The area enclosed by the curves x² + 4y² ≤ 4, y ≤ |x| − 1 and y ≥ 1 − |x| is:

Correct Answer: The correct option is 4 sin⁻¹(3/5) − 6/5.

We are required to find the area of the region bounded by the following curves:

$x^2+4y^2\le 4$

$y\le \left|x\right|-1$

$y\ge 1-\left|x\right|$

Step 1: Understand the geometric boundary conditions
1. The inequality $x^2+4y^2\le 4$ represents the region inside or on the ellipse $\frac{x^2}{2^2}+\frac{y^2}{1^2}=1$ with semi-major axis $a=2$ along the x-axis and semi-minor axis $b=1$ along the y-axis.
2. The inequality $y\ge 1-\left|x\right|$ rewrites as:
• For $x\ge 0$: $y\ge 1-x\Rightarrow x+y\ge 1$
• For $x<0$: $y\ge 1+x\Rightarrow -x+y\ge 1$
3. The inequality $y\le \left|x\right|-1$ rewrites as:
• For $x\ge 0$: $y\le x-1\Rightarrow x-y\ge 1$
• For $x<0$: $y\le -x-1\Rightarrow -x-y\ge 1$

Notice that $y\ge 1-\left|x\right|$ defines the region on or above the inverted V-shape with vertex at $(0,1)$, and $y\le \left|x\right|-1$ defines the region on or below the V-shape with vertex at $(0,-1)$.

Due to horizontal and vertical symmetry about the x-axis and y-axis, the region consists of four symmetrical segments in each of the four quadrants.

Step 2: Calculate the area in the first quadrant
In the first quadrant ($x\ge 0,y\ge 0$):
The region is bounded by the line $x+y=1$ (or $y=1-x$) below and the ellipse $x^2+4y^2=4$ above.

Let us find the point of intersection of the line $y=1-x$ and the ellipse $x^2+4y^2=4$:

$x^2+4{(1-x)}^2=4$

$x^2+4(1-2x+x^2)=4$

$5x^2-8x=0\Rightarrow x=0\text{or}x=\frac{8}{5}$

When $x=0$, $y=1$. When $x=\frac{8}{5}$, $y=1-\frac{8}{5}=-\frac{3}{5}$.

Integrating with respect to $y$ makes the calculation straightforward.
From the ellipse, $x=2\sqrt{1-y^2}$ and from the boundary line, $x=1-y$ for $y\ge 0$ and $x=1+y$ for $y<0$.
The region in the right half-plane ($x\ge 0$) extends for $y\in [0,1]$ as bounded between $x_{\text{inner}}=1-y$ and $x_{\text{outer}}=2\sqrt{1-y^2}$.

Due to four-fold symmetry, the total area $A$ is 4 times the area in one quadrant for $y\in [0,1]$:

$A=4{\int }_0^1(2\sqrt{1-y^2}-(1-y))dy$

Wait, let's double check if the line $x=1-y$ intersects the ellipse at $y=0$: at $y=0$, $x_{\text{ellipse}}=2$ and $x_{\text{line}}=1$.
So the region bounded by $y\ge 1-\left|x\right|$ and $y\le \left|x\right|-1$ inside the ellipse is split into two regions: upper ($y\ge 0$) and lower ($y\le 0$).
In the upper right region, $y$ ranges from $0$ to $1$ (or $x$ goes up to the intersection point). But since $y\ge 1-x$ and $y\ge 0$, for $y\in \left[\mathrm{0;}1\right]$, $x$ ranges from $1-y$ to $2\sqrt{1-y^2}$.

However, the lines $y=1-x$ and $y=x-1$ intersect at $(1,0)$.
The region defined by $y\ge 1-\left|x\right|$ and $y\le \left|x\right|-1$ inside the ellipse is bounded by the line $x-y=1$ and $x+y=1$ for $x\ge 0$.
Let's find the intersection of $y=1-x$ and $y=x-1$: $x=1,y=0$.
The region lying to the right of $x=1$ between and