Question

solve the integration​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\displaystyle\int_0^1\sf \: {x}^{7} {(1 - x)}^{6} \: dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\sf \: {x}^{7} {(1 - x)}^{6} \: dx$

We know the property, Beta function

$\rm :\longmapsto\:B(m, n) \: = \: \displaystyle\int_0^1\sf \: {x}^{m - 1} {(1 - x)}^{n - 1} \: dx \: where \: m \: and \: n \: > 0$

So, using this property, the above integral can be rewritten as

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\sf \: {x}^{8 - 1} {(1 - x)}^{7 - 1} \: dx$

$\bf\implies \:I = B(8, 7)$

We know,

$\green{ \boxed{ \bf \: B(m, n) = \dfrac{\Gamma \: m \: \Gamma \: n}{\Gamma \: (m + n)} }}$

So, using this,

$\rm :\longmapsto\:I = \dfrac{\Gamma \: 8 \: \Gamma \: 7}{\Gamma(8 + 7)}$

We know,

$\red{ \boxed{ \bf \: \Gamma \: n = (n - 1)! \: \: if \: n \: \in \: N}}$

So, using this,

$\rm :\longmapsto\:I \: = \: \dfrac{7! \: \times \: 6!}{14!}$

$\rm :\longmapsto\:I \: = \: \dfrac{7! \: \times \: 6 \times 5 \times 4 \times 3 \times 2}{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}$

$\bf :\implies\:I = \dfrac{1}{24024}$

Hence,

$\bf :\longmapsto\:\displaystyle\int_0^1\bf \: {x}^{7} {(1 - x)}^{6} \: dx = \dfrac{1}{24024}$

Additional Information :-

$\rm :\longmapsto\:B(m, n) = B(n, m) \: where \: m \: and \: n \: > 0$

$\rm :\longmapsto\:B(m, n) = \displaystyle\int_0^ \infty \sf \dfrac{ {x}^{m - 1} }{ {(1 + x)}^{m + n} } dx$

$\rm :\longmapsto\:\Gamma \: \dfrac{1}{2} = \sqrt{\pi}$

$\rm :\longmapsto\:\Gamma \: (1) = 1$

$\rm :\longmapsto\:\Gamma \: (x + 1) = x \: \Gamma \: (x)$