We need to find the area enclosed by the curves:

1. $y = -x^2 + 6$ (a downward-facing parabola)
2. $y = |x|$ (a V-shaped absolute value function)

Step 1: Find Intersection Points

To determine the limits of integration, we solve:

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

For $x \geq 0$ (i.e., $y = x$)

$-x^2 + 6 = x$ $-x^2 - x + 6 = 0$ Solving this quadratic equation:

$x^2 + x - 6 = 0$

Factoring:

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

So, $x = -3$ or $x = 2$. Since we're in $x \geq 0$, we take $x = 2$.

For $x \leq 0$ (i.e., $y = -x$)

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

Rewriting:

$x^2 - x - 6 = 0$

Factoring:

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

So, $x = 3$ or $x = -2$. Since we're in $x \leq 0$, we take $x = -2$.

Thus, the curves intersect at $x = -2$ and $x = 2$.

Step 2: Set Up the Integral

The area is symmetric about the y-axis, so we compute for $x \in [0,2]$ and double the result.

$A = 2 \int_0^2 [(-x^2 + 6) - x] \,dx$

Simplifying:

$A = 2 \int_0^2 (-x^2 + 6 - x) \,dx$

$A = 2 \int_0^2 (-x^2 - x + 6) \,dx$

Step 3: Compute the Integral

Breaking it down:

$\int (-x^2 - x + 6) \,dx = \int -x^2 \,dx + \int -x \,dx + \int 6 \,dx$

[ = \left[ -\frac{x^3}{3} \right] + \left[ -\frac{x^2}{2} \right] + \left[ 6x \right] \Big|_0^2 ]

Evaluating at $x = 2$:

$\left( -\frac{8}{3} - \frac{4}{2} + 6(2) \right) = \left( -\frac{8}{3} - 2 + 12 \right) = \left( \frac{-8 - 6 + 36}{3} \right) = \frac{22}{3}$

Evaluating at $x = 0$:

$0$

So,

$A = 2 \times \frac{22}{3} = \frac{44}{3}$

Final Answer:

$\frac{44}{3} \text{ square units}$

Would you like a graphical representation or further details?

Related Questions:

1. How do you find the enclosed area between two curves using definite integrals?
2. How does symmetry simplify area calculations in integrals?
3. What are the key steps in solving quadratic equations for intersection points?
4. How do modulus functions affect integration limits?
5. What is the geometric interpretation of this integral?

Tip:

When integrating to find the area, always check which function is on top and subtract accordingly to avoid negative areas.