To sketch the graph of each function, we can plot a few points and connect them smoothly.

For f(x) = x^2 + 3, we can start by finding the vertex. Since the coefficient of x^2 is positive, the parabola opens upward, and the vertex represents the minimum point. The x-coordinate of the vertex can be found by using the formula x = -b/(2a), where a and b are the coefficients of x^2 and x, respectively. In this case, a = 1 and b = 0, so x = 0. To find the corresponding y-coordinate, we can substitute x = 0 into the equation: f(0) = 0^2 + 3 = 3.

We can then plot the vertex (0, 3) and a few additional points, such as (1, 4) and (-1, 4). Connecting these points, we get a parabola that opens upward:

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For g(x) = 1, since it is a constant function, we can draw a horizontal line at y = 1.

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To find the area between the graphs of f and g, we need to find the x-values where the graphs intersect. From the graph, we can see that the graphs intersect at x = -1 and x = 2.

To calculate the area between the graphs, we need to find the definite integral of the difference between the functions over the interval [-1, 2].

∫[from -1 to 2] (f(x) - g(x)) dx = ∫[from -1 to 2] ((x^2 + 3) - 1) dx = ∫[from -1 to 2] (x^2 + 2) dx

To find the indefinite integral of x^2 + 2, we can use the power rule for integration:

∫ (x^n) dx = (x^(n+1))/(n+1), where n ≠ -1.

Applying the power rule to x^2 + 2, we get:
∫ (x^2 + 2) dx = (x^(2+1))/(2+1) + 2x = (x^3)/3 + 2x

Now, we can evaluate the definite integral from -1 to 2:

(2^3)/3 + 2(2) - ((-1)^3)/3 + 2(-1) = 8/3 + 4 - 1/3 - 2 = 8/3 + 11/3 - 6/3 = (8 + 11 - 6)/3 = 13/3

Therefore, the area between the graphs of f and g is 13/3 square units.

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