Gọi F(x) là nguyên hàm trên $\mathbb{R}$ của hàm số $f\left( x...

Ta có $F\left( x \right)=\int{f\left( x \right)dx=\int{{{x}^{2}}{{e}^{ax}}dx.}}$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right..$
$F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}\int{x}{{e}^{ax}}dx=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}{{F}_{1}}\left( x \right)$ với ${{F}_{1}}\left( x \right)=\int{x}{{e}^{ax}}dx$.
Đặt $\left\{ \begin{aligned}
& {{u}_{1}}=x \\
& d{{v}_{1}}={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& d{{u}_{1}}=dx \\
& {{v}_{1}}=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right.. $ Ta có $ {{F}_{1}}\left( x \right)=\dfrac{1}{a}x{{e}^{ax}}-\dfrac{1}{a}\int{{{e}^{ax}}dx}=\dfrac{1}{a}x{{e}^{ax}}-\dfrac{1}{{{a}^{2}}}{{e}^{ax}}+{{C}_{1}}.$
Vậy $F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}\left( \dfrac{1}{a}x{{e}^{ax}}-\dfrac{1}{{{a}^{2}}}{{e}^{ax}}+{{C}_{1}} \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{{{a}^{2}}}x{{e}^{ax}}+\dfrac{2}{{{a}^{3}}}{{e}^{ax}}+C.$
Khi đó $F\left( \dfrac{1}{a} \right)=F\left( 0 \right)+1\Leftrightarrow \dfrac{1}{{{a}^{3}}}e-\dfrac{2}{{{a}^{3}}}e+\dfrac{2}{{{a}^{3}}}e+C=\dfrac{2}{{{a}^{3}}}+C+1$
$\Leftrightarrow \dfrac{1}{{{a}^{3}}}e=\dfrac{2}{{{a}^{3}}}+1\Leftrightarrow e=2+{{a}^{3}}\Leftrightarrow {{a}^{3}}=e-2\Leftrightarrow a=\sqrt[3]{e-2}\approx 0,896$