Cho đồ thị hàm số

A

- Hướng dẫn giải

Đáp án A

Xét phương trình tương giao:

$$\begin{gathered} x^3=\sqrt{2-x^2}, 0<x<\sqrt{2} \\ \iff x^6=2-x^2\iff x^2=1 \iff x=\pm 1\Rightarrow x=1 \\ S=2S_{OIA} \\ {S}_{OIA}={\int }_0^1{x}^{3}dx+{\int }_1^{\sqrt{2}}\sqrt{2-x^2}dx=\frac{x^4}{4}\overline{)\begin{array}{c}1\\ 0\end{array}+I} \\ I: x=\sqrt{2}\sin t\Rightarrow dx=\sqrt{2}\cos tdt, t\in \left(-\frac{\mathrm{\pi }}{2};\frac{\mathrm{\pi }}{2}\right) \\ I={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}2{\cos }^{2}tdt={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}(1+\cos 2t)dt=(t+\frac{\sin 2t}{2})\overline{)\begin{array}{c}\frac{\mathrm{\pi }}{2}\\ \frac{\mathrm{\pi }}{4}\end{array}=\frac{\mathrm{\pi }}{4}=\frac{1}{2}} \\ \Rightarrow S_{OAI}=\frac{\mathrm{\pi }}{4}-\frac{1}{4}\Rightarrow S=\frac{\mathrm{\pi }}{2}-\frac{1}{2} \end{gathered}$$