The area bounded by `y=sqrt(1-x^(2))` and `y=x^(3)-x` is divided by y - axis in the ratio `lambda:1(AA lambda gt 1)`, then `lambda` is equal to

Options

A. `(pi+1)/(pi-1)`
B. `(pi)/(pi-1)`
C. `(pi+1)/(pi)`
D. `(pi+2)/(pi-2)`

Answer

To find the value of \(\lambda\) for the area bounded by the curves \(y = \sqrt{1 - x^2}\) and \(y = x^3 - x\), we will follow these steps:

### Step 1: Identify the curves and their intersection points
The curve \(y = \sqrt{1 - x^2}\) represents the upper half of a circle with radius 1, centered at the origin. The curve \(y = x^3 - x\) is a cubic function.

To find the points of intersection, we set:
\[ \sqrt{1 - x^2} = x^3 - x \]
Squaring both sides gives:
\[ 1 - x^2 = (x^3 - x)^2 \]
Expanding the right side:
\[ 1 - x^2 = x^6 - 2x^4 + x^2 \]
Rearranging gives:
\[ x^6 - 2x^4 + 2x^2 - 1 = 0 \]

### Step 2: Find the area between the curves
The area \(A\) between the curves from \(x = -1\) to \(x = 1\) can be computed as:
\[ A = \int_{-1}^{1} \left( \sqrt{1 - x^2} - (x^3 - x) \right) \, dx \]

### Step 3: Calculate the area
We can split the integral:
\[ A = \int_{-1}^{1} \sqrt{1 - x^2} \, dx - \int_{-1}^{1} (x^3 - x) \, dx \]

1. The first integral, \(\int_{-1}^{1} \sqrt{1 - x^2} \, dx\), represents the area of a semicircle with radius 1:
\[ \int_{-1}^{1} \sqrt{1 - x^2} \, dx = \frac{\pi}{2} \]

2. The second integral, \(\int_{-1}^{1} (x^3 - x) \, dx\), is zero because \(x^3 - x\) is an odd function:
\[ \int_{-1}^{1} (x^3 - x) \, dx = 0 \]

Thus, the total area \(A\) is:
\[ A = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 4: Divide the area by the y-axis
Since the area is symmetric about the y-axis, we can denote the area on the left side as \(A_1\) and the area on the right side as \(A_2\).

The area on the left side \(A_1\) is:
\[ A_1 = \frac{1}{2} A = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

The area on the right side \(A_2\) is also:
\[ A_2 = \frac{1}{2} A = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

### Step 5: Find the ratio \(\lambda:1\)
To find \(\lambda\), we need to express the areas in terms of the ratio:
\[ \frac{A_1}{A_2} = \frac{\frac{\pi}{4}}{\frac{\pi}{4}} = 1 \]

However, since we are asked for \(\lambda\) such that \(\lambda > 1\), we need to consider the areas more carefully.

The area \(A_1\) is the area under the semicircle minus the area under the cubic curve on the left side, and \(A_2\) is the area under the semicircle minus the area under the cubic curve on the right side.

### Final Calculation
After careful consideration, we find that:
\[ \lambda = \frac{A_1}{A_2} = \frac{\frac{\pi}{4} - \text{(area under cubic on left)}}{\frac{\pi}{4} - \text{(area under cubic on right)}} \]

Since the cubic function is symmetric, we can conclude that:
\[ \lambda = \frac{\frac{\pi}{4} - \frac{1}{4}}{\frac{\pi}{4} + \frac{1}{4}} = \frac{\pi - 1}{\pi + 1} \]

### Conclusion
Thus, the value of \(\lambda\) is:
\[ \lambda = \frac{\pi - 1}{\pi + 1} \]