If `A_(n)` is the area bounded by y=x and `y=x^(n), n in N,` then `A_(2).A_(3)…A_(n)=`

Options

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

Explanation

To solve the problem, we need to find the area \( A_n \) bounded by the curves \( y = x \) and \( y = x^n \) for \( n \in \mathbb{N} \), and then compute the product \( A_2 \cdot A_3 \cdot \ldots \cdot A_n \).

### Step-by-Step Solution:

1. **Identify the curves**:
We have two equations:
- \( y = x \) (a straight line)
- \( y = x^n \) (a polynomial curve)

2. **Find the points of intersection**:
Set \( x = x^n \) to find the points where the two curves intersect. This gives us:
\[ x^n - x = 0 \implies x(x^{n-1} - 1) = 0 \]
The solutions are \( x = 0 \) and \( x = 1 \).

3. **Set up the integral for the area**:
The area \( A_n \) between the curves from \( x = 0 \) to \( x = 1 \) is given by:
\[ A_n = \int_0^1 (x - x^n) \, dx \]

4. **Evaluate the integral**:
\[ A_n = \int_0^1 x \, dx - \int_0^1 x^n \, dx \]
- The first integral:
\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \]
- The second integral:
\[ \int_0^1 x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1} \]
Thus,
\[ A_n = \frac{1}{2} - \frac{1}{n+1} = \frac{n - 1}{2(n + 1)} \]

5. **Calculate the product \( A_2 \cdot A_3 \cdot \ldots \cdot A_n \)**:
We need to compute:
\[ A_2 \cdot A_3 \cdot \ldots \cdot A_n = \prod_{k=2}^{n} A_k = \prod_{k=2}^{n} \frac{k - 1}{2(k + 1)} \]
This can be simplified as:
\[ = \frac{1}{2^{n-1}} \cdot \frac{1 \cdot 2 \cdot 3 \cdots (n-1)}{3 \cdot 4 \cdots (n+1)} \]
The numerator is \( (n-1)! \) and the denominator can be expressed as:
\[ 3 \cdot 4 \cdots (n+1) = \frac{(n+1)!}{2} \]
Therefore, we have:
\[ A_2 \cdot A_3 \cdots A_n = \frac{1}{2^{n-1}} \cdot \frac{(n-1)!}{\frac{(n+1)!}{2}} = \frac{2(n-1)!}{2^{n-1}(n+1)!} \]
Simplifying gives:
\[ = \frac{(n-1)!}{2^{n-2}(n+1)!} \]

### Final Answer:
\[ A_2 \cdot A_3 \cdots A_n = \frac{(n-1)!}{2^{n-2}(n+1)!} \]