The area of the region bounded by the curve `x=2y+3` and the lines `y=1, y=-1` is

Options

A. `4" sq units"`
B. `3/2" sq units"`
C. `6" sq units"`
D. `8" sq unit"`

Explanation

To find the area of the region bounded by the curve \( x = 2y + 3 \) and the lines \( y = 1 \) and \( y = -1 \), we can follow these steps:

### Step 1: Identify the curve and lines
The curve is given by the equation \( x = 2y + 3 \). The lines that bound the area are \( y = 1 \) and \( y = -1 \).

### Step 2: Find the points of intersection
To find the area, we first need to determine the points where the curve intersects the lines \( y = 1 \) and \( y = -1 \).

1. For \( y = 1 \):
\[ x = 2(1) + 3 = 2 + 3 = 5 \]
So, the point is \( (5, 1) \).

2. For \( y = -1 \):
\[ x = 2(-1) + 3 = -2 + 3 = 1 \]
So, the point is \( (1, -1) \).

### Step 3: Set up the integral
The area \( A \) can be calculated using the integral with respect to \( y \) from \( y = -1 \) to \( y = 1 \):
\[ A = \int_{-1}^{1} (2y + 3) \, dy \]

### Step 4: Evaluate the integral
Now, we can evaluate the integral:
\[ A = \int_{-1}^{1} (2y + 3) \, dy = \int_{-1}^{1} 2y \, dy + \int_{-1}^{1} 3 \, dy \]

Calculating each part:
1. For \( \int_{-1}^{1} 2y \, dy \):
\[ \int 2y \, dy = y^2 \quad \text{(evaluating from -1 to 1)} \]
\[ = [1^2 - (-1)^2] = [1 - 1] = 0 \]

2. For \( \int_{-1}^{1} 3 \, dy \):
\[ \int 3 \, dy = 3y \quad \text{(evaluating from -1 to 1)} \]
\[ = [3(1) - 3(-1)] = [3 + 3] = 6 \]

### Step 5: Combine the results
Now, combining both parts:
\[ A = 0 + 6 = 6 \]

### Final Answer
The area of the region bounded by the curve \( x = 2y + 3 \) and the lines \( y = 1 \) and \( y = -1 \) is \( 6 \) square units.

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