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The function is $y = 2 + |x+1|$. We know that $|x+1|$ is a V-shaped graph with the vertex at $x = -1$. Adding 2 shifts the graph upwards by 2 units. Therefore, the vertex of the given graph is at $(-1, 2)$.
We can define the function as follows: $y = 2 + (x+1)$ for $x \ge -1$, which simplifies to $y = x + 3$. $y = 2 - (x+1)$ for $x < -1$, which simplifies to $y = -x + 1$.
We need to find the area bounded by the curve, $x = -4$, $x = 3$, and $y = 0$. Since $y$ is always positive, the area is given by the integral of $y$ with respect to $x$ from $-4$ to $3$. We need to split the integral at $x = -1$ because the function is defined differently on either side of this point. Area $ = \int_{-4}^{3} (2 + |x+1|) dx = \int_{-4}^{-1} (-x + 1) dx + \int_{-1}^{3} (x + 3) dx$
$\int_{-4}^{-1} (-x + 1) dx = [-\frac{x^2}{2} + x]_{-4}^{-1} = (-\frac{(-1)^2}{2} + (-1)) - (-\frac{(-4)^2}{2} + (-4)) = (-\frac{1}{2} - 1) - (-\frac{16}{2} - 4) = -\frac{3}{2} - (-8 - 4) = -\frac{3}{2} + 12 = \frac{21}{2}$
$\int_{-1}^{3} (x + 3) dx = [\frac{x^2}{2} + 3x]_{-1}^{3} = (\frac{3^2}{2} + 3(3)) - (\frac{(-1)^2}{2} + 3(-1)) = (\frac{9}{2} + 9) - (\frac{1}{2} - 3) = \frac{27}{2} - (-\frac{5}{2}) = \frac{27}{2} + \frac{5}{2} = \frac{32}{2} = 16$
Area $ = \frac{21}{2} + 16 = \frac{21}{2} + \frac{32}{2} = \frac{53}{2}$
Final Answer: $\frac{53}{2}$
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