A Region Lying Between Two Intersecting Graphs Find the area of the region bounded by the graphs of \( f(x)=2-x^{2} \) and \( g(x)=x \).

To find the area of the region bounded by the graphs of \( f(x)=2-x^{2} \) and \( g(x)=x \), we need to determine the intersection points of the two graphs and then calculate the area of the region between them. 1. Find the intersection points: To find the intersection points, we need to solve the equation \( f(x) = g(x) \). \[ 2 - x^{2} = x \] \[ x^{2} + x - 2 = 0 \] \[ (x + 2)(x - 1) = 0 \] \[ x = -2 \text{ or } x = 1 \] 2. Determine the limits of integration: The limits of integration are the intersection points we found in step 1. In this case, the limits are \( x = -2 \) and \( x = 1 \). 3. Calculate the area: The area of the region bounded by the graphs of \( f(x) \) and \( g(x) \) can be calculated using the definite integral: \[ \text{Area} = \int_{-2}^{1} (f(x) - g(x)) \, dx \] \[ \text{Area} = \int_{-2}^{1} (2 - x^{2} - x) \, dx \] Now, let's calculate the area. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-2}^{1} \left(2-x^{2}-x\right) dx\) - step1: Evaluate the integral: \(\int \left(2-x^{2}-x\right) dx\) - step2: Use properties of integrals: \(\int 2 dx+\int -x^{2} dx+\int -x dx\) - step3: Evaluate the integral: \(2x+\int -x^{2} dx+\int -x dx\) - step4: Evaluate the integral: \(2x-\frac{x^{3}}{3}+\int -x dx\) - step5: Evaluate the integral: \(2x-\frac{x^{3}}{3}-\frac{x^{2}}{2}\) - step6: Return the limits: \(\left(2x-\frac{x^{3}}{3}-\frac{x^{2}}{2}\right)\bigg |_{-2}^{1}\) - step7: Calculate the value: \(\frac{9}{2}\) The area of the region bounded by the graphs of \( f(x)=2-x^{2} \) and \( g(x)=x \) is \( \frac{9}{2} \) square units.