Find the area enclosed by the parabola $y = -x^2 - 4x$ and the x-axis.

1. Problem Description

Find the area enclosed by the parabola $y = -x^2 - 4x$ and the x-axis.

2. Solution Steps

First, find the points where the parabola intersects the x-axis. These points are where $y = 0$.

$-x^2 - 4x = 0$

$-x(x + 4) = 0$

$x = 0$ or $x = -4$

So the parabola intersects the x-axis at $x = 0$ and $x = -4$.

Next, we need to integrate the function $y = -x^2 - 4x$ between these two points to find the area. Since the parabola is below the x-axis in the interval $[-4, 0]$, we take the absolute value of the definite integral or integrate $-y$ instead.

The integral is:

$\int_{-4}^{0} (-x^2 - 4x) dx$

$= \int_{-4}^{0} (-x^2) dx + \int_{-4}^{0} (-4x) dx$

$= [-\frac{x^3}{3}]_{-4}^{0} + [-2x^2]_{-4}^{0}$

$= (-\frac{0^3}{3} - (-\frac{(-4)^3}{3})) + (-2(0)^2 - (-2(-4)^2))$

$= (0 - \frac{64}{3}) + (0 - (-32))$

$= -\frac{64}{3} + 32$

$= -\frac{64}{3} + \frac{96}{3}$

$= \frac{32}{3}$

Since the region is below the x-axis, we need to take the absolute value, or we can directly calculate $\int_{-4}^0 (-y) dx = \int_{-4}^0 (x^2 + 4x) dx = [\frac{x^3}{3} + 2x^2]_{-4}^0 = (0) - (\frac{(-4)^3}{3} + 2(-4)^2) = -(\frac{-64}{3} + 32) = \frac{64}{3} - 32 = \frac{64 - 96}{3} = -\frac{32}{3}$. Taking the absolute value yields $\frac{32}{3}$.

Alternatively, we can integrate $-( -x^2 - 4x ) = x^2 + 4x$ from $-4$ to $0$.

$\int_{-4}^{0} (x^2 + 4x) dx = [\frac{x^3}{3} + 2x^2]_{-4}^0$

$= (\frac{0^3}{3} + 2(0)^2) - (\frac{(-4)^3}{3} + 2(-4)^2)$

$= 0 - (\frac{-64}{3} + 32)$

$= -(\frac{-64+96}{3})$

$= -\frac{32}{3}$

The area is the absolute value of the integral, so the area is $\frac{32}{3}$.

3. Final Answer

The area is $\frac{32}{3}$.