Area enclosed by x² + 4y² ≤ 4, y ≤ |x| − 1, y ≥ 1 − |x| is

To find the area of the region enclosed by the curves:
1. $x^2+4y^2\le 4$ (which represents the region inside the ellipse $\frac{x^2}{4}+y^2\le 1$)
2. $y\le \left|x\right|-1$
3. $y\ge 1-\left|x\right|$

Let us analyze the symmetry of the region. Replacing $x$ with $-x$ keeps all three inequalities unchanged, indicating symmetry about the $y$-axis.
Replacing $y$ with $-y$ swaps the second and third inequalities, which means the combined region is also symmetric about the $x$-axis.
Therefore, the total area $A$ is equal to 4 times the area of the region in the first quadrant ($x\ge 0$, $y\ge 0$).

In the first quadrant, where $x\ge 0$ and $y\ge 0$, the boundary equations become:
- Ellipse: $x^2+4y^2=4\Rightarrow y=\frac{1}{2}\sqrt{4-x^2}$
- Line 1: $y\le x-1$
- Line 2: $y\ge 1-x$

Since $y\ge 0$, the condition $y\le x-1$ requires $x\ge 1$. For $x\ge 1$, the condition $y\ge 1-x$ is automatically satisfied since $1-x\le 0$ and $y\ge 0$.
Thus, the region in the first quadrant is bounded by the $x$-axis ($y=0$), the line $y=x-1$, and the ellipse $y=\frac{1}{2}\sqrt{4-x^2}$.

First, find the point of intersection of the line $y=x-1$ and the ellipse $x^2+4y^2=4$:
$x^2+4{(x-1)}^2=4$
$x^2+4(x^2-2x+1)=4$
$5x^2-8x=0$
Since $x\ge 1$, we have:
$x=\frac{8}{5}$
Correspondingly, $y=\frac{8}{5}-1=\frac{3}{5}$.

The area of the region in the first quadrant, $A_1$, is computed by integrating with respect to $x$:
$A_1={\int }_1^{8/5}(x-1)dx+{\int }_{8/5}^2\frac{1}{2}\sqrt{4-x^2}dx$

Let us evaluate the two integrals separately:
1. For the first integral:
$I_1={\left[\frac{{(x-1)}^2}{2}\right]}_1^{8/5}=\frac{1}{2}{(\frac{8}{5}-1)}^2=\frac{1}{2}{\left(\frac{3}{5}\right)}^2=\frac{9}{50}$

2. For the second integral, let $x=2sin\theta \Rightarrow dx=2cos\theta d\theta$:
When $x=\frac{8}{5}$, $sin\theta =\frac{4}{5}$, let this lower limit be ${\theta }_0={sin}^{-1}\left(\frac{4}{5}\right)$.
When $x=2$, $sin\theta =1\Rightarrow \theta =\frac{\pi }{2}$.
Substituting these values:
$I_2={\int }_{{\theta }_0}^{\pi /2}\frac{1}{2}(2cos\theta )(2cos\theta )d\theta ={\int }_{{\theta }_0}^{\pi /2}2{cos}^2\theta d\theta$
Using the identity $2{cos}^2\theta =1+cos2\theta$:
$I_2={[\theta +\frac{sin2\theta }{2}]}_{{\theta }_0}^{\pi /2}={[\theta +sin\theta cos\theta ]}_{{\theta }_0}^{\pi /2}$
$I_2=\frac{\pi }{2}-({\theta }_0+sin{\theta }_{0}cos{\theta }_0)$
Since $sin{\theta }_0=\frac{4}{5}$, we have $cos{\theta }_0=\frac{3}{5}$. Therefore:
$I_2=\frac{\pi }{2}-{\theta }_0-\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)={cos}^{-1}\left(\frac{4}{5}\right)-\frac{12}{25}={sin}^{-1}\left(\frac{3}{5}\right)-\frac{12}{25}$

Adding the two integrals together to find the quadrant area $A_1$:
$A_1=\frac{9}{50}+{sin}^{-1}\left(\frac{3}{5}\right)-\frac{24}{50}={sin}^{-1}\left(\frac{3}{5}\right)-\frac{15}{50}={sin}^{-1}\left(\frac{3}{5}\right)-\frac{3}{10}$

Finally, multiplying by 4 to get the total enclosed area $A$:
$A=4A_1=4{sin}^{-1}\left(\frac{3}{5}\right)-\frac{6}{5}$