Question

Use a graphing utility to graph the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=3-2 x-x^{2}, g(x)=0 $$

Short answer

The area of the region bounded by the graphs of the functions is 6 square units.

Step-by-step solution

Graph the Functions

The first step involves graphing both functions to visualize them and find the intersection points. By graphing the function \(f(x) = 3 - 2x - x^2\) and \(g(x) = 0\), it is evident that they intersect at x = -1 and x = 3.

Define the Interval

The region bound by the two curves is noticeable on the interval from x = -1 to x = 3. We can see that for any point in this interval, the value of the function \(f(x)\) is always greater than the value of the function \(g(x)=0\).

Calculate Area

We can calculate the area under the curve \(f(x)\) from x = -1 to x = 3 by using the formula for the definite integral: \[A = \int_{a}^{b} f(x) - g(x) \,dx = \int_{-1}^{3} (3 - 2x - x^2 - 0) \,dx\]. Hence, \[A = \left. [3x - x^2 - \frac{x^3}{3}]\right|_{-1}^{3} = 12 - 9 - 9 = -6\]. Since area cannot be negative, we take the absolute value to get the answer 6 square units.