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

Question image

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

  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 $x = -2$.

  1. 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 $x = -3$.

  1. 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:

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

  1. 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:

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

  1. 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:

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

  1. 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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