Let `I_(n) = int_(0)^(1)(1-x^(3))^(n)dx, (nin N)` then

Options

A. `3n I_(n) = (3n-1)I_(n-1) AA in ge 2`
B. `(3n-1)I_(n) = 3n I_(n-1) AA n ge 2`
C. `(3n-1)I_(n)= (3n+1)I_(n-1) AA n ge 2`
D. `(3n+1)I_(n)=3nI_(n-1) AA n ge 2`

Answer

To solve the problem, we need to evaluate the integral \( I_n = \int_0^1 (1 - x^3)^n \, dx \) for \( n \in \mathbb{N} \). We will use integration by parts to simplify the expression.

### Step-by-Step Solution:

1. **Integration by Parts Setup**:
We choose:
- \( u = (1 - x^3)^n \) (First function)
- \( dv = dx \) (Second function)

Then, we differentiate \( u \) and integrate \( dv \):
- \( du = -3nx^2(1 - x^3)^{n-1} \, dx \)
- \( v = x \)

2. **Applying Integration by Parts**:
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_n = \left[ x(1 - x^3)^n \right]_0^1 - \int_0^1 x \cdot (-3nx^2(1 - x^3)^{n-1}) \, dx \]

3. **Evaluating the Boundary Terms**:
Now we evaluate the boundary terms:
- At \( x = 1 \): \( 1 - 1^3 = 0 \), so \( x(1 - x^3)^n = 0 \).
- At \( x = 0 \): \( 0(1 - 0^3)^n = 0 \).

Therefore, the boundary term evaluates to \( 0 - 0 = 0 \).

4. **Simplifying the Integral**:
This gives us:
\[ I_n = 0 + 3n \int_0^1 x^3(1 - x^3)^{n-1} \, dx \]
We can rewrite this as:
\[ I_n = 3n \int_0^1 x^3(1 - x^3)^{n-1} \, dx \]

5. **Substituting the Integral**:
Let \( I_{n-1} = \int_0^1 (1 - x^3)^{n-1} \, dx \). We can express the integral in terms of \( I_{n-1} \):
\[ I_n = 3n \cdot \frac{1}{3} I_{n-1} = n I_{n-1} \]

6. **Final Recursion Relation**:
Thus, we have the recursion relation:
\[ I_n = n I_{n-1} \]

### Conclusion:
The integral \( I_n \) can be expressed recursively as:
\[ I_n = n I_{n-1} \]