سوال ۲۸ کنکور ارشد مهندسی کامپیوتر ۱۴۰۵

The two curves are the parabola $y=\dfrac{x^2}{4}$ (from $4y=x^2$) and the witch of Agnesi $y=\dfrac{8}{x^2+4}$ (peak value $2$ at $x=0$). We want the area they enclose on $[-2\sqrt3,\,2\sqrt3]$.

1. Intersection points. Set $\dfrac{x^2}{4}=\dfrac{8}{x^2+4}$. With $u=x^2$: $u(u+4)=32\;\Rightarrow\;u^2+4u-32=0\;\Rightarrow\;u=4\ (u=-8\text{ rejected}),$ so $x=\pm 2$.

2. Which curve is on top. At $x=0$ the parabola is $0$ and the curve is $2$, so on $(-2,2)$ the curve $\dfrac{8}{x^2+4}$ lies above the parabola; for $2<|x|\le 2\sqrt3$ the parabola is on top. The top/bottom curves swap at $x=\pm2$.

3. Antiderivatives. $\int\frac{8}{x^2+4}\,dx = 4\arctan\frac{x}{2},\qquad \int\frac{x^2}{4}\,dx=\frac{x^3}{12}.$

4. Integrate (both curves are even, so double the right half). $\int_0^2\!\left(\frac{8}{x^2+4}-\frac{x^2}{4}\right)dx = \big[4\arctan\tfrac{x}{2}-\tfrac{x^3}{12}\big]_0^2 = \pi-\frac{2}{3},$ $\int_2^{2\sqrt3}\!\left(\frac{x^2}{4}-\frac{8}{x^2+4}\right)dx = \left(2\sqrt3-\frac{2}{3}\right)-\frac{\pi}{3},$ using $\arctan 1=\tfrac{\pi}{4}$ and $\arctan\sqrt3=\tfrac{\pi}{3}$. Their sum is $\dfrac{2\pi}{3}+2\sqrt3-\dfrac{4}{3}$. Doubling: $A = \frac{4\pi}{3}+4\sqrt3-\frac{8}{3}=\frac{4\pi}{3}+4\left(\sqrt3-\frac{2}{3}\right).$

This is option 2.

Exam trick: The endpoints $\pm 2\sqrt3$ are a giveaway — $\arctan\sqrt3=\tfrac{\pi}{3}$ — and both curves are even, so integrate on $[0,2\sqrt3]$ and double. Don't forget the region splits at the intersections $x=\pm2$, where the upper and lower curves exchange.