Chọn phương án C.

Ta có \(\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x=F(1)-F(0)\).

\(\begin{aligned}\Rightarrow F(1)&=\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x+F(0)\\
&=\displaystyle\int\limits_0^1\left(2x+\mathrm{e}^x\right)\mathrm{\,d}x+2019\\
&=\left(x^2+\mathrm{e}^x\right)\bigg|_0^1+2019\\
&=\mathrm{e}+2019.\end{aligned}\)

Chọn phương án C.

\(\begin{aligned}F(x)&=\displaystyle\int f(x)\mathrm{\,d}x=\displaystyle\int\left(2x+\mathrm{e}^x\right)\mathrm{\,d}x\\
&=x^2+\mathrm{e}^x+C.\end{aligned}\)

\(\begin{aligned}\text{Mà }F(0)=2019\Leftrightarrow&\,0^2+\mathrm{e}^0+C=2019\\
\Leftrightarrow&\,C=2018.\end{aligned}\)

Vậy \(F(x)=x^2+\mathrm{e}^x+2018\).

Suy ra \(F(1)=\mathrm{e}+2019\).