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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( \sqrt{2} \right)=-2$ và ${f}'\left( x \right)=\dfrac{x}{\sqrt{6-{{x}^{2}}}}{{,}^{{}}}\forall x\in \left( -\sqrt{6};\sqrt{6} \right)$. Khi đó $\int\limits_{0}^{\sqrt{3}}{f\left( x \right)}dx$ bằng
A. $-\dfrac{3\pi }{4}$.
B. $\dfrac{3\pi +6}{4}$.
C. $\dfrac{\pi +2}{4}$.
D. $-\dfrac{3\pi +6}{4}$.
A. $-\dfrac{3\pi }{4}$.
B. $\dfrac{3\pi +6}{4}$.
C. $\dfrac{\pi +2}{4}$.
D. $-\dfrac{3\pi +6}{4}$.
Ta có $f\left( x \right)=\int{{f}'\left( x \right)}dx=\int{\dfrac{x}{\sqrt{6-{{x}^{2}}}}dx}=-\dfrac{1}{2}\int{\dfrac{d\left( 6-{{x}^{2}} \right)}{\sqrt{6-{{x}^{2}}}}}=-\sqrt{6-{{x}^{2}}}+C$
$f\left( \sqrt{2} \right)=-2$ $\Rightarrow C=0$ $\Rightarrow f\left( x \right)=-\sqrt{6-{{x}^{2}}}$
$\int\limits_{0}^{\sqrt{3}}{f\left( x \right)}dx=\int\limits_{0}^{\sqrt{3}}{-\sqrt{6-{{x}^{2}}}}dx=-\dfrac{3\pi +6}{4}$.
$f\left( \sqrt{2} \right)=-2$ $\Rightarrow C=0$ $\Rightarrow f\left( x \right)=-\sqrt{6-{{x}^{2}}}$
$\int\limits_{0}^{\sqrt{3}}{f\left( x \right)}dx=\int\limits_{0}^{\sqrt{3}}{-\sqrt{6-{{x}^{2}}}}dx=-\dfrac{3\pi +6}{4}$.
Đáp án D.