Let $\beta(\mathrm{m}, \mathrm{n})=\int_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}0$. If $\int_0^1\left(1-x^{10} ight)^{20} \mathrm{~d} x=\mathrm{a} imes \beta(\mathrm{b}, \mathrm{c})$, then $100(\mathrm{a}+\mathrm{b}+\mathrm{c})$ equals____

Question

Let $\beta(\mathrm{m}, \mathrm{n})=\int_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0$. If $\int_0^1\left(1-x^{10}ight)^{20} \mathrm{~d} x=\mathrm{a} 	imes \beta(\mathrm{b}, \mathrm{c})$, then $100(\mathrm{a}+\mathrm{b}+\mathrm{c})$ equals____

Aicharya · Accepted Answer

$I=\int_0^1 1 \cdot\left(1-x^{10}\right)^{20} d x$ $x^{10}=t$ $x=t^{1 / 10}$ $d x=\frac{1}{10}(t)^{-9 / 10} d t$ $I=\int_0^1(1-t)^{20} \frac{1}{10}(t)^{-9 / 10} d t$ $I=\frac{1}{10} \int_0^1 t^{-9 / 10}(1-t)^{20} d t$ $a=\frac{1}{10} \quad b=\frac{1}{10} \quad c=21$

Let $\beta(\mathrm{m}, \mathrm{n})=\int_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m},…

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