Answer :

Evaluating the definite integral at the upper and lower bounds, we have A = [-6³/3 - (15(6)²)/2 - 54(6)] - [(-(-3)³/3 - (15(-3)²)/2 - 54(-3))]. Simplifying, we find A = 729/2, which is the area bounded above by f(x) and below by g(x).

To find the area bounded above by f(x) = -2x² + 10x - 27 and below by g(x) = x² + 25x + 27, we need to determine the points of intersection between the two functions and calculate the definite integral between those points. The area between two curves is given by the integral of the difference of the upper and lower functions.

First, we need to find the points of intersection between f(x) and g(x) by setting them equal to each other: -2x² + 10x - 27 = x² + 25x + 27. Simplifying, we get -3x² + 15x + 54 = 0. Factoring, we have (x - 6)(-3x - 9) = 0, which gives us two possible solutions: x = 6 and x = -3.

Next, we need to determine which function is the upper and lower bound in the given interval. To do this, we can evaluate the functions at the points of intersection. Evaluating f(x) at x = 6, we get f(6) = -2(6)² + 10(6) - 27 = -99. Evaluating g(x) at x = 6, we get g(6) = 6² + 25(6) + 27 = 183. So, f(x) is the upper bound and g(x) is the lower bound.

Now we can calculate the area by taking the definite integral of the difference between f(x) and g(x) over the interval [x = -3, x = 6]. The area A is given by A = ∫[x=-3 to x=6] (f(x) - g(x)) dx. Integrating f(x) - g(x) gives us (-2x² + 10x - 27) - (x² + 25x + 27) = -3x² - 15x - 54.

Evaluating this integral over the given interval, we get A = ∫[-3 to 6] (-3x² - 15x - 54) dx = [-x³/3 - (15x²)/2 - 54x] from -3 to 6.

Evaluating the definite integral at the upper and lower bounds, we have A = [-6³/3 - (15(6)²)/2 - 54(6)] - [(-(-3)³/3 - (15(-3)²)/2 - 54(-3))].

Simplifying, we find A = 729/2, which is the area bounded above by f(x) and below by g(x).

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