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Ta có: $\dfrac{x^2}{4}=\dfrac{8}{x^2+4}\Leftrightarrow x^2=4 \Leftrightarrow x=\pm2$
Suy ra diện tích cần tìm là:
$S=\int\limits_{-2}^2\left|\dfrac{x^2}{4}-\dfrac{8}{x^2+4}\right|dx$
$=\int\limits_{-2}^2\dfrac{8}{x^2+4}dx-\int\limits_{-2}^2\dfrac{x^2}{4}dx$
Đặt: $x=2\tan t \Rightarrow dx=2(\tan^2t+1)dt$
Đổi cận: $x=-2\Rightarrow t=-\dfrac{\pi}{4}$
$x=2\Rightarrow t=\dfrac{\pi}{4}$
Suy ra:
$\int\limits_{-2}^2\dfrac{8}{x^2+4}dx=\int\limits_{-\pi/4}^{\pi/4}\dfrac{16(\tan^2t+1)dt}{4\tan^2t+4}$
$=\int\limits_{-\pi/4}^{\pi/4}4dt$
$=4t \left|\begin{array}{l}\dfrac{\pi}{4}\\\dfrac{-\pi}{4}\end{array}\right.=2\pi$
Mà ta có: $\int\limits_{-2}^2\dfrac{x^2}{4}dx=\dfrac{x^3}{12} \left|\begin{array}{l}2\\-2\end{array}\right.=\dfrac{4}{3}$
Suy ra: $S=2\pi-\dfrac{4}{3}$
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