The problem asks to find the area of the region bounded by the graph of $f(x) = |x| - |x-1|$ on the interval $[-1, 2]$.

1. Problem Description

The problem asks to find the area of the region bounded by the graph of $f(x) = |x| - |x-1|$ on the interval $[-1, 2]$.

2. Solution Steps

We need to analyze the function $f(x) = |x| - |x-1|$ in the interval $[-1, 2]$.

We consider the following cases:

Case 1: $x < 0$. Then $|x| = -x$ and $|x-1| = -(x-1) = 1-x$.

So, $f(x) = -x - (1-x) = -x - 1 + x = -1$.

This is valid for $-1 \le x < 0$.

Case 2: $0 \le x < 1$. Then $|x| = x$ and $|x-1| = -(x-1) = 1-x$.

So, $f(x) = x - (1-x) = x - 1 + x = 2x - 1$.

This is valid for $0 \le x < 1$.

Case 3: $x \ge 1$. Then $|x| = x$ and $|x-1| = x-1$.

So, $f(x) = x - (x-1) = x - x + 1 = 1$.

This is valid for $1 \le x \le 2$.

Thus, we have:

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

Now we find the area under the curve.

$Area = \int_{-1}^2 |f(x)| \, dx$.

Since $f(x)$ is sometimes negative, we need to take the absolute value.

Since $2x-1 < 0$ for $0 \le x < 1/2$, and $2x-1 \ge 0$ for $1/2 \le x < 1$, we need to split the integral from $0$ to $1$ into two parts.

$Area = \int_{-1}^0 |-1| \, dx + \int_0^{1/2} |2x-1| \, dx + \int_{1/2}^1 |2x-1| \, dx + \int_1^2 |1| \, dx$

$Area = \int_{-1}^0 1 \, dx + \int_0^{1/2} (1-2x) \, dx + \int_{1/2}^1 (2x-1) \, dx + \int_1^2 1 \, dx$

$Area = [x]_{-1}^0 + [x - x^2]_0^{1/2} + [x^2 - x]_{1/2}^1 + [x]_1^2$

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

$Area = 1 + (\frac{1}{4}) + 0 - (\frac{-1}{4}) + 1$

$Area = 1 + \frac{1}{4} + \frac{1}{4} + 1$

$Area = 2 + \frac{1}{2}$

$Area = \frac{5}{2}$

3. Final Answer

The area of the region is $\frac{5}{2}$.