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

The correct option is 6 sq. units.

Step-by-step Explanation:

We need to find the area of the region bounded by the curve:
$$x=2y+3$$
and the horizontal lines:
$$y=1$$
and
$$y=-1$$

Since the curve is expressed as $x$ in terms of $y$, and the boundary lines are given as horizontal lines $y=c$ and $y=d$, we can find the area $A$ by integrating with respect to $y$ from $y=-1$ to $y=1$. Let us first verify if $x\ge 0$ in this interval:
For $y=-1$, $x=2(-1)+3=1>0$.
For $y=1$, $x=2\left(1\right)+3=5>0$.
Since $x$ is positive throughout the interval $[-1,1]$, the bounded area is given by the definite integral:
$$A={\int }_{-1}^{1}xdy$$

Substitute the equation of the curve into the integral:
$$A={\int }_{-1}^1(2y+3)dy$$

Find the antiderivative:
$$\int (2y+3)dy=y^2+3y$$

Now, evaluate the definite integral by applying the limits from $-1$ to $1$:
$$A={[y^2+3y]}_{-1}^1$$

Substitute the upper limit ($y=1$):
$$(1^2+3(1\left)\right)=1+3=4$$

Substitute the lower limit ($y=-1$):
$$({(-1)}^2+3(-1\left)\right)=1-3=-2$$

Subtract the lower limit value from the upper limit value:
$$A=4-(-2)=4+2=6$$

Thus, the area of the bounded region is 6 sq. units.