Area of region {(x,y): 4x²+y²≤8 and y²≤4x} equals

Ellipse: $4x^2 + y^2 = 8 \Rightarrow \dfrac{x^2}{2} + \dfrac{y^2}{8} = 1$

Semi-axes: $b = \sqrt{2}$ (along $x$), $a = 2\sqrt{2}$ (along $y$)

Parabola: $y^2 = 4x$ (opens right, vertex at origin)

Substitute $y^2 = 4x$ into ellipse equation:

$4x^2 + 4x = 8$

$x^2 + x – 2 = 0$

$(x-1)(x+2) = 0$

Since $x \geq 0$ (parabola): $x = 1,\; y = \pm 2$

Intersection points: $(1, 2)$ and $(1, -2)$

By symmetry about $x$-axis:

$$A = 2\left[\int_0^1 2\sqrt{x}\,dx + \int_1^{\sqrt{2}} \sqrt{8-4x^2}\,dx\right]$$

For $0 \leq x \leq 1$: parabola gives $|y| \leq 2\sqrt{x}$

For $1 \leq x \leq \sqrt{2}$: ellipse gives $|y| \leq \sqrt{8-4x^2}$

$$I_1 = 2\int_0^1 2\sqrt{x}\,dx = 4\cdot\left[\frac{x^{3/2}}{3/2}\right]_0^1 = 4\cdot\frac{2}{3} = \frac{8}{3}$$

$$I_2 = 2\int_1^{\sqrt{2}} \sqrt{8-4x^2}\,dx = 4\int_1^{\sqrt{2}} \sqrt{2-x^2}\,dx$$

Use formula: $\int\sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\!\dfrac{x}{a}+C$

With $a = \sqrt{2}$:

$$4\left[\frac{x}{2}\sqrt{2-x^2}+\sin^{-1}\!\frac{x}{\sqrt{2}}\right]_1^{\sqrt{2}}$$

At $x=\sqrt{2}$: $\dfrac{\sqrt{2}}{2}\cdot 0 + \sin^{-1}(1) = \dfrac{\pi}{2}$

At $x=1$: $\dfrac{1}{2}\cdot 1 + \sin^{-1}\!\dfrac{1}{\sqrt{2}} = \dfrac{1}{2}+\dfrac{\pi}{4}$

$$I_2 = 4\left[\frac{\pi}{2} – \frac{1}{2} – \frac{\pi}{4}\right] = 4\left[\frac{\pi}{4}-\frac{1}{2}\right] = \pi – 2$$

$$A = I_1 + I_2 = \frac{8}{3} + (\pi – 2)$$

$$= \pi + \frac{8}{3} – 2 = \pi + \frac{8-6}{3} = \boxed{\pi + \frac{2}{3}}$$

✓ Answer: A) π + 2/3