The area bounded by the curves `x+y=2` and `y=x^2` above x-axis in the first quadrant is ,

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `(10)/(12)`
D. `(6)/(5)`

Explanation

To find the area bounded by the curves \(x + y = 2\) and \(y = x^2\) above the x-axis in the first quadrant, we can follow these steps:

### Step 1: Find the Points of Intersection
We start by solving the equations of the curves to find their points of intersection.

1. Set \(y = 2 - x\) (from \(x + y = 2\)) equal to \(y = x^2\):
\[ 2 - x = x^2 \]
Rearranging gives:
\[ x^2 + x - 2 = 0 \]

### Step 2: Solve the Quadratic Equation
Next, we solve the quadratic equation \(x^2 + x - 2 = 0\) using the factorization method.

1. Factor the equation:
\[ (x - 1)(x + 2) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = -2 \]

Since we are interested in the first quadrant, we take \(x = 1\).

### Step 3: Find the Corresponding y-Value
Now, we find the corresponding \(y\)-value when \(x = 1\).

1. Substitute \(x = 1\) into \(y = x^2\):
\[ y = 1^2 = 1 \]
So, the point of intersection in the first quadrant is \((1, 1)\).

### Step 4: Set Up the Integral for Area Calculation
The area bounded by the curves from \(x = 0\) to \(x = 1\) can be split into two parts:

1. From \(x = 0\) to \(x = 1\), the upper curve is \(y = x^2\).
2. From \(x = 1\) to \(x = 2\), the upper curve is \(y = 2 - x\).

### Step 5: Calculate the Area
The area \(A\) can be calculated as follows:
\[ A = \int_{0}^{1} (x^2) \, dx + \int_{1}^{2} (2 - x) \, dx \]

1. Calculate the first integral:
\[ \int_{0}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1^3}{3} - 0 = \frac{1}{3} \]

2. Calculate the second integral:
\[ \int_{1}^{2} (2 - x) \, dx = \left[2x - \frac{x^2}{2}\right]_{1}^{2} \]
Evaluating at the bounds:
\[ = \left(2(2) - \frac{2^2}{2}\right) - \left(2(1) - \frac{1^2}{2}\right) = (4 - 2) - (2 - 0.5) = 2 - 1.5 = 0.5 \]

### Step 6: Combine the Areas
Now, add the two areas together:
\[ A = \frac{1}{3} + 0.5 = \frac{1}{3} + \frac{3}{6} = \frac{1}{3} + \frac{2}{6} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
Thus, the area bounded by the curves \(x + y = 2\) and \(y = x^2\) above the x-axis in the first quadrant is:
\[ \frac{5}{6} \]