Cho hàm số \(f(x)\) có \(f'(x)=\left\{ \begin{align} & \frac{-x}{\sqrt{4-{{x}^{2}}}},\,\,0\le x\le 1 \\ & \frac{-\sqrt{3}}{3}x\,\,\,\,,x>1 \\\end{align} \right.\) và \(f(1)=\sqrt{3}\). Khi đó, kết quả \(\int\limits_{0}^{2}{f(x)dx}\) là:

A. \(\frac{\pi }{3}+\frac{23\sqrt{3}}{18}\).

B. \(\frac{\pi }{2}+\frac{7\sqrt{3}}{6}\).

C. \(\frac{\pi }{4}+\frac{4\sqrt{3}}{3}\).

D. \(\frac{\pi }{3}+\frac{8\sqrt{3}}{9}\).

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

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

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