Gọi F(x) là một nguyên hàm trên R của hàm f(x)= x^2.e^(ax)

* Hướng dẫn giải

Đáp án D

Ta có $F\left(x\right)=\int x^2{e}^{\alpha x}\text{d}x=\frac{1}{\alpha }\int x^2\text{d}\left(e^{\alpha x}\right)=\frac{1}{\alpha }\left[x^2{e}^{\alpha x}-\int e^{\alpha x}2x\text{d}x\right]$

$=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }\int x\text{d}\left(e^{\alpha x}\right)\right)=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }\left(x{e}^{\alpha x}-\int e^{\alpha x}\text{d}x\right)\right)$

$=\frac{1}{\alpha }\left(x^2{e}^{\alpha x}-\frac{2}{\alpha }x{e}^{\alpha x}+\frac{2}{{\alpha }^2}{e}^{\alpha x}\right)+C$

$F\left(\frac{1}{\alpha }\right)=\frac{1}{\alpha }\left(\frac{1}{{\alpha }^2}e-\frac{2}{{\alpha }^2}e+\frac{2}{{\alpha }^2}e\right)+C=\frac{1}{{\alpha }^3}e+C$

$F\left(0\right)=\frac{1}{\alpha }.\frac{2}{{\alpha }^2}+C$

Theo giả thiết $F\left(\frac{1}{\alpha }\right)-F\left(0\right)=\frac{e-2}{{\alpha }^3}=1\Rightarrow {\alpha }^3=e-2\Rightarrow \alpha =\sqrt[3]{e-2}\Rightarrow 0<\alpha \le 1$