For m, n > 0, let α(m, n)=∫0² tᵐ(1+3 t)ⁿ d t. If 11 α(10,6)+18 α(11,5)=p(14)⁶, then p is equal to ___________.

Solution

Maths sir · 4 steps

Integrate α(10,6) by parts once, and the α(11,5) term appears and cancels out.

1. Choose the parts. In α(10,6), differentiate (1+3t)⁶ and integrate t¹⁰ to get t¹¹/11. Differentiating (1+3t)⁶ gives 18(1+3t)⁵. That brings t¹¹(1+3t)⁵ into the new integral, which is exactly the integrand of α(11,5).

   α(10,6) = [t¹¹(1+3t)⁶/11]₀² - 18/11∫₀² t¹¹(1+3t)⁵ dt

2. Rearrange. The last integral is α(11,5). Multiply through by 11 to get 11 α(10,6) + 18 α(11,5) = [t¹¹(1+3t)⁶]₀².
3. Evaluate limits. At t = 2 the boundary term is 2¹¹ × 7⁶, and at t = 0 it is 0. Write 2¹¹ × 7⁶ = 2⁵ × (2 × 7)⁶ = 32 × 14⁶.
4. So. p(14)⁶ = 32 × 14⁶, which gives p = 32.

The idea to keep. When a question combines two integrals with shifted powers, one round of integration by parts usually links them. Pick the parts so the new integral matches the other term, and only the boundary value is left.

Answer from the NTA final answer key. Solution by PrepWiser.