1. We are given

$$I(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx, \qquad m,n>0.$$

We need to find

$$I(9,14)+I(10,13).$$

2. Write both integrals explicitly:

$$I(9,14)=\int_0^1 x^{8}(1-x)^{13}\,dx,$$

and

$$I(10,13)=\int_0^1 x^{9}(1-x)^{12}\,dx.$$

3. Add them:

$$I(9,14)+I(10,13) =\int_0^1 \left[x^8(1-x)^{13}+x^9(1-x)^{12}\right]dx.$$

4. Factor the integrand:

$$x^8(1-x)^{12}\big[(1-x)+x\big].$$

Since

$$(1-x)+x=1,$$

we get

$$x^8(1-x)^{13}+x^9(1-x)^{12}=x^8(1-x)^{12}.$$

Therefore,

$$I(9,14)+I(10,13)=\int_0^1 x^8(1-x)^{12}\,dx.$$

5. Compare with the definition of $I(m,n)$:

$$I(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx.$$

Here,

$$m-1=8 \Rightarrow m=9, \qquad n-1=12 \Rightarrow n=13.$$

So,

$$I(9,14)+I(10,13)=I(9,13).$$

6. Check options:

• A: $I(9,1)$ — incorrect
• B: $I(1,13)$ — incorrect
• C: $I(19,27)$ — incorrect
• D: $I(9,13)$ — correct

Hence the correct answer is

$$\boxed{I(9,13)}.$$