Gọi F(x) là nguyên hàm trên \(\mathbb{R}\) của hs \(f\left( x \right)={{x}^{2}}{{e}^{ax}}\left( a\ne 0 \right),\) sao cho \(F\left( \fr

* Hướng dẫn giải

Ta có \(F\left( x \right)=\int{f\left( x \right)dx=\int{{{x}^{2}}{{e}^{ax}}dx.}}\)

Đặt \(\left\{ \begin{array}{l} u = {x^2}\\ dv = {e^{ax}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = 2xdx\\ v = \frac{1}{a}{e^{ax}} \end{array} \right..\)

\(F\left( x \right)=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{a}\int{x}{{e}^{ax}}dx=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{a}{{F}_{1}}\left( x \right)\) với \({{F}_{1}}\left( x \right)=\int{x}{{e}^{ax}}dx\).

Đặt \(\left\{ \begin{array}{l} {u_1} = x\\ d{v_1} = {e^{ax}}dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} d{u_1} = dx\\ {v_1} = \frac{1}{a}{e^{ax}} \end{array} \right..\)

Ta có \({{F}_{1}}\left( x \right)=\frac{1}{a}x{{e}^{ax}}-\frac{1}{a}\int{{{e}^{ax}}dx}=\frac{1}{a}x{{e}^{ax}}-\frac{1}{{{a}^{2}}}{{e}^{ax}}+{{C}_{1}}.\)

Vậy \(F\left( x \right)=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{a}\left( \frac{1}{a}x{{e}^{ax}}-\frac{1}{{{a}^{2}}}{{e}^{ax}}+{{C}_{1}} \right)=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{{{a}^{2}}}x{{e}^{ax}}+\frac{2}{{{a}^{3}}}{{e}^{ax}}+C.\)

Khi đó \(F\left( \frac{1}{a} \right)=F\left( 0 \right)+1\Leftrightarrow \frac{1}{{{a}^{3}}}e-\frac{2}{{{a}^{3}}}e+\frac{2}{{{a}^{3}}}e+C=\frac{2}{{{a}^{3}}}+C+1\)

\(\Leftrightarrow \frac{1}{{{a}^{3}}}e=\frac{2}{{{a}^{3}}}+1\Leftrightarrow e=2+{{a}^{3}}\Leftrightarrow {{a}^{3}}=e-2\Leftrightarrow a=\sqrt[3]{e-2}\approx 0,896\)