Region A that on xy-plane is bounded by two (2) curves and a line. The curves are y=x^3-2x+3 and y=-x^2+3 while the line is x=0. It is located in the first quadrant of xy-plane. Determine the area of region A.

Make a sketch and then solve for the intersection of the two curves

x^2 - 2x + 3 = -x^2 + 3
.....
x = 0 or x = 1

so
area = [integral] (-x^2 + 3 -(x^2 - 2x+3)) dx from 0 to 1
= integral (-2x^2 + 2x)dx from 0 to 1
= (-2/3)x^3 + x^2 | from 0 to 1
= -2/3 + 1 - 0
= 1/3

To determine the area of region A, we need to find the points of intersection between the two curves and the line.

First, let's find the points of intersection between the curve y = x^3 - 2x + 3 and the line x = 0. When x = 0, the equation for the curve becomes y = 3. So one point of intersection is (0, 3).

Next, let's find the points of intersection between the curve y = -x^2 + 3 and the line x = 0. When x = 0, the equation for the curve becomes y = 3. So another point of intersection is (0, 3).

Now, we need to find the points of intersection between the two curves. To do this, we can set the two equations equal to each other:

x^3 - 2x + 3 = -x^2 + 3

Rearranging, we get:

x^3 + x^2 - 2x = 0

Factoring out an x, we get:

x(x^2 + x - 2) = 0

Setting each factor equal to zero, we have:

x = 0 (already one of the points of intersection)

x^2 + x - 2 = 0

Solving the quadratic equation, we find two more points of intersection: x = 1 and x = -2.

Now that we have the x-values of the points of intersection, we can find the corresponding y-values by substituting back into one of the original equations. Let's use the equation y = x^3 - 2x + 3.

For x = 1, y = 1^3 - 2(1) + 3 = 2.

For x = -2, y = (-2)^3 - 2(-2) + 3 = -3.

So the points of intersection are (0, 3), (1, 2), and (-2, -3).

To find the area of region A, we need to integrate the difference between the two curves with respect to x, from x = -2 to x = 1. So the area is given by:

Area = ∫(y2 - y1) dx

where y1 = -x^2 + 3 and y2 = x^3 - 2x + 3.

Calculating the integral, we have:

Area = ∫[(x^3 - 2x + 3) - (-x^2 + 3)] dx

Area = ∫(x^3 - 2x + 3 + x^2 - 3) dx

Area = ∫(x^3 + x^2 - 2x) dx

Integrating term by term, we get:

Area = (x^4/4 + x^3/3 - x^2) evaluated from x = -2 to x = 1

Area = [(1^4/4 + 1^3/3 - 1^2) - ((-2)^4/4 + (-2)^3/3 - (-2)^2)]

Simplifying, we find:

Area = [1/4 + 1/3 - 1 - (16/4 - 8/3 - 4)]

Area = [1/4 + 1/3 - 1 - (4 - 8/3 - 4)]

Area = [1/4 + 1/3 - 1 - (4 - 8/3 - 4)]

Area = 5/12

Therefore, the area of region A is 5/12 square units.