^{0}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-g\left( x \right) \right]\text{d}x}=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-2x-1 \right]\text{d}x=}\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-2}^{0}{\left[ f\left( x \right)-2x-1 \right]\text{d}x-}\int\limits_{0}^{1}{\left[ f\left( x \right)-2x-1 \right]\text{d}x=}\dfrac{37}{12}$
$\begin{aligned}
& \Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x-}\int\limits_{0}^{1}{f\left( x \right)\text{d}x=}\dfrac{37}{12}-4 \\
& \Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x-}\dfrac{19}{12}=\dfrac{37}{12}-4\Leftrightarrow \int\limits_{-2}^{0}{f\left( x \right)\text{d}x}=\dfrac{2}{3} \\
\end{aligned}$.
$I=\dfrac{1}{4}\left[ 2f\left( -2 \right)-\int\limits_{-2}^{0}{f\left( x \right)\text{d}x} \right]=\dfrac{1}{4}\left[ 2\left( -3 \right)-\dfrac{2}{3} \right]=-\dfrac{5}{3}$.
Đáp án A.