Find the area of the region above the curve y = x² - 6, below y = x, and above y = -x.
Understand the Problem
The question is asking to calculate the area of a specific region defined by the curves y = x² - 6, y = x, and y = -x. This involves determining where these curves intersect and setting up appropriate integrals to find the area between them.
Answer
The area of the region is \( \frac{56}{3} \).
Answer for screen readers
The area of the region is ( \frac{56}{3} ).
Steps to Solve
- 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 $x = -2$.
- 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 $x = -3$.
- 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:
$$ A = \int_{-2}^{2} (x + 6 - x^2) , dx $$
- 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:
$$ A = \left[ 6x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{2} $$
- 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:
$$ A(-2) = -12 + 2 + \frac{8}{3} = -10 + \frac{8}{3} = \frac{-30 + 8}{3} = \frac{-22}{3} $$
- Compute the Total Area
Therefore, the total area is:
$$ A = A(2) - A(-2) = \frac{34}{3} - \left(\frac{-22}{3}\right) = \frac{34 + 22}{3} = \frac{56}{3} $$
The area of the region is ( \frac{56}{3} ).
More Information
This area represents a bounded region formed by the intersection of a quadratic curve and two linear functions, providing insight into the geometric interpretation of integrals in finding areas between curves.
Tips
- Forgetting to find all intersection points can lead to incorrect limits for integration.
- Not properly setting up the integral in terms of the top and bottom functions may result in incorrect area calculations.
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