find the area of the region bounded by f(x)=-x-1 and g(x)=1-x^2 over the interval [-2,0]

To find the area of the region bounded by \( f(x) = -x - 1 \) and \( g(x) = 1 - x^2 \) over the interval \([-2, 0]\), we follow these steps:

### Step 1: Identify the boundaries of the curves
The curves are:
- \( f(x) = -x - 1 \)
- \( g(x) = 1 - x^2 \)

We need to determine the points where \( f(x) \) and \( g(x) \) intersect to find the limits of integration.

### Step 2: Find the points of intersection
Set \( f(x) = g(x) \):
\[
-x - 1 = 1 - x^2
\]
Rearrange this equation:
\[
x^2 - x - 2 = 0
\]
Solve this quadratic equation:
\[
(x - 2)(x + 1) = 0
\]
Thus, \( x = 2 \) and \( x = -1 \). However, since the interval of interest is \([-2, 0]\), the only relevant point of intersection is \( x = -1 \).

### Step 3: Set up the integrals for the area
We need to compute the area in two parts: from \([-2, -1]\) and \([-1, 0]\), since the curves cross each other at \(x = -1\).

- For \(x \in [-2, -1]\), \( f(x) = -x - 1 \) is above \( g(x) = 1 - x^2 \).
- For \(x \in [-1, 0]\), \( g(x) = 1 - x^2 \) is above \( f(x) = -x - 1 \).

Thus, the total area is:
\[
A = \int_{-2}^{-1} \left( (-x - 1) - (1 - x^2) \right) \, dx + \int_{-1}^{0} \left( (1 - x^2) - (-x - 1) \right) \, dx
\]

### Step 4: Simplify the integrals

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

2. Second integral from \([-1, 0]\):
\[
\int_{-1}^{0} \left( 1 - x^2 - (-x - 1) \right) \, dx = \int_{-1}^{0} \left( 1 - x^2 + x + 1 \right) \, dx = \int_{-1}^{0} \left( -x^2 + x + 2 \right) \, dx
\]

### Step 5: Calculate the integrals

#### First integral:
\[
\int_{-2}^{-1} (x^2 - x - 2) \, dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-2}^{-1}
\]
Evaluating at \(x = -1\) and \(x = -2\):
\[
\left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) \right) - \left( \frac{(-2)^3}{3} - \frac{(-2)^2}{2} - 2(-2) \right)
\]
\[
= \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) - \left( -\frac{8}{3} - 2 + 4 \right)
\]
\[
= \left( \frac{19}{6} \right) - \left( \frac{-2}{3} \right) = \frac{19}{6} + \frac{2}{3} = \frac{23}{6}
\]

#### Second integral:
\[
\int_{-1}^{0} (-x^2 + x + 2) \, dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{0}
\]
Evaluating at \(x = 0\) and \(x = -1\):
\[
\left( 0 \right) - \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) \right)
\]
\[
= 0 - \left( \frac{1}{3} + \frac{1}{2} - 2 \right) = 0 - \left( \frac{-7}{6} \right) = \frac{7}{6}
\]

### Step 6: Add the areas
The total area is:
\[
A = \frac{23}{6} + \frac{7}{6} = \frac{30}{6} = 5
\]

### Final Answer:
The area of the region bounded by \( f(x) = -x - 1 \) and \( g(x) = 1 - x^2 \) over the interval \([-2, 0]\) is \( 5 \).

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Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you find the points of intersection between two curves algebraically?
2. What is the geometric interpretation of the definite integral when calculating areas between curves?
3. Can areas between curves be negative? How do we interpret them in such cases?
4. How would the process change if the interval were different, say \([-3, 1]\)?
5. How do you calculate the area if one curve is always above the other on the entire interval?

**Tip:** Always sketch the graphs of functions before finding the area between curves to visually confirm which function is on top and bottom.

This problem focuses on finding the area of the region bounded by f(x) = -x - 1 and g(x) = 1 - x^2 over the interval [-2, 0]. Learn how to calculate the area using definite integrals and the fundamental theorem of calculus, ideal for high school students in Grades 11-12.

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