Cho $\int\limits_0^1 {\left( {1 + 3x} \right)f'\left( x \right)dx}  = 2019;$ $4f\left( 1 \right) - f\left( 0 \right) = 2020.$ Tính $\int\limits_0^{\frac{1}{3}} {f\left( {3x} \right)dx.}$

Xét $\int\limits_0^1 {\left( {1 + 3x} \right)f'\left( x \right)dx}  = 2019$

Đặt $\left\{ \begin{array}{l}1 + 3x = u\\f'\left( x \right)dx = dv\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3dx = du\\f\left( x \right) = v\end{array} \right.$

Suy ra $\int\limits_0^1 {\left( {1 + 3x} \right)f'\left( x \right)dx}  = \left. {\left( {1 + 3x} \right)f\left( x \right)} \right|_0^1 - 3\int\limits_0^1 {f\left( x \right)dx}$

$= 4f\left( 1 \right) - f\left( 0 \right) - 3\int\limits_0^1 {f\left( x \right)dx}  = 2020 - 3\int\limits_0^1 {f\left( x \right)dx}  = 2019$

$\Leftrightarrow \int\limits_0^1 {f\left( x \right)dx}  = \dfrac{1}{3}$

Xét $\int\limits_0^{\dfrac{1}{3}} {f\left( {3x} \right)dx}$, đặt $3x = t \Leftrightarrow 3dx = dt \Leftrightarrow dx = \dfrac{{dt}}{3}$.

Đổi cận : $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{1}{3} \Rightarrow t = 1\end{array} \right.$.

Suy ra $\int\limits_0^{\dfrac{1}{3}} {f\left( {3x} \right)dx}  = \dfrac{1}{3}\int\limits_0^1 {f\left( t \right)dt}  = \dfrac{1}{3}.\int\limits_0^1 {f\left( x \right)dx}  = \dfrac{1}{3}.\dfrac{1}{3} = \dfrac{1}{9}$