Ta thấy \(\left[xf\left(x\right)\right]'=xf'\left(x\right)+f\left(x\right)\) nên tích phân \(2\) vế trên \(\left[1;2\right]\)
\(\Rightarrow\int\limits^2_1\left[xf\left(x\right)\right]'dx=\int\limits^2_1\left[xf'\left(x\right)+f\left(x\right)\right]dx\)
\(\Rightarrow\int\limits^2_1\left[xf'\left(x\right)+f\left(x\right)\right]dx=\left[xf\left(x\right)\right]^2_1=\left[2.f\left(2\right)-1.f\left(1\right)\right]=2.20-4=36\)
Vậy \(\left(1\right)\) là \(36\)