$$\begin{array}{l}
\left\{ \begin{array}{l}
f'\left( x \right) = 6a{x^2} + 2b\\
f''\left( x \right) = 12ax\\
f'''\left( x \right) = 12a
\end{array} \right.\\
g'\left( x \right) = f'\left( x \right) + f\left( x \right) + f'''\left( x \right)\\
g\left( x \right) = f\left( x \right) + f'\left( x \right) + f''\left( x \right)\\
 \Rightarrow g'\left( x \right) - g\left( x \right) = 12a - f\left( x \right)\\
\int\limits_{ - 3}^{ - 2} {\dfrac{{12a - f\left( x \right)}}{{2{e^x}}}} dx = \int\limits_{ - 3}^{ - 2} {\dfrac{{g'\left( x \right) - g\left( x \right)}}{{2{e^x}}}dx} \\
 = \int\limits_{ - 3}^{ - 2} {\dfrac{{{e^x}g'\left( x \right) - {e^x}g\left( x \right)}}{{2{e^{2x}}}}dx} \\
 = \dfrac{1}{2}\int\limits_{ - 3}^2 {\left( {\dfrac{{g\left( x \right)}}{{{e^x}}}} \right)'dx} \\
 = \left. {\dfrac{1}{2}\dfrac{{g\left( x \right)}}{{{e^x}}}} \right|_{ - 3}^{ - 2} = \dfrac{1}{2}\dfrac{{g\left( { - 2} \right)}}{{{e^{ - 2}}}} - \dfrac{1}{2}\dfrac{{g\left( { - 3} \right)}}{{{e^{ - 3}}}}\\
 = \dfrac{{{e^2}\left( {1 + e} \right)}}{2}
\end{array}$$

Đáp án:

Giải thích các bước giải:

Dễ tính ra $$: f'''(x) = 12a$$

Nên ta có $$ : g'(x) - g(x) $$

$$ = [f'(x) + f"(x) + f"'(x)] - [f(x) + f'(x) + f"(x)]$$

$$ = 12a - f(x)$$

$$ ⇒ \dfrac{12a - f(x)}{2e^{x}} = \dfrac{g'(x) - g(x)}{2e^{x}} $$

$$ = \dfrac{g'(x).e^{x} - (e^{x})'.g(x)}{2(e^{x})²} = \dfrac{1}{2}(\dfrac{g(x)}{e^{x}})'$$

$$ ⇒ I = \dfrac{1}{2}.\dfrac{g(x)}{e^{x}}|^{-2}_{-3}$$
$$ = \dfrac{1}{2}[\dfrac{g(- 2)}{e^{- 2}} - \dfrac{g(- 3)}{e^{-3}}] = \dfrac{e²(1 + e)}{2}$$