Area of Region Bounded by Absolute Value Curve and Linear Inequalities

To find the total area of the bounded region defined by: $R = \left\{(x, y) \in \mathbb{R}^2 : y \ge \sqrt{|x+3|}, \; 5y \le x + 9 \le 15\right\}$

we first express the region in terms of inequalities for $y$: $\sqrt{|x+3|} \le y \le \frac{x+9}{5}$ along with the condition $x + 9 \le 15 \implies x \le 6$.

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Step 1: Determine the limits for $x$

For a valid $y$ to exist, the lower bound of $y$ must be less than or equal to its upper bound: $\sqrt{|x+3|} \le \frac{x+9}{5}$

Since $\sqrt{|x+3|} \ge 0$, we must have $\frac{x+9}{5} \ge 0 \implies x \ge -9$.

We solve this inequality by considering two cases based on the expression inside the absolute value, $x+3$:

Case I: $x \ge -3$

Here, $|x+3| = x+3$, so the inequality becomes: $\sqrt{x+3} \le \frac{x+9}{5}$

Squaring both sides (as both sides are non-negative): $x + 3 \le \frac{(x+9)^2}{25}$ $25(x+3) \le x^2 + 18x + 81$ $25x + 75 \le x^2 + 18x + 81$ $x^2 - 7x + 6 \ge 0$ $(x - 1)(x - 6) \ge 0$

The solutions to this inequality are $x \le 1$ or $x \ge 6$. Combining this with $-3 \le x \le 6$, we obtain: $x \in [-3, 1] \cup \{6\}$

Case II: $x < -3$

Here, $|x+3| = -(x+3)$, so the inequality becomes: $\sqrt{-(x+3)} \le \frac{x+9}{5}$

Squaring both sides: $-25(x+3) \le (x+9)^2$ $-25x - 75 \le x^2 + 18x + 81$ $x^2 + 43x + 156 \ge 0$ $(x + 39)(x + 4) \ge 0$

The solutions are $x \le -39$ or $x \ge -4$. Combining this with $-9 \le x < -3$, we obtain: $x \in [-4, -3)$

Combined $x$-interval:

Combining Case I and Case II, the region has non-zero area over the interval: $x \in [-4, 1]$ (Note: $x = 6$ represents a single intersection point between the curves, which contributes zero area.)

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Step 2: Set up and evaluate the area integral

The total bounded area $A$ is given by: $A = \int_{-4}^{1} \left( \frac{x+9}{5} - \sqrt{|x+3|} \right) dx$

We split the integral at $x = -3$: $A = \int_{-4}^{1} \frac{x+9}{5} \, dx - \int_{-4}^{-3} \sqrt{-(x+3)} \, dx - \int_{-3}^{1} \sqrt{x+3} \, dx$

1. Linear integral:

$\int_{-4}^{1} \frac{x+9}{5} \, dx = \frac{1}{5} \left[ \frac{x^2}{2} + 9x \right]_{-4}^{1} = \frac{1}{5} \left[ \left(\frac{1}{2} + 9\right) - \left(8 - 36\right) \right] = \frac{1}{5} \left( \frac{19}{2} + 28 \right) = \frac{1}{5} \left( \frac{75}{2} \right) = \frac{15}{2}$

2. First radical integral:

$\int_{-4}^{-3} \sqrt{-x-3} \, dx$ Using the substitution $u = -x-3$, $dx = -du$: $\int_{1}^{0} \sqrt{u} (-du) = \int_{0}^{1} u^{1/2} \, du = \left[ \frac{2}{3}u^{3/2} \right]_0^1 = \frac{2}{3}$

3. Second radical integral:

$\int_{-3}^{1} \sqrt{x+3} \, dx$ Using the substitution $v = x+3$, $dx = dv$: $\int_{0}^{4} v^{1/2} \, dv = \left[ \frac{2}{3}v^{3/2} \right]_0^4 = \frac{2}{3} (4)^{3/2} = \frac{16}{3}$

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Step 3: Compute the total area

Substituting the computed integrals back into the formula for $A$: $A = \frac{15}{2} - \left( \frac{2}{3} + \frac{16}{3} \right) = \frac{15}{2} - \frac{18}{3} = \frac{15}{2} - 6 = \frac{3}{2}$

Thus, the total area of the bounded region is $\frac{3}{2}$.

Correct Option: C