Đáp án và lời giải
Đáp án:D
Lời giải:Ta có \begin{align*} f'(x) = f(x) + x^2 \cdot \mathrm{e}^x + 1 & \Leftrightarrow f'(x) = \mathrm{e}^x\left(f(x)\mathrm{e}^{-x} + x^2 + \mathrm{e}^{-x}\right)\\ & \Leftrightarrow f'(x)\mathrm{e}^{-x} = f(x)\mathrm{e}^{-x} + x^2 + \mathrm{e}^{-x}\\ & \Leftrightarrow f'(x)\mathrm{e}^{-x} - f(x)\mathrm{e}^{-x} = x^2 + \mathrm{e}^{-x}\\ & \Leftrightarrow \left( f(x)\mathrm{e}^{-x} \right)' = x^2 + \mathrm{e}^{-x} \end{align*} Suy ra \begin{align*} & \displaystyle \int_{0}^{3} \left( f(x)\mathrm{e}^{-x} \right)' \mathrm{\,d}x = \int_{0}^{3} \left(x^2 + \mathrm{e}^{-x}\right) \mathrm{\,d}x \Leftrightarrow \left( f(x)\mathrm{e}^{-x} \right) \bigg|_{0}^{3} = \left(\dfrac{x^3}{3} - \mathrm{e}^{-x}\right) \bigg|_{0}^{3} \\ & \Leftrightarrow \dfrac{f(3)}{\mathrm{e}^3} - f(0) = \left(9 - \dfrac{1}{\mathrm{e}^3}\right) - \left(0 - 1\right) \Leftrightarrow f(3) = 9\mathrm{e}^3 - 1. \end{align*}
