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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{-1}^{1}{\dfrac{f\left( 3x \right)}{1+{{2}^{x}}}\text{d}x}=8$. Tính $\int\limits_{0}^{3}{f\left( x \right)\text{d}x}$.
A. $16$.
B. $24$.
C. $2$.
D. $4$.
A. $16$.
B. $24$.
C. $2$.
D. $4$.
Xét $\int\limits_{-1}^{1}{\dfrac{f\left( 3x \right)}{1+{{2}^{x}}}\text{d}x}=8\Leftrightarrow \int\limits_{-3}^{3}{\dfrac{f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}=24$.
Đặt $u=-x\Rightarrow du=-dx$, do đó: $24=I=\int\limits_{-3}^{3}{\dfrac{f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}=-\int\limits_{3}^{-3}{\dfrac{f\left( -u \right)}{1+{{\sqrt[3]{2}}^{-u}}}\text{d}u}=\int\limits_{-3}^{3}{\dfrac{{{\sqrt[3]{2}}^{u}}f\left( u \right)}{1+{{\sqrt[3]{2}}^{u}}}\text{d}u}$.
Khi đó: $2I=\int\limits_{-3}^{3}{\dfrac{f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}+\int\limits_{-3}^{3}{\dfrac{{{\sqrt[3]{2}}^{x}}f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}=\int\limits_{-3}^{3}{f\left( x \right)\text{d}x}=2\int\limits_{0}^{3}{f\left( x \right)\text{d}x}$ nên $\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=24$.
Đặt $u=-x\Rightarrow du=-dx$, do đó: $24=I=\int\limits_{-3}^{3}{\dfrac{f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}=-\int\limits_{3}^{-3}{\dfrac{f\left( -u \right)}{1+{{\sqrt[3]{2}}^{-u}}}\text{d}u}=\int\limits_{-3}^{3}{\dfrac{{{\sqrt[3]{2}}^{u}}f\left( u \right)}{1+{{\sqrt[3]{2}}^{u}}}\text{d}u}$.
Khi đó: $2I=\int\limits_{-3}^{3}{\dfrac{f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}+\int\limits_{-3}^{3}{\dfrac{{{\sqrt[3]{2}}^{x}}f\left( x \right)}{1+{{\sqrt[3]{2}}^{x}}}\text{d}x}=\int\limits_{-3}^{3}{f\left( x \right)\text{d}x}=2\int\limits_{0}^{3}{f\left( x \right)\text{d}x}$ nên $\int\limits_{0}^{3}{f\left( x \right)\text{d}x}=24$.
Đáp án B.