Area of the region {(x, y) ∈ R² : y ≥ √|x+3|, 5y ≤ x + 9 ≤ 15} is equal to (a) 1/6 (b) 4/3 (c) 3/2 (

Q: Wait, why do we have to split the integral at $x = -3$? Can't we just integrate the whole thing at once?

That is a fantastic question! The reason is the modulus function $|x+3|$. Calculus relies on the function being continuous and differentiable in a way that allows for a single antiderivative. Because $|x+3|$ behaves differently for $x < -3$ (where it becomes $-(x+3)$) and $x \ge -3$ (where it is $x+3$), the 'formula' for the curve changes at that point. If you try to integrate across the boundary without splitting, you are essentially trying to use one rule for two different physical realities. Always respect the modulus!

Q: I solved the quadratic for the left branch and got $x = -4$ and $x = -39$. Why do we reject $x = -39$?

Great catch on the algebra! When we square both sides of an equation like $\sqrt{f(x)} = g(x)$, we introduce 'extraneous roots'—solutions that satisfy the squared equation but not the original one. If you plug $x = -39$ back into the original equation $y = \frac{x+9}{5}$, you get $y = \frac{-39+9}{5} = -6$. But the curve $y = \sqrt{|x+3|}$ can never produce a negative value. Thus, $x = -39$ is mathematically valid for the squared equation but physically impossible for our curve. Always verify your roots!

Q: How do I know which function is the 'upper' one and which is the 'lower' one for the integral?

Visualizing is your best friend here. In the region of interest, the line $y = \frac{x+9}{5}$ is physically above the curve $y = \sqrt{|x+3|}$. If you are ever unsure, pick a test point inside your interval (like $x = 0$) and calculate the $y$-value for both the line and the curve. The one with the higher $y$-value is your 'upper' function. In this problem, the line stays above the curve between $x = -4$ and $x = 1$.

Question

Area of the region $\{(x, y) \in R^2 : y \ge \sqrt{|x+3|}, 5y \le x + 9 \le 15\}$ is equal to

Options:

(A) 1/6

(B) 4/3

(C) 3/2

(D) 5/3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The region is defined by the constraints $y \ge \sqrt{|x+3|}$ and $5y \le x + 9 \le 15$.

The first constraint, $y \ge \sqrt{|x+3|}$, represents a V-shaped curve with its vertex at $(-3, 0)$. For $x \ge -3$, the curve is $y = \sqrt{x+3}$, and for $x < -3$, it is $y = \sqrt{-(x+3)}$.

The second constraint, $5y \le x + 9 \le 15$, provides two boundaries:
1. The line $y \le \frac{x+9}{5}$
2. The vertical boundary $x+9 \le 15$, which simplifies to $x \le 6$.

### The Algebra of Intersections

To find the area, we must determine the intersection points of the line $y = \frac{x+9}{5}$ with the two branches of the curve.

For the right branch ($x \ge -3$):
$$\sqrt{x+3} = \frac{x+9}{5}$$
$$25(x+3) = (x+9)^2$$
$$x^2 - 7x + 6 = 0 \Rightarrow (x-1)(x-6) = 0$$
The intersections occur at $x = 1$ and $x = 6$.

For the left branch ($x < -3$):
$$\sqrt{-(x+3)} = \frac{x+9}{5}$$
$$25(-x-3) = (x+9)^2$$
$$x^2 + 43x + 156 = 0 \Rightarrow (x+4)(x+39) = 0$$
Testing these, $x = -39$ is extraneous as it yields a negative $y$-value. Thus, the valid intersection is $x = -4$.

### The Calculus: Integrating the Region

The area $A$ is bounded between $x = -4$ and $x = 1$. Because the curve definition changes at $x = -3$, we split the integral:

$$A = \int_{-4}^{-3} \left( \frac{x+9}{5} - \sqrt{-(x+3)} ight) dx + \int_{-3}^{1} \left( \frac{x+9}{5} - \sqrt{x+3} ight) dx$$

First, we calculate the integral of the line $\frac{x+9}{5}$ from $-4$ to $1$:

$$\int_{-4}^{1} \left( \frac{x}{5} + \frac{9}{5} ight) dx = \left[ \frac{x^2}{10} + \frac{9x}{5} ight]_{-4}^{1}$$
$$= \left( \frac{1}{10} + \frac{9}{5} ight) - \left( \frac{16}{10} - \frac{36}{5} ight) = 1.9 - (-5.6) = 7.5$$

Next, we integrate the curve components:

$$\int_{-4}^{-3} \sqrt{-(x+3)} dx = \left[ -\frac{2}{3}(-(x+3))^{3/2} ight]_{-4}^{-3} = 0 - \left( -\frac{2}{3} ight) = \frac{2}{3}$$
$$\int_{-3}^{1} \sqrt{x+3} dx = \left[ \frac{2}{3}(x+3)^{3/2} ight]_{-3}^{1} = \frac{2}{3}(4)^{3/2} - 0 = \frac{16}{3}$$

### Final Calculation

The total area under the curve is the sum of these two integrals:

$$	ext{Area}_{	ext{curve}} = \frac{2}{3} + \frac{16}{3} = \frac{18}{3} = 6$$

Subtracting this from the area under the line:

$$	ext{Area} = 7.5 - 6 = 1.5$$

The final area of the region is **$\frac{3}{2}$**.

Animated Solution for Mathematics - Definite Integration: Area of the region ${(x, y) \in R^{2} : y \geq ∣ x + 3∣, 5 y \leq x + 9 \leq 15}$ is equal to

Select Answer:

Visualized Solution

Analyze the Inequalities

The Curve $y = ∣ x + 3∣$

The Line and the Boundary

Intersections for $x \geq - 3$

Solving for Right Intersections

Intersections for $x < - 3$

Solving for Left Intersections

Identifying the Bounded Region

Setting up the Integral

Splitting the Integral

Integrating the Line Part

Integrating the Curve Part

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Algebra of Intersections

The Calculus

Final Calculation

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