If I _ 1 = ∫ _ 0 ^ 1 (1 - x ^ 50 ) ^ 100 dx and I _ 2 = ∫ _ 0 ^ 1...

Given

$\alpha = \dfrac{{I}_{2}}{{I}_{1}} = \dfrac{{∫}_{0}^{1}{\left(1 - {x}^{50}\right)}^{101}dx}{{∫}_{0}^{1}{\left(1 - {x}^{50}\right)}^{100}dx}$
Now,

${I}_{2} = {∫}_{0}^{1}\left(1 - {x}^{50}\right){\left(1 - {x}^{50}\right)}^{100}dx$
${I}_{2} = {I}_{1} - {∫}_{0}^{1}{x}_{.}{x}^{49}{\left(1 - {x}^{50}\right)}^{100}dx$
${I}_{2} = {I}_{1} - {\left[\dfrac{- x{\left(1 - {x}^{50}\right)}^{101}}{5050}\right]}_{0}^{1} - {∫}_{0}^{1}\dfrac{{\left(1 - {x}^{50}\right)}^{101}}{5050}dx$
${I}_{2} = {I}_{1} - \dfrac{{I}_{2}}{5050}$
$⇒ \alpha = \dfrac{{I}_{2}}{{I}_{1}} = \dfrac{5050}{5051}$