\(F'\left(x\right)=\left(x^2\right)'.cos\sqrt{x^2}=2x.cosx\)
43.
\(F'\left(x\right)=x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
- Với \(x=0\Rightarrow F\left(x\right)=\int\limits^0_1\left(t^2+t\right)dt=-\frac{5}{6}\)
- Với \(x=-1\Rightarrow F\left(x\right)=\int\limits^{-1}_1\left(t^2+t\right)dt=-\frac{2}{3}\)
- Với \(x=1\Rightarrow F\left(x\right)=\int\limits^1_1\left(t^2+t\right)dt=0\)
Vậy \(F\left(x\right)_{min}=-\frac{5}{6}\) khi \(x=0\)