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: Understand the boundaries
The area we want to find is bounded by the curve \( x = 2y + 3 \) and the horizontal lines \( y = 1 \) and \( y = -1 \).

### Step 2: Set up the integral
Since we are integrating with respect to \( y \), we need to express the area as an integral from the lower limit \( y = -1 \) to the upper limit \( y = 1 \). The function \( x = 2y + 3 \) gives us the right boundary of the area.

The area \( A \) can be expressed as:
\[ A = \int_{-1}^{1} (2y + 3) \, dy \]

### Step 3: Calculate the integral
Now we will compute the integral:
\[ A = \int_{-1}^{1} (2y + 3) \, dy \]

Breaking it down:
\[ A = \int_{-1}^{1} 2y \, dy + \int_{-1}^{1} 3 \, dy \]

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

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

Now, combining both results:
\[ A = 0 + 6 = 6 \]

### Step 4: Conclusion
Thus, the area of the region bounded by the curve and the lines is:
\[ \text{Area} = 6 \text{ square units} \]

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