For m, n 0,let alpha(m, n)=int_0^2 t^m(1+3 | vedclass

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(B) Given $\alpha(m, n) = \int_0^2 t^m(1+3t)^n dt$.We need to evaluate $11\alpha(10, 6) + 18\alpha(11, 5)$.Consider the integral $I = \int_0^2 t^{10}(

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$32$

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(B) Given $\alpha(m, n) = \int_0^2 t^m(1+3t)^n dt$.We need to evaluate $11\alpha(10, 6) + 18\alpha(11, 5)$.Consider the integral $I = \int_0^2 t^{10}(1+3t)^6 dt$.Using integration by parts,let $u = (1+3t)^6$ and $dv = t^{10} dt$.Then $du = 6(1+3t)^5 \cdot 3 dt = 18(1+3t)^5 dt$ and $v = \frac{t^{11}}{11}$.So,$\alpha(10, 6) = \left[ \frac{t^{11}}{11}(1+3t)^6 \right]_0^2 - \int_0^2 \frac{t^{11}}{11} \cdot 18(1+3t)^5 dt$.Multiplying by $11$:$11\alpha(10, 6) = \left[ t^{11}(1+3t)^6 \right]_0^2 - 18 \int_0^2 t^{11}(1+3t)^5 dt$.$11\alpha(10, 6) = 2^{11}(1+3(2))^6 - 0 - 18\alpha(11, 5)$.$11\alpha(10, 6) + 18\alpha(11, 5) = 2^{11}(7)^6$.$11\alpha(10, 6) + 18\alpha(11, 5) = 2^5 \cdot 2^6 \cdot 7^6 = 32 \cdot (2 \cdot 7)^6 = 32(14)^6$.Comparing this with $p(14)^6$,we get $p = 32$.

For $m, n > 0$,let $\alpha(m, n)=\int_0^2 t^m(1+3 t)^n d t$. If $11 \alpha(10,6)+18 \alpha(11,5)= p (14)^6$,then $p$ is equal to $......$.

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