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Câu hỏi: Cho $\int\limits_{0}^{2}{f\left( x \right)}\text{d}x=2$ và $\int\limits_{1}^{0}{g\left( x \right)}\text{d}x=1$, khi đó $\int\limits_{0}^{1}{\left[ f\left( 2x \right)-3g\left( x \right) \right]}\text{d}x$ bằng
A. $4.$
B. $1.$
C. $7.$
D. $-2.$
A. $4.$
B. $1.$
C. $7.$
D. $-2.$
$\begin{aligned}
& \int\limits_{0}^{1}{\left[ f\left( 2x \right)-3g\left( x \right) \right]}\text{d}x=\int\limits_{0}^{1}{f\left( 2x \right)}\text{d}x-3\int\limits_{0}^{1}{g\left( x \right)}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)}\text{dt+}3\int\limits_{1}^{0}{g\left( x \right)}\text{d}x \\
& =\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)}\text{d}x\text{+}3\int\limits_{1}^{0}{g\left( x \right)}\text{d}x=\dfrac{1}{2}.2+3.1=4. \\
\end{aligned}$
& \int\limits_{0}^{1}{\left[ f\left( 2x \right)-3g\left( x \right) \right]}\text{d}x=\int\limits_{0}^{1}{f\left( 2x \right)}\text{d}x-3\int\limits_{0}^{1}{g\left( x \right)}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( t \right)}\text{dt+}3\int\limits_{1}^{0}{g\left( x \right)}\text{d}x \\
& =\dfrac{1}{2}\int\limits_{0}^{2}{f\left( x \right)}\text{d}x\text{+}3\int\limits_{1}^{0}{g\left( x \right)}\text{d}x=\dfrac{1}{2}.2+3.1=4. \\
\end{aligned}$
Đáp án A.