The area bounded by `|x|+|y|=1 and y ge x^(2)` in the first quadrant is `a-(a^(2))/(2)-(a^(3))/(3)` sq. units, then the value of `(2a+1)^(2)` is equal to

Answer

To solve the problem, we need to find the area bounded by the equations \( |x| + |y| = 1 \) and \( y \geq x^2 \) in the first quadrant, and then determine the value of \( (2a + 1)^2 \) where the area is given in the form \( a - \frac{a^2}{2} - \frac{a^3}{3} \).

### Step 1: Understand the equations
The equation \( |x| + |y| = 1 \) describes a square in the coordinate plane with vertices at \( (1, 0) \), \( (0, 1) \), \( (-1, 0) \), and \( (0, -1) \). In the first quadrant, this simplifies to the line \( x + y = 1 \).

The equation \( y = x^2 \) describes a parabola opening upwards. In the first quadrant, we are interested in the area above this parabola.

### Step 2: Find the intersection points
To find the area of interest, we need to find the intersection points of the line \( y = 1 - x \) and the parabola \( y = x^2 \).

Set the equations equal to each other:
\[ 1 - x = x^2 \]
Rearranging gives:
\[ x^2 + x - 1 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

Since we are in the first quadrant, we take the positive root:
\[ x = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Calculate the area
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{-1 + \sqrt{5}}{2} \) can be calculated using the integral:
\[ A = \int_0^{\frac{-1 + \sqrt{5}}{2}} ((1 - x) - x^2) \, dx \]

### Step 5: Evaluate the integral
Calculating the integral:
\[ A = \int_0^{\frac{-1 + \sqrt{5}}{2}} (1 - x - x^2) \, dx \]
The antiderivative is:
\[ x - \frac{x^2}{2} - \frac{x^3}{3} \]
Evaluating from \( 0 \) to \( \frac{-1 + \sqrt{5}}{2} \):
\[ A = \left[ \left( \frac{-1 + \sqrt{5}}{2} - \frac{1}{2} \left( \frac{-1 + \sqrt{5}}{2} \right)^2 - \frac{1}{3} \left( \frac{-1 + \sqrt{5}}{2} \right)^3 \right) - 0 \right] \]

### Step 6: Substitute and simplify
Let \( a = \frac{-1 + \sqrt{5}}{2} \). Then, the area can be expressed as:
\[ A = a - \frac{a^2}{2} - \frac{a^3}{3} \]

### Step 7: Find \( (2a + 1)^2 \)
Now, we need to calculate \( (2a + 1)^2 \):
\[ 2a = 2 \cdot \frac{-1 + \sqrt{5}}{2} = -1 + \sqrt{5} \]
Thus,
\[ 2a + 1 = \sqrt{5} \]
Now squaring gives:
\[ (2a + 1)^2 = (\sqrt{5})^2 = 5 \]

### Final Answer
The value of \( (2a + 1)^2 \) is \( \boxed{5} \).