Đáp án đúng: D
Giải chi tiết:Ta có: \(\int{\frac{-x}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{d\left( 4-{{x}^{2}} \right)}{2\sqrt{4-{{x}^{2}}}}}=\sqrt{4-{{x}^{2}}}+{{C}_{1}}\), \(\int{\frac{-\sqrt{3}}{3}xdx}=-\frac{\sqrt{3}}{6}{{x}^{2}}+{{C}_{2}}\)
\(f'(x) = \left\{ \begin{array}{l}\frac{{ - x}}{{\sqrt {4 - {x^2}} }},\,\,0 \le x \le 1\\\frac{{ - \sqrt 3 }}{3}x\,\,\,\,,x > 1\end{array} \right. \Rightarrow f(x) = \left\{ \begin{array}{l}\sqrt {4 - {x^2}} + {C_1}\,\,\,\,,\,\,0 \le x \le 1\\ - \frac{{\sqrt 3 }}{6}{x^2} + {C_2}\,\,\,\,,x > 1\end{array} \right.\)
\(f(1) = \sqrt 3 \Rightarrow \left\{ \begin{array}{l}\sqrt 3 + {C_1} = \sqrt 3 \\ - \frac{{\sqrt 3 }}{6} + {C_2} = \sqrt 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{C_1} = 0\\{C_2} = \frac{{7\sqrt 3 }}{6}\end{array} \right.\)
\( \Rightarrow f(x) = \left\{ \begin{array}{l}\sqrt {4 - {x^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,0 \le x \le 1\\ - \frac{{\sqrt 3 }}{6}{x^2} + \frac{{7\sqrt 3 }}{6}\,\,\,\,,x > 1\end{array} \right.\)
\(I=\int\limits_{0}^{2}{f(x)dx}=\int\limits_{0}^{1}{\sqrt{4-{{x}^{2}}}dx}+\int\limits_{1}^{2}{\left( -\frac{\sqrt{3}}{3}{{x}^{2}}+\frac{7\sqrt{3}}{6} \right)dx}\)
+) Tính \({{I}_{1}}=\int\limits_{0}^{1}{\sqrt{4-{{x}^{2}}}dx}\)
Đặt \(x=2\sin t,\,\,t\in \left[ -\frac{\pi }{2};\frac{\pi }{2} \right]\), \(dx=2\cos tdt\), đổi cận: \(x=0\to t=0,\,\,x=1\to t=\frac{\pi }{6}\)
\(\begin{align} & {{I}_{1}}=\int\limits_{0}^{1}{\sqrt{4-{{x}^{2}}}dx}=\int\limits_{0}^{\frac{\pi }{6}}{\sqrt{4-4{{\sin }^{2}}t}.2\cos tdt=}4\int\limits_{0}^{\frac{\pi }{6}}{\left| \cos t \right|.\cos tdt=}4\int\limits_{0}^{\frac{\pi }{6}}{{{\cos }^{2}}tdt}=2\int\limits_{0}^{\frac{\pi }{6}}{\left( 1+\cos 2t \right)dt}=\left. 2t \right|_{0}^{\frac{\pi }{6}}+\left. \sin 2t \right|_{0}^{\frac{\pi }{6}} \\ & =\left( \frac{\pi }{3}-0 \right)+\left( \frac{\sqrt{3}}{2}-0 \right)=\frac{\pi }{3}+\frac{\sqrt{3}}{2} \\\end{align}\)
+) Tính \({{I}_{2}}=\int\limits_{1}^{2}{\left( -\frac{\sqrt{3}}{3}{{x}^{2}}+\frac{7\sqrt{3}}{6} \right)dx}=\left. \left( -\frac{\sqrt{3}}{9}{{x}^{3}}+\frac{7\sqrt{3}}{6}x \right) \right|_{1}^{2}=\left( -\frac{8\sqrt{3}}{9}+\frac{7\sqrt{3}}{3} \right)-\left( -\frac{\sqrt{3}}{9}+\frac{7\sqrt{3}}{6} \right)=\frac{7\sqrt{3}}{18}\)
\(\Rightarrow I={{I}_{1}}+{{I}_{2}}=\frac{\pi }{3}+\frac{\sqrt{3}}{2}+\frac{7\sqrt{3}}{18}=\frac{\pi }{3}+\frac{8\sqrt{3}}{9}\)
Chọn: D
