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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=6; \int\limits_{1}^{3}{f\left( x \right)\text{d}x}=4$. Tính tích phân $I=\int\limits_{-1}^{1}{f\left( \left| 2x-1 \right| \right)\text{d}x}$.
A. $I=3$.
B. $I=8$.
C. $I=14$.
D. $I=6$.
A. $I=3$.
B. $I=8$.
C. $I=14$.
D. $I=6$.
Ta có: $I=\int\limits_{-1}^{1}{f\left( \left| 2x-1 \right| \right)\text{d}x}=\int\limits_{-1}^{\dfrac{1}{2}}{f\left( 1-2x \right)\text{d}x}+\int\limits_{\dfrac{1}{2}}^{1}{f\left( 2x-1 \right)\text{d}x}={{I}_{1}}+{{I}_{2}}$
Tính ${{I}_{1}}=\int\limits_{-1}^{\dfrac{1}{2}}{f\left( 1-2x \right)\text{d}x}$. Đặt $u=1-2x\Rightarrow \text{d}u=-2\text{d}x$. Đổi cận: $\left\{ \begin{aligned}
& x=-1\Rightarrow u=3 \\
& x=\dfrac{1}{2}\Rightarrow u=0 \\
\end{aligned} \right.$.
$\Rightarrow {{I}_{1}}=\dfrac{-1}{2}\int\limits_{3}^{0}{f\left( u \right) \text{d}u}=\dfrac{1}{2}\int\limits_{0}^{3}{f\left( u \right) \text{d}u}=\dfrac{1}{2}\int\limits_{0}^{3}{f\left( x \right) \text{d}}x= \dfrac{1}{2}\left[ \int\limits_{0}^{1}{f\left( x \right) \text{d}}x+ \int\limits_{1}^{3}{f\left( x \right) \text{d}}x \right] =5$
Tính ${{I}_{2}}=\int\limits_{\dfrac{1}{2}}^{1}{f\left( 2x-1 \right)\text{d}x}$. Đặt $v=2x-1\Rightarrow \text{d}v=2 \text{d}x$. Đổi cận: $\left\{ \begin{aligned}
& x=1\Rightarrow v=1 \\
& x=\dfrac{1}{2}\Rightarrow v=0 \\
\end{aligned} \right.$.
$\Rightarrow {{I}_{2}}=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( v \right)\text{d}}v=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)\text{d}x= }3.$ Vậy $I={{I}_{1}}+{{I}_{2}}=8$.
Tính ${{I}_{1}}=\int\limits_{-1}^{\dfrac{1}{2}}{f\left( 1-2x \right)\text{d}x}$. Đặt $u=1-2x\Rightarrow \text{d}u=-2\text{d}x$. Đổi cận: $\left\{ \begin{aligned}
& x=-1\Rightarrow u=3 \\
& x=\dfrac{1}{2}\Rightarrow u=0 \\
\end{aligned} \right.$.
$\Rightarrow {{I}_{1}}=\dfrac{-1}{2}\int\limits_{3}^{0}{f\left( u \right) \text{d}u}=\dfrac{1}{2}\int\limits_{0}^{3}{f\left( u \right) \text{d}u}=\dfrac{1}{2}\int\limits_{0}^{3}{f\left( x \right) \text{d}}x= \dfrac{1}{2}\left[ \int\limits_{0}^{1}{f\left( x \right) \text{d}}x+ \int\limits_{1}^{3}{f\left( x \right) \text{d}}x \right] =5$
Tính ${{I}_{2}}=\int\limits_{\dfrac{1}{2}}^{1}{f\left( 2x-1 \right)\text{d}x}$. Đặt $v=2x-1\Rightarrow \text{d}v=2 \text{d}x$. Đổi cận: $\left\{ \begin{aligned}
& x=1\Rightarrow v=1 \\
& x=\dfrac{1}{2}\Rightarrow v=0 \\
\end{aligned} \right.$.
$\Rightarrow {{I}_{2}}=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( v \right)\text{d}}v=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)\text{d}x= }3.$ Vậy $I={{I}_{1}}+{{I}_{2}}=8$.
Đáp án B.