In order to discover the area of the region bounded by the curve x = 2y + 3, the lines y = 1, y = -1, and the y-axis, we should set up the definite integral.

Step 1: Determine the x-intercepts of the curve
The curve is defined by the equation x = 2y + 3. For the y-axis, we let x = 0:
0 = 2y + 3

Solve for y:

So the curve crosses the y-axis at y = -3/2.

Step 2: Put the integral
The area is bounded by y = 1, y = -1, and the curve. We’ll have to integrate the function w.r.t to y in the range from y = -1 to y = 1 for finding the area,

A = ∫₋₁¹ (2y + 3) dy

Step 3: Evaluate the integral
Integrate the expression 2y + 3 first.

∫ (2y + 3) dy = y² + 3y

Now, substitute these values into this from y = -1 to y = 1:

A = [y² + 3y]₋₁¹

When y = 1:
 
1² + 3(1) = 1 + 3 = 4

When y = -1:
 
(-1)² + 3(-1) = 1 – 3 = -2

Thus the area is,

A = 4 – (-2) = 4 + 2 = 6

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