To find the area between the curves $y = 4 - x^2$ and $y = 2 - x$ from $x = -2$ to $x = 2$, we proceed step by step:

Step 1: Determine the points of intersection

First, we set the two functions equal to each other to find where they intersect:

$4 - x^2 = 2 - x$

Rearranging the equation:

$4 - 2 = x^2 - x$

$2 = x^2 - x$

$x^2 - x - 2 = 0$

Now, solve this quadratic equation using factoring:

$(x - 2)(x + 1) = 0$

Thus, the points of intersection are $x = 2$ and $x = -1$.

Step 2: Set up the integral

To find the total area, we need to break the problem into two parts: one for the region between $x = -2$ and $x = -1$, and the other between $x = -1$ and $x = 2$.

1. For $x = -2$ to $x = -1$, the curve $y = 4 - x^2$ is above $y = 2 - x$.
2. For $x = -1$ to $x = 2$, the curve $y = 2 - x$ is above $y = 4 - x^2$.

The total area is the sum of the areas of these two regions:

[ \text{Area} = \int_{-2}^{-1} \left[(4 - x^2) - (2 - x)\right] , dx + \int_{-1}^{2} \left[(2 - x) - (4 - x^2)\right] , dx ]

Step 3: Evaluate the integrals

1. First integral: $\int_{-2}^{-1} \left[(4 - x^2) - (2 - x)\right] \, dx = \int_{-2}^{-1} (2 - x^2 + x) \, dx$

2. Second integral: $\int_{-1}^{2} \left[(2 - x) - (4 - x^2)\right] \, dx = \int_{-1}^{2} (-2 + x^2 + x) \, dx$

Let me calculate the result.The total area between the curves $y = 4 - x^2$ and $y = 2 - x$ from $x = -2$ to $x = 2$ is $\frac{10}{3}$ square units.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you solve systems of equations by substitution to find intersection points?
2. What is the general formula for finding the area between two curves?
3. Can we use a different method, like the trapezoidal rule, to approximate the area?
4. How would the limits of integration change if the curves were symmetric about the y-axis?
5. What if we wanted to calculate the volume by rotating these curves around the x-axis?

Tip: Always check which curve is on top within the integration limits when calculating areas between curves!