We are asked to find the area S enclosed by the given parabolas and the x-axis. (1) $y = -x^2 - 4x$ and the x-axis. (2) $y = x^2 + 4x + 3$ and the x-axis.

1. Problem Description

We are asked to find the area S enclosed by the given parabolas and the x-axis.

(1) $y = -x^2 - 4x$ and the x-axis.

(2) $y = x^2 + 4x + 3$ and the x-axis.

2. Solution Steps

(1) The x-intercepts of $y = -x^2 - 4x$ are found by setting $y=0$:

$-x^2 - 4x = 0$

$-x(x+4) = 0$

$x = 0, -4$

The area S is given by the integral:

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

$S = [-\frac{1}{3}x^3 - 2x^2]_{-4}^{0}$

$S = (0) - (-\frac{1}{3}(-4)^3 - 2(-4)^2)$

$S = -(-\frac{1}{3}(-64) - 2(16))$

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

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

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

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

(2) The x-intercepts of $y = x^2 + 4x + 3$ are found by setting $y=0$:

$x^2 + 4x + 3 = 0$

$(x+1)(x+3) = 0$

$x = -1, -3$

The area S is given by the integral:

$S = \left| \int_{-3}^{-1} (x^2 + 4x + 3) dx \right|$

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

$S = \left| (\frac{1}{3}(-1)^3 + 2(-1)^2 + 3(-1)) - (\frac{1}{3}(-3)^3 + 2(-3)^2 + 3(-3)) \right|$

$S = \left| (-\frac{1}{3} + 2 - 3) - (\frac{1}{3}(-27) + 2(9) - 9) \right|$

$S = \left| (-\frac{1}{3} - 1) - (-9 + 18 - 9) \right|$

$S = \left| -\frac{4}{3} - 0 \right|$

$S = \left| -\frac{4}{3} \right|$

$S = \frac{4}{3}$

3. Final Answer

(1) $\frac{32}{3}$

(2) $\frac{4}{3}$