The area (in sq. units) of the region bounded by the curves `y=2-x^(2) and y=|x|` is k, then the value of 3k is

Answer

To find the area of the region bounded by the curves \( y = 2 - x^2 \) and \( y = |x| \), we will follow these steps:

### Step 1: Identify the Points of Intersection
To find the area between the curves, we first need to determine the points where they intersect. We set the equations equal to each other:
\[ 2 - x^2 = |x| \]
This gives us two cases to consider:

**Case 1:** \( x \geq 0 \) (where \( |x| = x \))
\[ 2 - x^2 = x \implies x^2 + x - 2 = 0 \]
Factoring:
\[ (x - 1)(x + 2) = 0 \]
Thus, \( x = 1 \) (since \( x \geq 0 \)).

**Case 2:** \( x < 0 \) (where \( |x| = -x \))
\[ 2 - x^2 = -x \implies x^2 - x - 2 = 0 \]
Factoring:
\[ (x - 2)(x + 1) = 0 \]
Thus, \( x = -1 \) (since \( x < 0 \)).

The points of intersection are \( (-1, 1) \) and \( (1, 1) \).

### Step 2: Set Up the Area Integral
The area \( A \) between the curves from \( x = -1 \) to \( x = 1 \) can be calculated as:
\[ A = \int_{-1}^{1} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
From the graphs, we see that:
- For \( x \in [-1, 0] \), \( y_{\text{top}} = 2 - x^2 \) and \( y_{\text{bottom}} = -x \).
- For \( x \in [0, 1] \), \( y_{\text{top}} = 2 - x^2 \) and \( y_{\text{bottom}} = x \).

Thus, we can express the area as:
\[ A = \int_{-1}^{0} ((2 - x^2) - (-x)) \, dx + \int_{0}^{1} ((2 - x^2) - x) \, dx \]

### Step 3: Calculate the Area
Calculating the first integral:
\[ \int_{-1}^{0} (2 - x^2 + x) \, dx = \int_{-1}^{0} (2 + x - x^2) \, dx \]
Calculating:
\[ = \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{0} \]
Evaluating at the limits:
\[ = \left( 0 + 0 - 0 \right) - \left( -2 + \frac{1}{2} + \frac{1}{3} \right) \]
\[ = 0 + 2 - \frac{1}{2} - \frac{1}{3} = 2 - \frac{3}{6} - \frac{2}{6} = 2 - \frac{5}{6} = \frac{12}{6} - \frac{5}{6} = \frac{7}{6} \]

Now for the second integral:
\[ \int_{0}^{1} (2 - x^2 - x) \, dx = \int_{0}^{1} (2 - x - x^2) \, dx \]
Calculating:
\[ = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \]
Evaluating at the limits:
\[ = \left( 2 - \frac{1}{2} - \frac{1}{3} \right) - (0) \]
\[ = 2 - \frac{3}{6} - \frac{2}{6} = 2 - \frac{5}{6} = \frac{12}{6} - \frac{5}{6} = \frac{7}{6} \]

### Step 4: Total Area
Combining both areas:
\[ A = \frac{7}{6} + \frac{7}{6} = \frac{14}{6} = \frac{7}{3} \]

### Step 5: Calculate \( 3k \)
Since \( k = \frac{7}{3} \), we find:
\[ 3k = 3 \times \frac{7}{3} = 7 \]

Thus, the final answer is:
\[ \boxed{7} \]