If `m,n in N,` then the value of `int_a^b(x-a)^m(b-x)^n dx` is equal to

Options

A. `((b-a)^(m+n). M!n!)/((m+n))!`
B. `((b-a)^(m+n+1).m!n!)/((m+n+1)!`
C. `((b-a)^(m).m!)/(m!)`
D. None of these

Answer

To solve the integral \( I = \int_a^b (x-a)^m (b-x)^n \, dx \), where \( m, n \in \mathbb{N} \), we will use integration by parts and properties of definite integrals. Let's go through the steps systematically.

### Step 1: Define the Integral
Let:
\[ I = \int_a^b (x-a)^m (b-x)^n \, dx \]

### Step 2: Use Integration by Parts
We will apply integration by parts. We can choose:
- \( u = (b-x)^n \) (which we will differentiate)
- \( dv = (x-a)^m \, dx \) (which we will integrate)

Then, we find:
- \( du = -n(b-x)^{n-1} \, dx \)
- \( v = \frac{(x-a)^{m+1}}{m+1} \)

### Step 3: Apply the Integration by Parts Formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = \left[ (b-x)^n \cdot \frac{(x-a)^{m+1}}{m+1} \right]_a^b - \int_a^b \frac{(x-a)^{m+1}}{m+1} \cdot (-n(b-x)^{n-1}) \, dx \]

### Step 4: Evaluate the Boundary Terms
Evaluating the boundary terms:
1. At \( x = b \): \( (b-b)^n \cdot \frac{(b-a)^{m+1}}{m+1} = 0 \)
2. At \( x = a \): \( (b-a)^n \cdot \frac{(a-a)^{m+1}}{m+1} = 0 \)

Thus, both boundary terms are zero:
\[ I = 0 - \int_a^b \frac{(x-a)^{m+1}}{m+1} \cdot (-n(b-x)^{n-1}) \, dx \]
\[ I = \frac{n}{m+1} \int_a^b (x-a)^{m+1} (b-x)^{n-1} \, dx \]

### Step 5: Set Up a Recursive Relation
Let:
\[ I_{m,n} = \int_a^b (x-a)^m (b-x)^n \, dx \]
Then we have:
\[ I_{m,n} = \frac{n}{m+1} I_{m+1,n-1} \]

### Step 6: Continue the Process
We can continue this process recursively. Each time we apply integration by parts, we reduce the power of \( n \) until \( n = 0 \):
\[ I_{m,0} = \int_a^b (x-a)^m (b-x)^0 \, dx = \int_a^b (x-a)^m \, dx \]
This integral can be computed directly:
\[ I_{m,0} = \frac{(b-a)^{m+1}}{m+1} \]

### Step 7: General Formula
Continuing the recursion leads to:
\[ I_{m,n} = \frac{n!}{(m+1)(m+2)(m+3)\ldots(m+n)} \cdot (b-a)^{m+n+1} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{m! n! (b-a)^{m+n+1}}{(m+n+1)!} \]