We need to find the area of the shaded region enclosed by the curves $y = 2x^2 + 2$ and $y = 2x + 6$.

1. Problem Description

We need to find the area of the shaded region enclosed by the curves $y = 2x^2 + 2$ and $y = 2x + 6$.

2. Solution Steps

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

$2x^2 + 2 = 2x + 6$

$2x^2 - 2x - 4 = 0$

$x^2 - x - 2 = 0$

$(x - 2)(x + 1) = 0$

So, the intersection points are $x = -1$ and $x = 2$.

The area of the shaded region is given by the integral of the difference between the two functions, from the lower limit to the upper limit:

$Area = \int_{-1}^{2} (2x + 6 - (2x^2 + 2)) dx$

$Area = \int_{-1}^{2} (2x + 6 - 2x^2 - 2) dx$

$Area = \int_{-1}^{2} (-2x^2 + 2x + 4) dx$

Now, we find the antiderivative:

$Area = [-\frac{2}{3}x^3 + x^2 + 4x]_{-1}^{2}$

$Area = (-\frac{2}{3}(2)^3 + (2)^2 + 4(2)) - (-\frac{2}{3}(-1)^3 + (-1)^2 + 4(-1))$

$Area = (-\frac{16}{3} + 4 + 8) - (\frac{2}{3} + 1 - 4)$

$Area = (-\frac{16}{3} + 12) - (\frac{2}{3} - 3)$

$Area = (-\frac{16}{3} + \frac{36}{3}) - (\frac{2}{3} - \frac{9}{3})$

$Area = \frac{20}{3} - (-\frac{7}{3})$

$Area = \frac{20}{3} + \frac{7}{3}$

$Area = \frac{27}{3}$

$Area = 9$

3. Final Answer

The area of the shaded region is 9.