The area of the region, bounded by the curves `y=sin^(-1)x+x(1-x) and y=sin^(-1)x-x(1-x)` in the first quadrant (in sq. units), is

Options

A. 1
B. `(1)/(2)`
C. `(1)/(3)`
D. `(1)/(4)`

Answer

To find the area of the region bounded by the curves \( y = \sin^{-1}(x) + x(1-x) \) and \( y = \sin^{-1}(x) - x(1-x) \) in the first quadrant, we can follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y_1 = \sin^{-1}(x) + x(1-x) \)
2. \( y_2 = \sin^{-1}(x) - x(1-x) \)

### Step 2: Find the points of intersection
To find the points where the curves intersect, we set \( y_1 = y_2 \):
\[ \sin^{-1}(x) + x(1-x) = \sin^{-1}(x) - x(1-x) \]
Subtracting \( \sin^{-1}(x) \) from both sides gives:
\[ x(1-x) + x(1-x) = 0 \]
This simplifies to:
\[ 2x(1-x) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = 1 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_{0}^{1} (y_1 - y_2) \, dx \]
Substituting \( y_1 \) and \( y_2 \):
\[ A = \int_{0}^{1} \left( \left( \sin^{-1}(x) + x(1-x) \right) - \left( \sin^{-1}(x) - x(1-x) \right) \right) dx \]
This simplifies to:
\[ A = \int_{0}^{1} \left( 2x(1-x) \right) dx \]

### Step 4: Evaluate the integral
Now we need to evaluate the integral:
\[ A = 2 \int_{0}^{1} x(1-x) \, dx \]
We can expand \( x(1-x) \):
\[ x(1-x) = x - x^2 \]
Thus,
\[ A = 2 \int_{0}^{1} (x - x^2) \, dx \]
Now, we can integrate term by term:
\[ \int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3} \]
Evaluating from 0 to 1:
\[ A = 2 \left[ \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \right] \]
This simplifies to:
\[ A = 2 \left( \frac{1}{2} - \frac{1}{3} \right) \]
Finding a common denominator (which is 6):
\[ A = 2 \left( \frac{3}{6} - \frac{2}{6} \right) = 2 \left( \frac{1}{6} \right) = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
The area of the region bounded by the curves in the first quadrant is:
\[ \boxed{\frac{1}{3}} \text{ square units} \]