1. We are given

$$\alpha(m,n)=\int_0^2 t^m(1+3t)^n\,dt$$

and need to evaluate

$$11\alpha(10,6)+18\alpha(11,5).$$

2. Substitute the definition of $\alpha$:

$$11\alpha(10,6)+18\alpha(11,5) =11\int_0^2 t^{10}(1+3t)^6dt+18\int_0^2 t^{11}(1+3t)^5dt.$$

Combine into one integral:

$$=\int_0^2 \left[11t^{10}(1+3t)^6+18t^{11}(1+3t)^5\right]dt.$$

Factor common terms:

$$=\int_0^2 t^{10}(1+3t)^5\left[11(1+3t)+18t\right]dt.$$

Now simplify the bracket:

$$11(1+3t)+18t=11+33t+18t=11+51t.$$

So,

$$11\alpha(10,6)+18\alpha(11,5)=\int_0^2 t^{10}(1+3t)^5(11+51t)\,dt.$$

3. Notice a derivative pattern:

$$\frac{d}{dt}\left[t^{11}(1+3t)^6\right] =11t^{10}(1+3t)^6+t^{11}\cdot 6(1+3t)^5\cdot 3.$$

Thus,

$$\frac{d}{dt}\left[t^{11}(1+3t)^6\right] =11t^{10}(1+3t)^6+18t^{11}(1+3t)^5.$$

This is exactly the integrand before combining. Therefore,

$$11\alpha(10,6)+18\alpha(11,5) =\int_0^2 \frac{d}{dt}\left[t^{11}(1+3t)^6\right]dt.$$

4. Apply the Fundamental Theorem of Calculus:

$$=\left[t^{11}(1+3t)^6\right]_0^2.$$

At $t=2$,

$$2^{11}(1+3\cdot 2)^6=2^{11}\cdot 7^6.$$

At $t=0$,

$$0^{11}(1+0)^6=0.$$

Hence,

$$11\alpha(10,6)+18\alpha(11,5)=2^{11}7^6.$$

5. We are told

$$11\alpha(10,6)+18\alpha(11,5)=p(14)^6.$$

Since

$$14^6=(2\cdot 7)^6=2^6\cdot 7^6,$$

we get

$$p\cdot 14^6=p\cdot 2^6\cdot 7^6=2^{11}7^6.$$

Cancel $7^6$:

$$p\cdot 2^6=2^{11}.$$

So,

$$p=2^{11-6}=2^5=32.$$

6. Therefore, the required integer is

$$\boxed{32}.$$