Evaluate `int_(0)^(1)(tx+1-x)^(n)dx`, where n is a positive integer and t is a parameter independent of x . Hence , show that ∫ 0 1 ​ x k (1−x) n−k dx= [ n C k ​ (n+1)] P ​ fork=0,1,......n, then P=

Answer

To evaluate the integral

\[ I = \int_0^1 (tx + 1 - x)^n \, dx, \]

where \( n \) is a positive integer and \( t \) is a parameter independent of \( x \), we can follow the steps below:

### Step 1: Rewrite the Integral
We can rewrite the expression inside the integral:

\[ I = \int_0^1 (tx + (1 - x))^n \, dx = \int_0^1 ((t - 1)x + 1)^n \, dx. \]

### Step 2: Use the Binomial Theorem
Using the binomial theorem, we can expand \(((t - 1)x + 1)^n\):

\[ I = \int_0^1 \sum_{k=0}^{n} \binom{n}{k} (t - 1)^k x^k \, dx. \]

### Step 3: Interchange the Integral and Summation
We can interchange the integral and summation (justified by uniform convergence):

\[ I = \sum_{k=0}^{n} \binom{n}{k} (t - 1)^k \int_0^1 x^k \, dx. \]

### Step 4: Evaluate the Integral
The integral \(\int_0^1 x^k \, dx\) can be evaluated as:

\[ \int_0^1 x^k \, dx = \frac{1}{k+1}. \]

### Step 5: Substitute Back
Substituting this result back into the expression for \( I \):

\[ I = \sum_{k=0}^{n} \binom{n}{k} (t - 1)^k \cdot \frac{1}{k+1}. \]

### Step 6: Simplify the Expression
We can express this sum in a more compact form. Notice that:

\[ I = \frac{1}{n+1} \sum_{k=0}^{n} \binom{n}{k} (t - 1)^k. \]

### Step 7: Use the Binomial Theorem Again
Using the binomial theorem again, we have:

\[ \sum_{k=0}^{n} \binom{n}{k} (t - 1)^k = (1 + (t - 1))^n = t^n. \]

### Step 8: Final Expression for \( I \)
Thus, we can write:

\[ I = \frac{t^n}{n+1}. \]

### Step 9: Show the Required Result
Now, we need to show that:

\[ \int_0^1 x^k (1-x)^{n-k} \, dx = \frac{1}{n+1} \binom{n}{k}. \]

Using the result from \( I \):

\[ \int_0^1 x^k (1-x)^{n-k} \, dx = \frac{1}{n+1} \cdot \binom{n}{k}, \]

which matches the required form.

### Conclusion
The value of \( P \) is \( 1 \).