Gọi F(x) là nguyên hàm trên \(\mathbb{R}\) của hàm số \(f\left( x \right)={{x}^{2}}{{e}^{ax}}\left( a\ne 0 \right)\), sao cho \(F\left( \frac{1}{a} \right)=F\left( 0 \right)+1.\) C...

* Hướng dẫn giải

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

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

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

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

\( \Rightarrow A = \frac{1}{a}x{e^{ax}} - \frac{1}{a}\int {{e^{ax}}dx} \,\,\left( 2 \right)\)

Từ (1) và (2) suy ra \(F\left( x \right)=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{{{a}^{2}}}x{{e}^{ax}}+\frac{2}{{{a}^{2}}}\int{{{e}^{ax}}dx}=\frac{1}{a}{{x}^{2}}{{e}^{ax}}-\frac{2}{{{a}^{2}}}x{{e}^{ax}}+\frac{2}{{{a}^{3}}}{{e}^{ax}}+C.\)

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

\(\Rightarrow {{a}^{3}}=e-2\Rightarrow a=\sqrt[3]{e-2}\Rightarrow 0<a\le 1.\)