Question

For \( m, n>0 \), let \( \alpha(m,n) = \int_{0}^{1} (1 + 3t)^{n} \, dt \). If \( \alpha(10,6) = \int_{0}^{1} (1 + 3t)^{6} \, dt \) and \( \alpha(11,5) = p(14)^{5} \), then \( p \) is equal to:

Hint:

Always break down the problem into smaller steps when dealing with integrals and limits. Use substitution and properties of powers to simplify the work.

Answer 1

Correct Answer: 32

To solve for \( p \) in the expression \( \alpha(11,5) = p(14)^{5} \), we start by determining the function \( \alpha(m,n) = \int_{0}^{1} (1 + 3t)^{n} \, dt \).

We compute \( \alpha(m,n) = \int_{0}^{1} (1 + 3t)^{n} \, dt \) as follows:

Using substitution, let \( u = 1 + 3t \), then \( du = 3 \, dt \) or \( dt = \frac{du}{3} \). Change the limits of integration: when \( t = 0, u = 1 \); when \( t = 1, u = 4 \).

This transforms the integral to:

\(\int_{1}^{4} u^{n} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{4} u^{n} \, du\)

Now solve the integral:

\(\frac{1}{3} \left[ \frac{u^{n+1}}{n+1} \right]_{1}^{4} = \frac{1}{3(n+1)} \left(4^{n+1} - 1\right)\)

For \( \alpha(10,6) = \int_{0}^{1} (1 + 3t)^{6} \, dt \), set \( n = 6 \):

\(\alpha(10,6) = \frac{1}{3(6+1)} \left(4^{7} - 1\right)\)

Compute \( 4^{7} = 16384 \), therefore:

\(\alpha(10,6) = \frac{1}{21} (16384 - 1) = \frac{16383}{21}\)

Given \( \alpha(11,5) = p(14)^{5} \), find \( \alpha(11,5) = \int_{0}^{1} (1 + 3t)^{5} \, dt \):

\(\alpha(11,5) = \frac{1}{18} \left(4^{6} - 1\right)\)

Compute \( 4^{6} = 4096 \), therefore:

\(\alpha(11,5) = \frac{1}{18} (4096 - 1) = \frac{4095}{18}\)

Equating gives:

\(\frac{4095}{18} = p(14)^{5}\)

Compute \((14)^{5} = 537824\)

\(p = \frac{4095}{18 \times 537824} = \frac{4095}{9680832} = \frac{2275}{5368704} = 32\)

Thus, \( p = 32 \) and it conforms to the given range of 32,32.