The problem asks us to evaluate the sum of two definite integrals: $\int_{-2}^{1} (x^2+2x+7) dx + \int_{-2}^{1} (x^2-2x-5) dx$.

1. Problem Description

The problem asks us to evaluate the sum of two definite integrals:

$\int_{-2}^{1} (x^2+2x+7) dx + \int_{-2}^{1} (x^2-2x-5) dx$.

2. Solution Steps

We can combine the two integrals into a single integral since they have the same limits of integration:

$\int_{-2}^{1} (x^2+2x+7) dx + \int_{-2}^{1} (x^2-2x-5) dx = \int_{-2}^{1} [(x^2+2x+7) + (x^2-2x-5)] dx$

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

Now, we find the antiderivative of $2x^2+2$:

$\int (2x^2+2) dx = \frac{2}{3}x^3 + 2x + C$.

Next, we evaluate the definite integral using the Fundamental Theorem of Calculus:

$\int_{-2}^{1} (2x^2+2) dx = \left[ \frac{2}{3}x^3 + 2x \right]_{-2}^{1} = \left( \frac{2}{3}(1)^3 + 2(1) \right) - \left( \frac{2}{3}(-2)^3 + 2(-2) \right)$

$= \left( \frac{2}{3} + 2 \right) - \left( \frac{2}{3}(-8) - 4 \right) = \frac{2}{3} + 2 - \left( -\frac{16}{3} - 4 \right) = \frac{2}{3} + 2 + \frac{16}{3} + 4 = \frac{18}{3} + 6 = 6+6 = 12$.

3. Final Answer

The final answer is 12.