toán 12 chắc ko ai biết =))

$\begin{array}{l}
\sqrt x  = t \Rightarrow x = {t^2} \Rightarrow dx = 2tdt\\
x = 0 \Rightarrow t = 0,x = 4 \Rightarrow t = 2\\
I = \int\limits_0^2 {f'\left( t \right).2tdt}  = 2\int\limits_0^2 {f'\left( t \right).tdt}  = 2{I_1}\\
\left\{ \begin{array}{l}
t = u\\
f'\left( t \right)dt = dv
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
dt = du\\
f\left( t \right) = v
\end{array} \right.\\
{I_1} = \left. {tf\left( t \right)} \right|_0^2 - \int\limits_0^2 {f\left( t \right)dt}  = 2f\left( 2 \right) - 3 = 4 - 3 = 1\\
 \Rightarrow I = 2.1 = 2
\end{array}$