The area of the region satisfying the inequalities $|\mathrm{x}|-\mathrm{y} \leq 1, \mathrm{y} \geq 0$ and $\mathrm{y} \leq 1$ is

We need to find the area of the region where all three conditions are satisfied:

$|x| - y \leq 1$, $y \geq 0$, and $y \leq 1$

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Rewrite the first inequality as $y \geq |x| - 1$.

Since $|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0$:

• When $x \geq 0$: $y \geq x - 1$
• When $x < 0$: $y \geq -x - 1$

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Find the key intersection points by solving where boundary lines meet:

Where $y = x - 1$ meets $y = 0$: $0 = x - 1$, so $x = 1$

Point: $(1, 0)$

Where $y = -x - 1$ meets $y = 0$: $0 = -x - 1$, so $x = -1$

Point: $(-1, 0)$

Where $y = x - 1$ meets $y = 1$: $1 = x - 1$, so $x = 2$

Point: $(2, 1)$

Where $y = -x - 1$ meets $y = 1$: $1 = -x - 1$, so $x = -2$

Point: $(-2, 1)$

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From $y \geq |x| - 1$, we get $|x| \leq y + 1$

This means $-(y + 1) \leq x \leq (y + 1)$

For any $y$ between $0$ and $1$, the width of the region is $2(y + 1)$

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The region forms a trapezoid:

At $y = 0$: width $= 2(0 + 1) = 2$ (from $x = -1$ to $x = 1$)

At $y = 1$: width $= 2(1 + 1) = 4$ (from $x = -2$ to $x = 2$)

Height $= 1 - 0 = 1$

Area of trapezoid:

$\text{Area} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

$\text{Area} = \dfrac{1}{2} \times (2 + 4) \times 1 = \dfrac{1}{2} \times 6 \times 1 = 3$

The area of the region is $3$ square units.