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 integral \( I_n = \int_0^1 (1 - x^3)^n \, dx \), where \( n \in \mathbb{N} \), we can use integration by parts. Here’s the step-by-step solution:

### Step 1: Set Up Integration by Parts
We will use integration by parts, which states that:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = (1 - x^3)^n \)
- \( dv = dx \)

### Step 2: Differentiate and Integrate
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -3nx^2(1 - x^3)^{n-1} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply Integration by Parts
Now we can apply the integration by parts formula:
\[ I_n = \left[ x(1 - x^3)^n \right]_0^1 - \int_0^1 x \cdot (-3nx^2(1 - x^3)^{n-1}) \, dx \]

### Step 4: Evaluate the Boundary Terms
Evaluate the boundary terms:
\[ \left[ x(1 - x^3)^n \right]_0^1 = (1)(1 - 1)^n - (0)(1 - 0)^n = 0 - 0 = 0 \]
Thus, the first term is zero.

### Step 5: Simplify the Integral
Now we simplify the integral:
\[ I_n = 3n \int_0^1 x^3(1 - x^3)^{n-1} \, dx \]

### Step 6: Change of Variable
Let \( u = x^3 \), then \( du = 3x^2 \, dx \) or \( dx = \frac{du}{3x^2} = \frac{du}{3u^{2/3}} \):
- When \( x = 0 \), \( u = 0 \)
- When \( x = 1 \), \( u = 1 \)

Thus, the integral becomes:
\[ I_n = 3n \int_0^1 u(1 - u)^{n-1} \cdot \frac{du}{3u^{2/3}} = n \int_0^1 u^{1/3}(1 - u)^{n-1} \, du \]

### Step 7: Recognize the Beta Function
The integral \( \int_0^1 u^{1/3}(1 - u)^{n-1} \, du \) can be recognized as a Beta function:
\[ \int_0^1 u^{a-1}(1-u)^{b-1} \, du = B(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \]
Here, \( a = \frac{4}{3} \) and \( b = n \):
\[ I_n = n B\left(\frac{4}{3}, n\right) = n \cdot \frac{\Gamma\left(\frac{4}{3}\right) \Gamma(n)}{\Gamma\left(n + \frac{4}{3}\right)} \]

### Step 8: Conclusion
Thus, we have expressed \( I_n \) in terms of the Beta function. The final answer can be simplified further based on the properties of the Gamma function if needed.