We need to find the area of the region limited by the graphs of the following functions: a) $f(x) = |x| - |x-1|$ in the interval $[-1, 2]$. b) $f(x) = x(x-1)(x-2)$ and the x-axis. c) $y = x^2 - 9$, $x^2 + (y-3)^2 = 9$, and $y = 3 - x$.

1. Problem Description

We need to find the area of the region limited by the graphs of the following functions:

a) $f(x) = |x| - |x-1|$ in the interval $[-1, 2]$.

b) $f(x) = x(x-1)(x-2)$ and the x-axis.

c) $y = x^2 - 9$, $x^2 + (y-3)^2 = 9$, and $y = 3 - x$.

2. Solution Steps

a) $f(x) = |x| - |x-1|$ in the interval $[-1, 2]$.

We need to consider the different intervals where the absolute values change their signs.

- If $x < 0$, then $|x| = -x$ and $|x-1| = -(x-1) = 1-x$. So $f(x) = -x - (1-x) = -1$.

- If $0 \le x < 1$, then $|x| = x$ and $|x-1| = -(x-1) = 1-x$. So $f(x) = x - (1-x) = 2x-1$.

- If $x \ge 1$, then $|x| = x$ and $|x-1| = x-1$. So $f(x) = x - (x-1) = 1$.

Therefore,

$f(x) = \begin{cases} -1, & -1 \le x < 0 \\ 2x-1, & 0 \le x < 1 \\ 1, & 1 \le x \le 2 \end{cases}$

The area is given by:

$A = \int_{-1}^{2} |f(x)| dx = \int_{-1}^{0} |-1| dx + \int_{0}^{1} |2x-1| dx + \int_{1}^{2} |1| dx$

$A = \int_{-1}^{0} 1 dx + \int_{0}^{1/2} (1-2x) dx + \int_{1/2}^{1} (2x-1) dx + \int_{1}^{2} 1 dx$

$A = [x]_{-1}^{0} + [x - x^2]_{0}^{1/2} + [x^2 - x]_{1/2}^{1} + [x]_{1}^{2}$

$A = (0 - (-1)) + (1/2 - 1/4) - (0-0) + (1-1) - (1/4 - 1/2) + (2-1)$

$A = 1 + 1/4 + 1/4 + 1 = 2 + 1/2 = 5/2$

b) $f(x) = x(x-1)(x-2)$ and the x-axis.

$f(x) = x(x-1)(x-2) = x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x$

We need to find the roots of $f(x) = 0$, which are $x = 0, 1, 2$.

We need to find the area between the curve and the x-axis.

$A = \int_{0}^{2} |f(x)| dx = \int_{0}^{1} f(x) dx - \int_{1}^{2} f(x) dx$

$A = \int_{0}^{1} (x^3 - 3x^2 + 2x) dx - \int_{1}^{2} (x^3 - 3x^2 + 2x) dx$

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

$A = (\frac{1}{4} - 1 + 1) - (0) - [(\frac{16}{4} - 8 + 4) - (\frac{1}{4} - 1 + 1)]$

$A = \frac{1}{4} - (4 - 8 + 4 - \frac{1}{4}) = \frac{1}{4} - (0 - \frac{1}{4}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

c) $y = x^2 - 9$, $x^2 + (y-3)^2 = 9$, and $y = 3 - x$.

$x^2 + (y-3)^2 = 9$ is a circle centered at $(0, 3)$ with radius $3$.

$y = x^2 - 9$ is a parabola.

$y = 3 - x$ is a straight line.

Intersection of the circle and the line:

$x^2 + (3-x - 3)^2 = 9$

$x^2 + x^2 = 9$

$2x^2 = 9$

$x^2 = 9/2$

$x = \pm \frac{3}{\sqrt{2}}$

So the points are $(\frac{3}{\sqrt{2}}, 3-\frac{3}{\sqrt{2}})$ and $(-\frac{3}{\sqrt{2}}, 3+\frac{3}{\sqrt{2}})$.

Intersection of the parabola and the line:

$3 - x = x^2 - 9$

$x^2 + x - 12 = 0$

$(x+4)(x-3) = 0$

$x = -4, 3$

So the points are $(-4, 7)$ and $(3, 0)$.

Intersection of the parabola and the circle:

$x^2 + (x^2 - 9 - 3)^2 = 9$

$x^2 + (x^2 - 12)^2 = 9$

$x^2 + x^4 - 24x^2 + 144 = 9$

$x^4 - 23x^2 + 135 = 0$

Let $u = x^2$.

$u^2 - 23u + 135 = 0$

$u = \frac{23 \pm \sqrt{23^2 - 4(135)}}{2} = \frac{23 \pm \sqrt{529 - 540}}{2} = \frac{23 \pm \sqrt{-11}}{2}$

No real solutions. The parabola and circle do not intersect.

We need to find where the line intersects the circle. From previous calculations, these occur at $x = \pm 3/\sqrt{2}$.

We need to find where the line intersects the parabola. From previous calculations, these occur at $x = -4$ and $x = 3$.

The area calculation is not simple given that we are working with a circle, parabola, and a line.

It involves setting up proper integrals and determining the limits. However, it is difficult to determine the accurate region needed based on the equations alone. Without additional context, I will not provide a solution for part (c).

3. Final Answer

a) $5/2$

b) $1/2$

c) Cannot determine the final answer.