Cho $F\left( x \right)=\left( a{{x}^{2}}+bx-c \right){{e}^{2x}}$ là một nguyên hàm của hàm số $f\left(x \right)=\left( 2020{{x}^{2}}+2022x-1...

Xét $F\left( x \right)=\int\limits_{{}}^{{}}{\left( 2020{{x}^{2}}+2022x-1 \right){{e}^{2x}}dx}$.
Đặt $\left\{ \begin{aligned}
& u=2020{{x}^{2}}+2022x-1 \\
& dv={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 4040x+2022 \right)dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Do đó $F\left( x \right)=\dfrac{1}{2}\left( 2020{{x}^{2}}+2022x-1 \right){{e}^{2x}}-\dfrac{1}{2}\int\limits_{{}}^{{}}{\left( 4040x+2022 \right){{e}^{2x}}dx}+C.$
Đặt $I=\int\limits_{{}}^{{}}{\left( 4040x+2022 \right){{e}^{2x}}dx}$.
Đặt $\left\{ \begin{aligned}
& {{u}_{1}}=4040x+2022 \\
& d{{v}_{1}}={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& d{{u}_{1}}=4040dx \\
& {{v}_{1}}=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
Do đó
$I=\dfrac{1}{2}\left( 4040x+2022 \right){{e}^{2x}}-2020\int\limits_{{}}^{{}}{{{e}^{2x}}dx}=\dfrac{1}{2}\left( 4040x+2022 \right){{e}^{2x}}-1010{{e}^{2x}}=\left( 2020x+1 \right){{e}^{2x}}.$
$F\left( x \right)=\dfrac{1}{2}\left( 2020{{e}^{2x}}+2022x-1 \right){{e}^{2x}}-\dfrac{1}{2}\left( 2020x+1 \right){{e}^{2x}}+C$
$=1010{{x}^{2}}{{e}^{2x}}+1011x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}-1010x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+C$
$=1010{{x}^{2}}{{e}^{2x}}+x{{e}^{2x}}-{{e}^{2x}}+C$
$=\left( 1010{{x}^{2}}+x-1 \right){{e}^{2x}}+C.$
Theo đề bài, ta có $a=1010,b=1,c=1,C=0.$ Vậy $T=1010-2+4=1012.$