Cho hàm số y = f(x) liên tục trên $ℝ$ và $\underset{-1}{\overset{1}{\int }}\frac{f\left(3x\right)}{1+2^x}\text{d}x=8$ . Tính $\underset{0}{\overset{3}{\int }}f\left(x\right)\text{d}x$ .

Đáp án đúng là: B

Xét $\underset{-1}{\overset{1}{\int }}\frac{f\left(3x\right)}{1+2^x}\text{d}x=8\iff \underset{-3}{\overset{3}{\int }}\frac{f\left(x\right)}{1+{\sqrt[3]{2}}^x}\text{d}x=24$ .

Đặt $u=-x\Rightarrow du=-dx$ , do đó: $24=I=\underset{-3}{\overset{3}{\int }}\frac{f\left(x\right)}{1+{\sqrt[3]{2}}^x}\text{d}x=-\underset{3}{\overset{-3}{\int }}\frac{f\left(-u\right)}{1+{\sqrt[3]{2}}^{-u}}\text{d}u=\underset{-3}{\overset{3}{\int }}\frac{{\sqrt[3]{2}}^{u}f\left(u\right)}{1+{\sqrt[3]{2}}^u}\text{d}u$

Khi đó:  $2I=\underset{-3}{\overset{3}{\int }}\frac{f\left(x\right)}{1+{\sqrt[3]{2}}^x}\text{d}x+\underset{-3}{\overset{3}{\int }}\frac{{\sqrt[3]{2}}^{x}f\left(x\right)}{1+{\sqrt[3]{2}}^x}\text{d}x=\underset{-3}{\overset{3}{\int }}f\left(x\right)\text{d}x=2\underset{0}{\overset{3}{\int }}f\left(x\right)\text{d}x$ nên $\underset{0}{\overset{3}{\int }}f\left(x\right)\text{d}x=24$ .