giúp ạ...................

$\displaystyle\int\limits^0_{6} f(x) \, dx+\displaystyle\int\limits^6_{0} f(x) \, dx=0\\ \Rightarrow \displaystyle\int\limits^0_{6} f(x) \, dx=-12\\ u=-\dfrac{1}{3}x\\ du=-\dfrac{1}{3}dx \Rightarrow dx=-3du\\ \begin{array}{|c|c|} \hline x&-18&0\\\hline u&6&0\\\hline \end{array} \\ \displaystyle\int\limits^0_{-18} f\left(-\dfrac{1}{3}x\right) \, dx\\ =-3\displaystyle\int\limits^0_{6} f\left(u\right) \, du\\ =36$