Consider `f(x)=x^2-3x+2` The area bounded by `|y|=|f(|x|)|,xge1` is A, then find the value of `3A+2`.

Answer

To solve the problem, we need to find the area bounded by the curves defined by \( |y| = |f(|x|)| \) for \( x \geq 1 \), where \( f(x) = x^2 - 3x + 2 \).

### Step-by-step Solution:

1. **Find the Roots of \( f(x) \)**:
We start by finding the roots of the function \( f(x) = x^2 - 3x + 2 \).
\[ f(x) = 0 \implies x^2 - 3x + 2 = 0 \]
Factoring gives:
\[ (x-1)(x-2) = 0 \implies x = 1, 2 \]

2. **Determine the Vertex of the Parabola**:
The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \).
Here, \( a = 1 \) and \( b = -3 \):
\[ x = \frac{3}{2} \]
Substituting \( x = \frac{3}{2} \) back into \( f(x) \):
\[ f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = -\frac{1}{4} \]

3. **Sketch the Graph of \( f(x) \)**:
The graph of \( f(x) \) is a parabola opening upwards with roots at \( x = 1 \) and \( x = 2 \) and a vertex at \( \left(\frac{3}{2}, -\frac{1}{4}\right) \).

4. **Consider the Absolute Values**:
For \( |y| = |f(|x|)| \), we need to consider the graph of \( f(x) \) for \( x \geq 1 \) and reflect it in the x-axis for negative values of \( y \).

5. **Calculate the Area for \( x \geq 1 \)**:
We need to find the area between the curve and the x-axis from \( x = 1 \) to \( x = 2 \):
\[ A = \int_{1}^{2} f(x) \, dx \]
First, we compute the integral:
\[ A = \int_{1}^{2} (x^2 - 3x + 2) \, dx \]
Evaluating the integral:
\[ = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_{1}^{2} \]
Calculating at the bounds:
\[ = \left( \frac{2^3}{3} - \frac{3 \cdot 2^2}{2} + 2 \cdot 2 \right) - \left( \frac{1^3}{3} - \frac{3 \cdot 1^2}{2} + 2 \cdot 1 \right) \]
\[ = \left( \frac{8}{3} - 6 + 4 \right) - \left( \frac{1}{3} - \frac{3}{2} + 2 \right) \]
\[ = \left( \frac{8}{3} - \frac{18}{3} + \frac{12}{3} \right) - \left( \frac{1}{3} - \frac{9}{6} + \frac{12}{6} \right) \]
\[ = \left( \frac{2}{3} \right) - \left( \frac{1}{3} + \frac{3}{6} \right) \]
\[ = \frac{2}{3} - \left( \frac{1}{3} + \frac{1}{2} \right) = \frac{2}{3} - \frac{5}{6} = \frac{4}{6} - \frac{5}{6} = -\frac{1}{6} \]
Since we are interested in the area, we take the absolute value:
\[ A = \frac{1}{6} \]

6. **Calculate \( 3A + 2 \)**:
Finally, we compute \( 3A + 2 \):
\[ 3A + 2 = 3 \cdot \frac{1}{6} + 2 = \frac{3}{6} + 2 = \frac{1}{2} + 2 = \frac{5}{2} \]

### Final Answer:
\[ \boxed{\frac{5}{2}} \]