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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa $\int\limits_{0}^{1}{f\left( 2x \right)dx}=2$ và $\int\limits_{0}^{2}{f\left( 6x \right)dx}=14$. Tính $\int\limits_{-2}^{2}{f\left( 5\left| x \right|+2 \right)dx}$.
A. 30.
B. 32.
C. 34.
D. 36.
Nên $2=\int\limits_{0}^{1}{f\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( u \right)du}\Rightarrow \int\limits_{0}^{2}{f\left( u \right)du}=4.$
Nên $14=\int\limits_{0}^{2}{f\left( 6x \right)dx}=\dfrac{1}{6}\int\limits_{0}^{12}{f\left( v \right)dv}\Rightarrow \int\limits_{0}^{12}{f\left( v \right)dv}=84.$
Khi $-2<x<0$ thì $t=-5x+2\Rightarrow dt=-5dx;\ x=-2\Rightarrow t=12;\ x=0\Rightarrow t=2.$
${{I}_{1}}=\dfrac{-1}{5}\int\limits_{12}^{2}{f\left( t \right)dt}=\dfrac{1}{5}\left[ \int\limits_{0}^{12}{f\left( t \right)dt}-\int\limits_{0}^{2}{f\left( t \right)dt} \right]=\dfrac{1}{5}\left( 84-4 \right)=16.$
+ Tính ${{I}_{2}}=\int\limits_{0}^{2}{f\left( 5\left| x \right|+2 \right)dx}$ : Đặt $t=5\left| x \right|+2$.
Khi $0<x<2$ thì $t=5x+2\Rightarrow dt=5dx;\ x=2\Rightarrow t=12;\ x=0\Rightarrow t=2.$
${{I}_{2}}=\dfrac{1}{5}\int\limits_{2}^{12}{f\left( t \right)dt}=\dfrac{1}{5}\left[ \int\limits_{0}^{12}{f\left( t \right)dt}-\int\limits_{0}^{2}{f\left( t \right)dt} \right]=\dfrac{1}{5}\left( 84-4 \right)=16.$
Vậy $\int\limits_{-2}^{2}{f\left( 5\left| x \right|+2 \right)dx}=32.$
A. 30.
B. 32.
C. 34.
D. 36.
- Xét $\int\limits_{0}^{1}{f\left( 2x \right)dx}=2:$
Nên $2=\int\limits_{0}^{1}{f\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( u \right)du}\Rightarrow \int\limits_{0}^{2}{f\left( u \right)du}=4.$
- Xét $\int\limits_{0}^{2}{f\left( 6x \right)dx}=14:$
Nên $14=\int\limits_{0}^{2}{f\left( 6x \right)dx}=\dfrac{1}{6}\int\limits_{0}^{12}{f\left( v \right)dv}\Rightarrow \int\limits_{0}^{12}{f\left( v \right)dv}=84.$
- Xét $\int\limits_{-2}^{2}{f\left( 5\left| x \right|+2 \right)dx}=\int\limits_{-2}^{0}{f\left( 5\left| x \right|+2 \right)dx}+\int\limits_{0}^{2}{f\left( 5\left| x \right|+2 \right)dx}$ :
Khi $-2<x<0$ thì $t=-5x+2\Rightarrow dt=-5dx;\ x=-2\Rightarrow t=12;\ x=0\Rightarrow t=2.$
${{I}_{1}}=\dfrac{-1}{5}\int\limits_{12}^{2}{f\left( t \right)dt}=\dfrac{1}{5}\left[ \int\limits_{0}^{12}{f\left( t \right)dt}-\int\limits_{0}^{2}{f\left( t \right)dt} \right]=\dfrac{1}{5}\left( 84-4 \right)=16.$
+ Tính ${{I}_{2}}=\int\limits_{0}^{2}{f\left( 5\left| x \right|+2 \right)dx}$ : Đặt $t=5\left| x \right|+2$.
Khi $0<x<2$ thì $t=5x+2\Rightarrow dt=5dx;\ x=2\Rightarrow t=12;\ x=0\Rightarrow t=2.$
${{I}_{2}}=\dfrac{1}{5}\int\limits_{2}^{12}{f\left( t \right)dt}=\dfrac{1}{5}\left[ \int\limits_{0}^{12}{f\left( t \right)dt}-\int\limits_{0}^{2}{f\left( t \right)dt} \right]=\dfrac{1}{5}\left( 84-4 \right)=16.$
Vậy $\int\limits_{-2}^{2}{f\left( 5\left| x \right|+2 \right)dx}=32.$
Đáp án B.