Đáp án:
$\begin{array}{l}
Dkxd:12 - 3{x^2} \ge 0\\
\Rightarrow {x^2} \le 4\\
\Rightarrow - 2 \le x \le 2\\
\sqrt {12 - 3{x^2}} .\sin \left( {2x + \dfrac{\pi }{4}} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
3{x^2} = 12\\
\sin \left( {2x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 4\\
2x + \dfrac{\pi }{4} = k\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 2\left( {tm} \right)\\
x = \dfrac{{ - \pi }}{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
Do: - 2 \le x \le 2\\
\Rightarrow - 2 \le \dfrac{{ - \pi }}{8} + \dfrac{{k\pi }}{2} \le 2\\
\Rightarrow \dfrac{{ - 16 + \pi }}{8} \le \dfrac{{k\pi }}{2} \le \dfrac{{16 + \pi }}{8}\\
\Rightarrow \dfrac{{ - 16 + \pi }}{{4\pi }} \le k \le \dfrac{{16 + \pi }}{{4\pi }}\\
\Rightarrow - 1,02 \le k \le 1,52\\
\Rightarrow \left[ \begin{array}{l}
k = - 1\\
k = 0\\
k = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5\pi }}{8}\\
x = \dfrac{{ - \pi }}{8}\\
x = \dfrac{{3\pi }}{8}
\end{array} \right.\\
\Rightarrow S = \left( { - 2} \right) + 2 + \dfrac{{ - 5\pi }}{8} + \dfrac{{ - \pi }}{8} + \dfrac{{3\pi }}{8} = \dfrac{{ - 3\pi }}{8}
\end{array}$
