Find the area of the region above the curve y = x² - 6, below y = x, and above y = -x.

Steps to Solve

1. Find the Points of Intersection

We need to determine where the curves intersect by setting the equations equal to each other. First, solve for the intersection between $y = x^2 - 6$ and $y = x$:

$$ x^2 - 6 = x $$

Rearranging gives:

$$ x^2 - x - 6 = 0 $$

Factoring this equation:

$$ (x - 3)(x + 2) = 0 $$

Thus, the points of intersection are $x = 3$ and $x = -2$.

2. Verify Intersection with $y = -x$

Next, we confirm the intersection points with $y = -x$. Set $x^2 - 6 = -x$:

$$ x^2 + x - 6 = 0 $$

Factoring yields:

$$ (x - 2)(x + 3) = 0 $$

So, the intersection points are $x = 2$ and $x = -3$.

3. Determine the Relevant Area

Next, identify the region defined by the curves. We need to consider the area between the curves from $x = -2$ to $x = 2$.

The area between the curves can be expressed as:

$$ A = \int_{-2}^{2} \left((x) - (x^2 - 6)\right) , dx $$

This simplifies to:

$$ A = \int_{-2}^{2} (x + 6 - x^2) , dx $$

4. Evaluate the Integral

Now, compute the integral:

$$ A = \int_{-2}^{2} (6 + x - x^2) , dx $$

This can be evaluated as follows:

First, find the antiderivative:

• The integral of $6$ is $6x$.
• The integral of $x$ is $\frac{x^2}{2}$.
• The integral of $-x^2$ is $-\frac{x^3}{3}$.

Thus:

$$ A = \left[ 6x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{2} $$

5. Substituting Limits

Substituting $x = 2$:

$$ A(2) = 6(2) + \frac{(2)^2}{2} - \frac{(2)^3}{3} $$

Calculate:

$$ A(2) = 12 + 2 - \frac{8}{3} = 14 - \frac{8}{3} = \frac{42 - 8}{3} = \frac{34}{3} $$

Now substitute $x = -2$:

$$ A(-2) = 6(-2) + \frac{(-2)^2}{2} - \frac{(-2)^3}{3} $$

Calculate:

$$ A(-2) = -12 + 2 + \frac{8}{3} = -10 + \frac{8}{3} = \frac{-30 + 8}{3} = \frac{-22}{3} $$

6. Compute the Total Area

Therefore, the total area is:

$$ A = A(2) - A(-2) = \frac{34}{3} - \left(\frac{-22}{3}\right) = \frac{34 + 22}{3} = \frac{56}{3} $$