Key Concepts and Formulas

• Integration by Parts (IBP): For definite integrals, $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$. This is used to integrate products of functions.
• Properties of Exponents: $a^m \cdot a^n = a^{m+n}$, $\frac{a^m}{a^n} = a^{m-n}$, $(ab)^n = a^n b^n$.
• Chain Rule for Differentiation: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

Step 1: Understand the Problem and Define $\alpha(m, n)$ We are given a function $\alpha(m, n)$ defined by a definite integral: \alpha(m, n)=\int_\limits{0}^{2} t^{m}(1+3 t)^{n} d t We are also provided with an equation involving $\alpha(10,6)$ and $\alpha(11,5)$: $11 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6}$ Our objective is to find the value of $p$. Substituting the definition of $\alpha(m, n)$ into the given equation yields: 11 \int_\limits{0}^{2} t^{10}(1+3 t)^{6} d t + 18 \int_\limits{0}^{2} t^{11}(1+3 t)^{5} d t = p(14)^{6}

Step 2: Apply Integration by Parts to one of the Integrals Observe the structure of the two integrals. The powers of $t$ and $(1+3t)$ are related in a way that suggests Integration by Parts can be used to transform one integral into the other. Let's apply IBP to the first integral, \int_\limits{0}^{2} t^{10}(1+3 t)^{6} d t, with the aim of reducing the power of $(1+3t)$ from 6 to 5 and increasing the power of $t$ from 10 to 11.

Let I_1 = \int_\limits{0}^{2} t^{10}(1+3 t)^{6} d t. We choose:

• $u = (1+3t)^6$ (This will simplify upon differentiation)
• $dv = t^{10} dt$ (This is easily integrable)

Now, we compute $du$ and $v$:

• $du = \frac{d}{dt}((1+3t)^6) dt = 6(1+3t)^5 \cdot 3 dt = 18(1+3t)^5 dt$
• $v = \int t^{10} dt = \frac{t^{11}}{11}$

Applying the Integration by Parts formula for definite integrals: I_1 = \left[ u \cdot v \right]_0^2 - \int_\limits{0}^{2} v \, du I_1 = \left[ (1+3t)^6 \cdot \frac{t^{11}}{11} \right]_0^2 - \int_\limits{0}^{2} \frac{t^{11}}{11} \cdot 18(1+3t)^5 dt I_1 = \left[ \frac{t^{11}(1+3t)^6}{11} \right]_0^2 - \frac{18}{11} \int_\limits{0}^{2} t^{11}(1+3t)^5 dt

Step 3: Substitute the Result of IBP back into the Original Equation Now, substitute the expression for $I_1$ back into the equation from Step 1: 11 \left( \left[ \frac{t^{11}(1+3t)^6}{11} \right]_0^2 - \frac{18}{11} \int_\limits{0}^{2} t^{11}(1+3t)^5 dt \right) + 18 \int_\limits{0}^{2} t^{11}(1+3t)^5 d t = p(14)^{6} Distribute the $11$ into the parentheses: 11 \cdot \left[ \frac{t^{11}(1+3t)^6}{11} \right]_0^2 - 11 \cdot \frac{18}{11} \int_\limits{0}^{2} t^{11}(1+3t)^5 dt + 18 \int_\limits{0}^{2} t^{11}(1+3t)^5 d t = p(14)^{6} Simplify the terms: \left[ t^{11}(1+3t)^6 \right]_0^2 - 18 \int_\limits{0}^{2} t^{11}(1+3t)^5 dt + 18 \int_\limits{0}^{2} t^{11}(1+3t)^5 d t = p(14)^{6} The two integral terms cancel each other out: $\left[ t^{11}(1+3t)^6 \right]_0^2 = p(14)^{6}$

Step 4: Evaluate the Remaining Definite Term Now, we evaluate the definite term $\left[ t^{11}(1+3t)^6 \right]_0^2$:

• At the upper limit $t=2$: $(2)^{11}(1+3 \cdot 2)^6 = 2^{11}(1+6)^6 = 2^{11} \cdot 7^6$
• At the lower limit $t=0$: $(0)^{11}(1+3 \cdot 0)^6 = 0^{11} \cdot 1^6 = 0$ So, the definite term evaluates to $2^{11} \cdot 7^6 - 0 = 2^{11} \cdot 7^6$.

Step 5: Solve for p Equate the evaluated definite term to the right side of the equation: $2^{11} \cdot 7^6 = p(14)^{6}$ We know that $14 = 2 \times 7$, so $14^6 = (2 \times 7)^6 = 2^6 \cdot 7^6$. Substitute this into the equation: $2^{11} \cdot 7^6 = p \cdot (2^6 \cdot 7^6)$ To find $p$, divide both sides by $2^6 \cdot 7^6$: $p = \frac{2^{11} \cdot 7^6}{2^6 \cdot 7^6}$ Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $p = 2^{11-6} = 2^5$ Calculate the value of $2^5$: $p = 32$

Common Mistakes & Tips

• Incorrectly applying the Chain Rule: When differentiating $(1+3t)^6$, remember to multiply by the derivative of the inner function, which is 3. Missing this factor is a common error.
• Algebraic errors with exponents: Be careful when simplifying expressions involving exponents, especially when dealing with products and quotients like $(2 \times 7)^6 = 2^6 \times 7^6$.
• Choosing $u$ and $dv$ incorrectly: If the choice of $u$ and $dv$ does not lead to simplification or cancellation, consider swapping them. In this case, choosing $u=t^{10}$ would increase the power of $t$ further, making it less likely to cancel.

Summary The problem was solved by strategically applying the Integration by Parts technique to one of the given integrals. By carefully selecting $u$ and $dv$, we were able to transform the integral \int_\limits{0}^{2} t^{10}(1+3 t)^{6} d t into a form that, when multiplied by 11, canceled out the second integral term. This simplification led to a single definite expression that was evaluated. Equating this result to $p(14)^6$ and using exponent properties allowed us to determine the value of $p$.

The final answer is $\boxed{32}$.