The problem asks to find the area $S$ enclosed by the parabola $y = -x^2 - 4x$ and the x-axis. Also, find the area enclosed by the curves $y = 5x - x^2$ and $y = x^2 - x + 4$.

1. Problem Description

The problem asks to find the area $S$ enclosed by the parabola $y = -x^2 - 4x$ and the x-axis. Also, find the area enclosed by the curves $y = 5x - x^2$ and $y = x^2 - x + 4$.

2. Solution Steps

(1) Find the intersection points of $y = -x^2 - 4x$ and the x-axis ($y=0$).

$-x^2 - 4x = 0$

$-x(x+4) = 0$

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

The area is given by the definite integral:

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

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

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

$S = -(\frac{64}{3} - 32) = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}$

(4) Find the intersection points of $y = 5x - x^2$ and $y = x^2 - x + 4$.

$5x - x^2 = x^2 - x + 4$

$0 = 2x^2 - 6x + 4$

$0 = x^2 - 3x + 2$

$0 = (x-1)(x-2)$

$x=1$ or $x=2$

The area is given by the definite integral:

$S = \int_{1}^{2} (5x - x^2 - (x^2 - x + 4)) dx$

$S = \int_{1}^{2} (5x - x^2 - x^2 + x - 4) dx$

$S = \int_{1}^{2} (-2x^2 + 6x - 4) dx$

$S = [-\frac{2}{3}x^3 + 3x^2 - 4x]_{1}^{2}$

$S = (-\frac{2}{3}(2)^3 + 3(2)^2 - 4(2)) - (-\frac{2}{3}(1)^3 + 3(1)^2 - 4(1))$

$S = (-\frac{16}{3} + 12 - 8) - (-\frac{2}{3} + 3 - 4)$

$S = (-\frac{16}{3} + 4) - (-\frac{2}{3} - 1)$

$S = -\frac{16}{3} + 4 + \frac{2}{3} + 1$

$S = -\frac{14}{3} + 5 = \frac{-14 + 15}{3} = \frac{1}{3}$

3. Final Answer

(1) The area enclosed by $y = -x^2 - 4x$ and the x-axis is $\frac{32}{3}$.

(4) The area enclosed by $y = 5x - x^2$ and $y = x^2 - x + 4$ is $\frac{1}{3}$.