If `m gt 0, n gt 0`, the definite integral `l=int_(0)^(1)x^(m-1)(1-x)^(n-1)dx` depends upon the vlaues of m and n and is denoted by `beta(m,n)`, called the beta function. E.g.
`int_(0)^(1)x^(4)(1-x)^(5)dx=int_(0)^(1)x^(5-1)(1-x)^(6-1)dx=beta(5, 6) and int_(0)^(1)x^(5//2)(1-x)^(-1//2)dx=int_(0)^(1)x^(7//2-1)(1-x)^(1//2-1)dx=beta((7)/(2),(1)/(2))`. Obviously, `beta(n, m)=beta(m, n)`.
If `int_(0)^(n)(1-(x)/(n))^(n)x^(k-1)dx=R beta(k, n+1)`, then R is equal to

Options

A. n
B. `n^(kn)`
C. `n^(k)`
D. none of these

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{0}^{n} \left(1 - \frac{x}{n}\right)^{n} x^{k-1} \, dx \]

and express it in terms of the Beta function.

### Step 1: Change of Variables
We will perform a change of variables to simplify the integral. Let

\[ t = \frac{x}{n} \implies x = nt \quad \text{and} \quad dx = n \, dt. \]

The limits of integration change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = n \), \( t = 1 \)

Substituting these into the integral gives:

\[ I = \int_{0}^{1} \left(1 - t\right)^{n} (nt)^{k-1} n \, dt. \]

### Step 2: Simplifying the Integral
Now, we can simplify the integral:

\[ I = n^k \int_{0}^{1} (1 - t)^{n} t^{k-1} \, dt. \]

### Step 3: Recognizing the Beta Function
The integral

\[ \int_{0}^{1} (1 - t)^{n} t^{k-1} \, dt \]

is recognized as the Beta function \( \beta(k, n + 1) \). Therefore, we can express \( I \) as:

\[ I = n^k \beta(k, n + 1). \]

### Step 4: Relating to the Given Expression
According to the problem, we have:

\[ I = R \beta(k, n + 1). \]

From our expression for \( I \), we can equate:

\[ R \beta(k, n + 1) = n^k \beta(k, n + 1). \]

### Step 5: Solving for \( R \)
To find \( R \), we can divide both sides by \( \beta(k, n + 1) \) (assuming \( \beta(k, n + 1) \neq 0 \)):

\[ R = n^k. \]

Thus, the value of \( R \) is

\[ \boxed{n^k}. \]