for each problem, find the area of the region enclosed by the curves. you may use the provided graph to…

Explanation:

Step1: Find intersection points of the curves

Set $-x^3 + 6x = -x^2$ and solve for $x$: $$x^3 - x^2 - 6x = 0 \implies x(x - 3)(x + 2) = 0 \implies x = -2, 0, 3$$

Step2: Determine relative positions of curves

On $[-2, 0]$, $-x^2 \ge -x^3 + 6x$. On $[0, 3]$, $-x^3 + 6x \ge -x^2$.

Step3: Set up the definite integrals

$$A = \int_{-2}^{0} (-x^2 - (-x^3 + 6x)) , dx + \int_{0}^{3} (-x^3 + 6x - (-x^2)) , dx$$

Step4: Simplify the integrands

$$A = \int_{-2}^{0} (x^3 - x^2 - 6x) , dx + \int_{0}^{3} (-x^3 + x^2 + 6x) , dx$$

Step5: Evaluate the first integral

$$\left[ \frac{x^4}{4} - \frac{x^3}{3} - 3x^2 \right]_{-2}^{0} = 0 - \left( \frac{16}{4} - \frac{-8}{3} - 3(4) \right) = -\left( 4 + \frac{8}{3} - 12 \right) = \frac{16}{3}$$

Step6: Evaluate the second integral

$$\left[ -\frac{x^4}{4} + \frac{x^3}{3} + 3x^2 \right]_{0}^{3} = \left( -\frac{81}{4} + \frac{27}{3} + 3(9) \right) - 0 = -\frac{81}{4} + 9 + 27 = \frac{63}{4}$$

Step7: Sum the areas

$$A = \frac{16}{3} + \frac{63}{4} = \frac{64 + 189}{12} = \frac{253}{12}$$

Answer:

$\frac{253}{12}$ units squared (approximately $21.083$)