determine the area enclosed by the given boundaries: y=10-x^2and y=x^2+2.

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The two parabolas should be easy to graph to see what you are finding.

You will first need to find their intersection,

x^2 + 2 = 10 - x^2
2x^2 = 8
x = ± 2

The height of the region is (10-x^2) - (x^2+2)
= 8 - 2x^2

Because of the symmetry we will find the area from 0 to 2 and then double it.

Area = 2[integral] ( 8-2x^2) dx from 0 to 2
= 2 (8x - 2x^3/3)| 0 to 2
= 2[16 - 16/3 - 0]
= 64/3

To determine the area enclosed by the given boundaries y = 10 - x^2 and y = x^2 + 2, we need to find the points of intersection between the two curves.

First, let's set the two equations equal to each other to find the x-values of the points of intersection:

10 - x^2 = x^2 + 2

Rearranging this equation, we get:

2x^2 + x - 8 = 0

Now, we can solve this quadratic equation for x by factoring or using the quadratic formula. Factoring the equation, we have:

(2x - 4)(x + 2) = 0

Setting each factor equal to zero, we have:

2x - 4 = 0 or x + 2 = 0

Solving these equations, we find two x-values:

x = 2/2 = 1 or x = -2

Now, substitute these x-values back into either of the original equations to find the corresponding y-values:

For x = 1, using y = 10 - x^2:
y = 10 - (1)^2 = 10 - 1 = 9
So, one point of intersection is (1, 9).

For x = -2, using y = 10 - x^2:
y = 10 - (-2)^2 = 10 - 4 = 6
So, the second point of intersection is (-2, 6).

To determine the area enclosed, we need to find the definite integral of the curves between these two points of intersection.

The integral is given by:

integral (from x = -2 to x = 1) of (y_2 - y_1) dx

Where y_2 is the upper curve (y = 10 - x^2) and y_1 is the lower curve (y = x^2 + 2).

So, the area enclosed is:

integral (from x = -2 to x = 1) of ((10 - x^2) - (x^2 + 2)) dx

Simplifying this expression, we have:

integral (from x = -2 to x = 1) of (10 - 2x^2 - 2) dx

Now, we can integrate each term separately:

integral (from x = -2 to x = 1) of 10 dx = 10x | from x = -2 to x = 1 = (10 * 1) - (10 * -2) = 10 + 20 = 30

integral (from x = -2 to x = 1) of -2x^2 dx = (-2/3) * x^3 | from x = -2 to x = 1 = (-2/3) * (1^3 - (-2)^3) = (-2/3) * (1 + 8) = (-2/3) * 9 = -6

Lastly, we subtract the two integrals:

Area = 30 - (-6) = 30 + 6 = 36 square units

Therefore, the area enclosed by the given boundaries y = 10 - x^2 and y = x^2 + 2 is 36 square units.

To determine the area enclosed by the given boundaries, we first need to find the points of intersection between the two curves. Let's set the two equations equal to each other and solve for x:

10 - x^2 = x^2 + 2

Combining like terms:

2x^2 = 8

Dividing both sides by 2:

x^2 = 4

Taking the square root of both sides:

x = ±2

So, the curves intersect at x = -2 and x = 2.

Now, let's integrate to find the area enclosed between the curves. We need to choose which curve is the upper bound and which is the lower bound. Looking at the equations, we can see that y = 10 - x^2 is above y = x^2 + 2 for values of x between -2 and 2. Therefore, we will use y = 10 - x^2 as the upper bound and y = x^2 + 2 as the lower bound.

The area between two curves can be found by taking the integral of the difference between the upper curve and the lower curve with respect to x over the interval where they intersect:

Area = ∫[a, b] (upper curve - lower curve) dx

In this case, the boundaries of integration are -2 and 2:

Area = ∫[-2, 2] (10 - x^2) - (x^2 + 2) dx

Simplifying the integrand:

Area = ∫[-2, 2] (10 - 2x^2 - 2) dx

Area = ∫[-2, 2] (8 - 2x^2) dx

Now, we can find the antiderivative of the integrand and evaluate it at the upper and lower limits of integration:

Area = [8x - (2/3)x^3] evaluated from -2 to 2

Substituting the upper limit into the antiderivative expression:

Area = [8(2) - (2/3)(2)^3] - [8(-2) - (2/3)(-2)^3]

Simplifying:

Area = [16 - (2/3)(8)] - [-16 - (2/3)(-8)]

Area = [16 - 16/3] - [-16 + 16/3]

Area = (48/3 - 16/3) - (-48/3 + 16/3)

Area = 32/3 - (-32/3)

Area = 64/3

Therefore, the area enclosed by the given boundaries is 64/3 square units.