If the area bounded by `y = ||x|^(2) - 4|x| + 3||` and the x-axis from `x = 1` to `x = 3` is `p/q` (where, p & q are coprime) then the value of `p + q ` is

Answer

To solve the problem of finding the area bounded by the curve \( y = ||x|^2 - 4|x| + 3| \) and the x-axis from \( x = 1 \) to \( x = 3 \), we will follow these steps:

### Step 1: Simplify the Expression Inside the Modulus
First, we simplify the expression inside the modulus:
\[ f(x) = |x^2 - 4|x| + 3| \]
Since we are interested in the interval from \( x = 1 \) to \( x = 3 \), we note that for \( x \geq 0 \), \( |x| = x \). Thus, we can rewrite the function as:
\[ f(x) = |x^2 - 4x + 3| \]

### Step 2: Factor the Quadratic
Next, we factor the quadratic:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]
The roots of the quadratic are \( x = 1 \) and \( x = 3 \).

### Step 3: Determine the Sign of the Quadratic in the Interval
We need to determine the sign of \( x^2 - 4x + 3 \) in the interval \( [1, 3] \):
- At \( x = 1 \): \( f(1) = 0 \)
- At \( x = 2 \): \( f(2) = 2^2 - 4 \cdot 2 + 3 = 1 \) (positive)
- At \( x = 3 \): \( f(3) = 0 \)

Since the quadratic opens upwards and is negative between its roots, we find that:
\[ x^2 - 4x + 3 < 0 \quad \text{for } x \in (1, 3) \]
Thus, in the interval \( [1, 3] \), we have:
\[ f(x) = -(x^2 - 4x + 3) = -x^2 + 4x - 3 \]

### Step 4: Set Up the Integral for Area
The area \( A \) bounded by the curve and the x-axis from \( x = 1 \) to \( x = 3 \) can be calculated using the integral:
\[ A = \int_{1}^{3} -(-x^2 + 4x - 3) \, dx = \int_{1}^{3} (x^2 - 4x + 3) \, dx \]

### Step 5: Evaluate the Integral
Now we compute the integral:
\[ A = \int_{1}^{3} (x^2 - 4x + 3) \, dx \]
Calculating the integral:
\[ = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{1}^{3} \]

Calculating at the upper limit \( x = 3 \):
\[ = \frac{3^3}{3} - 2(3^2) + 3(3) = 9 - 18 + 9 = 0 \]
Calculating at the lower limit \( x = 1 \):
\[ = \frac{1^3}{3} - 2(1^2) + 3(1) = \frac{1}{3} - 2 + 3 = \frac{1}{3} + 1 = \frac{4}{3} \]

Now, substituting the limits into the integral:
\[ A = 0 - \frac{4}{3} = \frac{4}{3} \]

### Step 6: Identify \( p \) and \( q \)
The area can be expressed as \( \frac{p}{q} \) where \( p = 4 \) and \( q = 3 \). Since 4 and 3 are coprime, we can conclude that:
\[ p + q = 4 + 3 = 7 \]

### Final Answer
The value of \( p + q \) is \( \boxed{7} \).