Google
Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và đường thẳng $\left( d \right) : y=ax+b$ có đồ thị như hình vẽ.
Biết diện tích phần tô đậm bằng $\dfrac{37}{12}$ và $\int\limits_{-1}^{0}{f\left( x \right)\text{d}x}=\dfrac{5}{12}$. Tích phân $\int\limits_{0}^{1}{x{f}'\left( 2x \right)\text{d}x}$ bằng
A. $\dfrac{35}{8}$.
B. $\dfrac{13}{3}$.
C. $\dfrac{20}{3}$.
D. $\dfrac{50}{3}$.
A. $\dfrac{35}{8}$.
B. $\dfrac{13}{3}$.
C. $\dfrac{20}{3}$.
D. $\dfrac{50}{3}$.
Từ đồ thị suy ra được phương trình đường thẳng $\left( d \right) : y=2x+1$.
+) Diện tích phần tô đậm bằng $\dfrac{37}{12}$, suy ra:
$\int\limits_{-1}^{2}{\left| f\left( x \right)-\left( 2x+1 \right) \right|}\text{d}x=\dfrac{37}{12}\Leftrightarrow \int\limits_{-1}^{0}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]}\text{d}x-\int\limits_{0}^{2}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]}\text{d}x=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-1}^{0}{f\left( x \right)}\text{d}x-\int\limits_{-1}^{0}{\left( 2x+1 \right)}\text{d}x-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x+\int\limits_{0}^{2}{\left( 2x+1 \right)}\text{d}x=\dfrac{37}{12}\Leftrightarrow \dfrac{5}{12}-0-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x+6=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)}\text{d}x=\dfrac{10}{3}$.
+) Xét tích phân $I=\int\limits_{0}^{1}{x{f}'\left( 2x \right)\text{d}x}.$ Đặt $2x=t\Rightarrow \text{d}x=\dfrac{\text{d}t}{2}$. Đổi cận: $x=0\Rightarrow t=0; x=1\Rightarrow t=2.$
Khi đó $I=\int\limits_{0}^{2}{\dfrac{t}{2}{f}'\left( t \right)\dfrac{\text{dt}}{2}}=\dfrac{1}{4}\int\limits_{0}^{2}{t{f}'\left( t \right)\text{dt}=}\dfrac{1}{4}\int\limits_{0}^{2}{x{f}'\left( x \right)\text{d}x.}$ Đặt $\left\{ \begin{aligned}
& x=u \\
& {f}'\left( x \right)\text{d}x=\text{d}v \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}x=\text{d}u \\
& f\left( x \right)=v \\
\end{aligned} \right.$
$\Rightarrow I=xf\left( x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=2f\left( 2 \right)-\dfrac{10}{3}=2.5-\dfrac{10}{3}=\dfrac{20}{3}.$
+) Diện tích phần tô đậm bằng $\dfrac{37}{12}$, suy ra:
$\int\limits_{-1}^{2}{\left| f\left( x \right)-\left( 2x+1 \right) \right|}\text{d}x=\dfrac{37}{12}\Leftrightarrow \int\limits_{-1}^{0}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]}\text{d}x-\int\limits_{0}^{2}{\left[ f\left( x \right)-\left( 2x+1 \right) \right]}\text{d}x=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{-1}^{0}{f\left( x \right)}\text{d}x-\int\limits_{-1}^{0}{\left( 2x+1 \right)}\text{d}x-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x+\int\limits_{0}^{2}{\left( 2x+1 \right)}\text{d}x=\dfrac{37}{12}\Leftrightarrow \dfrac{5}{12}-0-\int\limits_{0}^{2}{f\left( x \right)}\text{d}x+6=\dfrac{37}{12}$
$\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)}\text{d}x=\dfrac{10}{3}$.
+) Xét tích phân $I=\int\limits_{0}^{1}{x{f}'\left( 2x \right)\text{d}x}.$ Đặt $2x=t\Rightarrow \text{d}x=\dfrac{\text{d}t}{2}$. Đổi cận: $x=0\Rightarrow t=0; x=1\Rightarrow t=2.$
Khi đó $I=\int\limits_{0}^{2}{\dfrac{t}{2}{f}'\left( t \right)\dfrac{\text{dt}}{2}}=\dfrac{1}{4}\int\limits_{0}^{2}{t{f}'\left( t \right)\text{dt}=}\dfrac{1}{4}\int\limits_{0}^{2}{x{f}'\left( x \right)\text{d}x.}$ Đặt $\left\{ \begin{aligned}
& x=u \\
& {f}'\left( x \right)\text{d}x=\text{d}v \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}x=\text{d}u \\
& f\left( x \right)=v \\
\end{aligned} \right.$
$\Rightarrow I=xf\left( x \right)\left| \begin{aligned}
& 2 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=2f\left( 2 \right)-\dfrac{10}{3}=2.5-\dfrac{10}{3}=\dfrac{20}{3}.$
Đáp án C.