Cho hàm số $f\left( x \right)=\left\{ \begin{aligned} &...

Xét các phương trình hoành độ giao điểm:
$\begin{aligned}
& 4-{{x}^{2}}=0\Leftrightarrow \left[ \begin{aligned}
& x=2 \\
& x=-2\notin \left( 1;+\infty \right) \\
\end{aligned} \right.\Leftrightarrow x=2; \\
& 7-4{{x}^{2}}=0\Leftrightarrow x=\pm \dfrac{\sqrt{7}}{2}\notin \left[ 0;1 \right] \\
\end{aligned}$
Suy ra $\begin{aligned}
& S=\int\limits_{0}^{1}{\left| 7-4{{x}^{2}} \right|dx}+\int\limits_{1}^{2}{\left| 4-{{x}^{2}} \right|dx}+\int\limits_{2}^{3}{\left| 4-{{x}^{2}} \right|dx}=\int\limits_{0}^{1}{\left( 7-4{{x}^{2}} \right)dx}+\int\limits_{1}^{2}{\left( 4-{{x}^{2}} \right)dx}+\int\limits_{2}^{3}{\left( {{x}^{2}}-4 \right)dx} \\
& \ \ \ =\left( 7x-\dfrac{4}{3}{{x}^{3}} \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.+\left( 4x-\dfrac{{{x}^{3}}}{3} \right)\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.+\left( \dfrac{{{x}^{3}}}{3}-4x \right)\left| \begin{aligned}
& ^{3} \\
& _{2} \\
\end{aligned} \right.=7-\dfrac{4}{3}+\dfrac{16}{3}-\dfrac{11}{3}-3+\dfrac{16}{3}=10. \\
\end{aligned}$