Đặt \(2x-1=t\Rightarrow x=\dfrac{1}{2}t+\dfrac{1}{2}\Rightarrow dx=\dfrac{1}{2}dt\)
\(\left\{{}\begin{matrix}x=1\Rightarrow t=1\\x=2\Rightarrow t=3\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^3_1f\left(t\right).\left(\dfrac{1}{2}dt\right)=\dfrac{1}{2}\int\limits^3_1f\left(t\right)dt=\dfrac{1}{2}.5=\dfrac{5}{2}\)

giải giúp e câu 2,4,5,6 vs ạ.


