Question

find the area of the region bounded by the graphs of the given equations.

$y = -2x - x^2$, $y = -3$

the area is \boxed{}.
(type an integer or a simplified fraction.)

Explanation:

Step1: Find Intersection Points

Set $-2x - x^2 = -3$. Rearrange to $x^2 + 2x - 3 = 0$. Factor: $(x + 3)(x - 1) = 0$. So roots are $x = -3$ and $x = 1$.

Step2: Determine Upper and Lower Functions

For $x \in [-3, 1]$, compare $y = -2x - x^2$ and $y = -3$. The parabola $y = -x^2 - 2x$ can be rewritten as $y = -(x^2 + 2x) = -(x + 1)^2 + 1$, which has vertex at $(-1, 1)$. So $-2x - x^2 \geq -3$ on $[-3, 1]$.

Step3: Set Up Integral for Area

Area $A = \int_{-3}^{1} [(-2x - x^2) - (-3)] dx = \int_{-3}^{1} (-x^2 - 2x + 3) dx$.

Step4: Integrate the Function

Integrate term by term: $\int (-x^2 - 2x + 3) dx = -\frac{x^3}{3} - x^2 + 3x + C$.

Step5: Evaluate Definite Integral

Evaluate from $-3$ to $1$:
At $x = 1$: $-\frac{1}{3} - 1 + 3 = \frac{5}{3}$.
At $x = -3$: $-\frac{(-3)^3}{3} - (-3)^2 + 3(-3) = 9 - 9 - 9 = -9$.
Subtract: $\frac{5}{3} - (-9) = \frac{5}{3} + 9 = \frac{5 + 27}{3} = \frac{32}{3}$.

Answer:

$\frac{32}{3}$