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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ có $\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=8$ và $\int\limits_{0}^{5}{f\left( x \right)d\text{x}}=4$. Giá trị của $\int\limits_{-1}^{1}{f\left( \left| 4\text{x}-1 \right| \right)d\text{x}}$ bằng
A. 3
B. 6
C. $\dfrac{9}{4}$
D. $\dfrac{11}{4}$
A. 3
B. 6
C. $\dfrac{9}{4}$
D. $\dfrac{11}{4}$
Ta có: $I=\int\limits_{-1}^{1}{f\left( \left| 4\text{x}-1 \right| \right)d\text{x}}=\int\limits_{-1}^{\dfrac{1}{4}}{f\left( -4\text{x}+1 \right)d\text{x}}+\int\limits_{\dfrac{1}{4}}^{1}{f\left( 4\text{x}-1 \right)d\text{x}}$.
Xét ${{I}_{1}}=\int\limits_{-1}^{\dfrac{1}{4}}{f\left( -4\text{x}+1 \right)d\text{x}}$. Đặt $-4\text{x}+1=t\Rightarrow dt=-4\text{dx}$. Đổi cận: $\left\{ \begin{aligned}
& x=-1\Rightarrow t=5 \\
& x=\dfrac{1}{4}\Rightarrow t=0 \\
\end{aligned} \right.$
${{I}_{1}}=-\dfrac{1}{4}\int\limits_{5}^{0}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{5}{f\left( t \right)dt}=\dfrac{1}{4}.4=1$.
Xét ${{I}_{2}}=\int\limits_{\dfrac{1}{4}}^{1}{f\left( 4\text{x}-1 \right)d\text{x}}$. Đặt $4\text{x}-1=t\Rightarrow dt=4\text{dx}$. Đổi cận: $\left\{ \begin{aligned}
& x=1\Rightarrow t=3 \\
& x=\dfrac{1}{4}\Rightarrow t=0 \\
\end{aligned} \right.$
${{I}_{2}}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=\dfrac{1}{4}.8=2$.
Vậy $I={{I}_{1}}+{{I}_{2}}=1+2=3$.
Xét ${{I}_{1}}=\int\limits_{-1}^{\dfrac{1}{4}}{f\left( -4\text{x}+1 \right)d\text{x}}$. Đặt $-4\text{x}+1=t\Rightarrow dt=-4\text{dx}$. Đổi cận: $\left\{ \begin{aligned}
& x=-1\Rightarrow t=5 \\
& x=\dfrac{1}{4}\Rightarrow t=0 \\
\end{aligned} \right.$
${{I}_{1}}=-\dfrac{1}{4}\int\limits_{5}^{0}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{5}{f\left( t \right)dt}=\dfrac{1}{4}.4=1$.
Xét ${{I}_{2}}=\int\limits_{\dfrac{1}{4}}^{1}{f\left( 4\text{x}-1 \right)d\text{x}}$. Đặt $4\text{x}-1=t\Rightarrow dt=4\text{dx}$. Đổi cận: $\left\{ \begin{aligned}
& x=1\Rightarrow t=3 \\
& x=\dfrac{1}{4}\Rightarrow t=0 \\
\end{aligned} \right.$
${{I}_{2}}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( t \right)dt}=\dfrac{1}{4}\int\limits_{0}^{3}{f\left( x \right)d\text{x}}=\dfrac{1}{4}.8=2$.
Vậy $I={{I}_{1}}+{{I}_{2}}=1+2=3$.
Đáp án A.