$$ \text{If } I(m,n)=\int_{0}^{1} x^{m-1}(1-x)^{\,n-1}\,dx,\quad m,n>0, \text{ then } I(9,14)+I(10,13) \text{ is:} $$

$$I(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}dx = B(m,n)$$ (Beta function). Find $$I(9,14) + I(10,13)$$.

We start by recalling the Beta function identity:

$$B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

Applying this to the integral with parameters (9,14) gives

$$I(9,14) = \frac{\Gamma(9)\Gamma(14)}{\Gamma(23)} = \frac{8! \cdot 13!}{22!}$$

Similarly, for (10,13) we have

$$I(10,13) = \frac{\Gamma(10)\Gamma(13)}{\Gamma(23)} = \frac{9! \cdot 12!}{22!}$$

Adding these two expressions yields

$$I(9,14) + I(10,13) = \frac{8! \cdot 13! + 9! \cdot 12!}{22!} = \frac{8! \cdot 12!(13 + 9)}{22!} = \frac{8! \cdot 12! \cdot 22}{22!}$$

This simplifies further as

$$\frac{22 \cdot 8! \cdot 12!}{22!}$$

On the other hand, using the same Beta function formula for (9,13) gives

$$I(9,13) = \frac{8! \cdot 12!}{21!}$$

Hence,

$$\frac{22 \cdot 8! \cdot 12!}{22!} = \frac{22 \cdot 8! \cdot 12!}{22 \cdot 21!} = \frac{8! \cdot 12!}{21!} = I(9,13)$$

The correct answer is Option 4: I(9, 13).