Mathematics - Definite Integration Question with Solution | TestHub

Given,

$\alpha \left(m,n\right)={\int }_0^2{t}^m{\left(1+3t\right)}^{n}dt$

Now using integration by parts we get,

$\Rightarrow \alpha \left(m,n\right)={\left.{\left(1+3t\right)}^n\cdot \frac{t^{m+1}}{m+1}\right|}_0^2-{\int }_0^{2}n{\left(1+3t\right)}^{n-1}\times 3\cdot \frac{t^{m+1}}{m+1}dt$

$\Rightarrow \left(m+1\right)\alpha \left(m,n\right)={\left.{\left(1+3t\right)}^n\cdot \left(t^{m+1}\right)\right|}_0^2-3n\left(m+1\right){\int }_0^2{\left(1+3t\right)}^{n-1}\times t^{m+1}dt$

$\Rightarrow \left(m+1\right)\alpha \left(m,n\right)={\left(1+3\times 2\right)}^n\cdot \left(2^{m+1}\right)-3n\left(m+1\right)\alpha \left(m+1,n-1\right)$

$\Rightarrow \left(m+1\right)\alpha \left(m,n\right)=7^n·2^{m+1}-3n\alpha \left(m+1,n-1\right)$

$\Rightarrow \left(m+1\right)\alpha \left(m,n\right)+3n\alpha \left(m+1,n-1\right)=7^n·2^{m+1}$

Now put $m=10, n=6$ in above equation we get,

$11\alpha \left(10,6\right)+18\alpha \left(11,5\right)=7^6·2^{11}=32\times {\left(14\right)}^6$

Hence, on comparing with $11\alpha \left(10,6\right)+18\alpha \left(11,5\right)=p\times {\left(14\right)}^6$ we get,

$\Rightarrow p=32$