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Step-by-Step Solution
Step 1: Identify the Given Integral and Expression
We are given:
\alpha(m, n) = \int_{0}^{2} t^{m} \, (1 + 3t)^{n} \, dt
and the expression:
11 \,\alpha(10, 6) \;+\; 18 \,\alpha(11, 5) \;=\; p \,(14)^6.
Our goal is to find the value of p .
Step 2: Write Out the Given Expression in Integral Form
From the definition of \alpha(m, n) , we can rewrite:
11 \int_{0}^{2} t^{10} (1 + 3t)^{6} \, dt
\;+\;
18 \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt
\;=\;
p \,(14)^6.
Step 3: Analyze the First Integral
Consider the first integral
\int_{0}^{2} t^{10} \,(1 + 3t)^{6} \, dt.
Notice that t^{10} = \frac{d}{dt} \Bigl( \frac{t^{11}}{11} \Bigr).
Hence,
\int_{0}^{2} t^{10} \,(1 + 3t)^{6} \, dt
=
\int_{0}^{2} \frac{d}{dt}\Bigl(\frac{t^{11}}{11}\Bigr) (1 + 3t)^{6} \, dt.
Using integration by parts (or recognizing the product of a derivative with a function):
\int_{0}^{2} t^{10} \,(1 + 3t)^{6} \, dt
=
\left[ \frac{t^{11}}{11}\,(1 + 3t)^{6} \right]_{0}^{2}
\;-\;
\int_{0}^{2} \frac{t^{11}}{11} \,\frac{d}{dt}\bigl((1 + 3t)^{6}\bigr) \, dt.
Since \frac{d}{dt} \bigl((1 + 3t)^{6}\bigr) = 6 \cdot (1 + 3t)^{5} \cdot 3 = 18 (1 + 3t)^{5},
we have
\int_{0}^{2} t^{10} \,(1 + 3t)^{6} \, dt
=
\left[ \frac{t^{11}}{11}\,(1 + 3t)^{6} \right]_{0}^{2}
\;-\;
\frac{18}{11} \int_{0}^{2} t^{11} \,(1 + 3t)^{5} \, dt.
Evaluating the boundary term:
\left[ \frac{t^{11}}{11}\,(1 + 3t)^{6} \right]_{0}^{2}
=
\frac{2^{11}}{11} \,(1 + 3 \times 2)^{6}
=
\frac{2^{11}}{11}\,7^{6}.
Therefore,
\int_{0}^{2} t^{10} \,(1 + 3t)^{6} \, dt
=
\frac{2^{11}\,7^{6}}{11}
\;-\;
\frac{18}{11} \int_{0}^{2} t^{11} \,(1 + 3t)^{5} \, dt.
Step 4: Substitute into the Original Sum
The original sum is
11 \int_{0}^{2} t^{10} (1 + 3t)^{6} \, dt
\;+\;
18 \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt
\;=\;
p \,(14)^6.
Substitute the expression for the first integral:
11 \Bigl(\frac{2^{11}\,7^6}{11} \;-\; \frac{18}{11} \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt \Bigr)
\;+\;
18 \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt
=
p \,(14)^6.
Distributing the factor of 11:
11 \times \frac{2^{11}\,7^6}{11} = 2^{11}\,7^6,
and
-\,11 \times \frac{18}{11} \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt
=
-\,18 \int_{0}^{2} t^{11} (1 + 3t)^{5} \, dt.
These terms combine with the remaining integral piece:
-\,18 \int_{0}^{2} (\dots) + 18 \int_{0}^{2} (\dots) = 0.
Thus, the entire expression simplifies to
2^{11}\,7^6 = p \,(14)^6.
Step 5: Solve for p
Observe that
(14)^6 = (2 \times 7)^6 = 2^6 \cdot 7^6.
Therefore,
2^{11}\,7^6 = p \,\bigl(2^6 \cdot 7^6\bigr).
Dividing both sides by 2^6 \,7^6 , we get
p = \frac{2^{11}\,7^6}{2^6 \, 7^6} = 2^{11-6} = 2^5 = 32.
Final Answer
The value of p is 32.
