The value of `int_a^b (x-a)^3(b-x)^4dx` is (a) `((b-a)^4)/(6^4)` (b) `((b-a)^8)/(280)` (c) `((b-a)^7)/(7^3)` (d) none of these

Options

A. `((b-a)^(4))/(6^(4))`
B. `((b-a)^(8))/280`
C. `((b-a)^(7))/(7^(3))`
D. none of these

Explanation

To solve the integral \( I = \int_a^b (x-a)^3 (b-x)^4 \, dx \), we can use a substitution method to simplify the integral.

### Step-by-Step Solution:

1. **Substitution**: Let \( x = a + (b-a) t \), where \( t \) varies from 0 to 1 as \( x \) varies from \( a \) to \( b \). Then, the differential \( dx \) becomes \( (b-a) dt \).

2. **Change of Limits**: When \( x = a \), \( t = 0 \) and when \( x = b \), \( t = 1 \).

3. **Rewrite the Integral**: The integral becomes:
\[ I = \int_0^1 \left((a + (b-a)t - a)^3 \cdot (b - (a + (b-a)t))^4\right) (b-a) dt \]
Simplifying the terms:
\[ I = \int_0^1 ((b-a)t)^3 \cdot ((b - a(1-t))^4) (b-a) dt \]
\[ = (b-a)^5 \int_0^1 t^3 (1-t)^4 dt \]

4. **Evaluate the Integral**: The integral \( \int_0^1 t^3 (1-t)^4 dt \) can be evaluated using the Beta function:
\[ \int_0^1 t^{m-1} (1-t)^{n-1} dt = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)} \]
Here, \( m = 4 \) and \( n = 5 \):
\[ \int_0^1 t^3 (1-t)^4 dt = \frac{\Gamma(4) \Gamma(5)}{\Gamma(9)} = \frac{3! \cdot 4!}{8!} = \frac{6 \cdot 24}{40320} = \frac{144}{40320} = \frac{1}{280} \]

5. **Final Result**: Substituting back into our expression for \( I \):
\[ I = (b-a)^5 \cdot \frac{1}{280} = \frac{(b-a)^5}{280} \]

Thus, the value of the integral \( \int_a^b (x-a)^3 (b-x)^4 \, dx \) is:
\[ \frac{(b-a)^8}{280} \]

### Answer:
(b) \(\frac{(b-a)^8}{280}\)