Let \beta(\mathrm{m}, \mathrm{n})=\int\limits_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}0. If \int\limits_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\limits_0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0. If \int\limits_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 _________.

Accepted Answer

Answer: D. 2120 First, let's rewrite the given integral using the given form of the Beta function. The given integral is: \int\limits_0^1\left(1-x^{10}ight)^{20} \mathrm{~d} x To use the Beta function, let us make a substitution. Let x^{10} = t . Then, dx = \frac{1}{10}t^{-\frac{9}{10}} dt or dx = \frac{1}{10} t^{-\frac{9}{10}} dt . The limits of integration change as follows: when x = 0 , t = 0 , and when x = 1 , t = 1 . Substituting these into the integral, we have: \int\limits_0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt which simplifies to: \frac{1}{10} \int\limits_0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt We recognize this integral as a Beta function \beta(m, n) where m = 1 - \frac{9}{10} = \frac{1}{10} and n = 20 + 1 = 21 . Therefore, we can write this as: \frac{1}{10} \beta \left( \frac{1}{10}, 21 ight) Comparing this to a 	imes \beta(b, c) , we have a = \frac{1}{10} , b = \frac{1}{10} , and c = 21 . Now we calculate 100(a + b + c) : 100 \left( \frac{1}{10} + \frac{1}{10} + 21 ight) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 ight) = 100 \left( \frac{1}{5} + 21 ight) = 100 \left( \frac{1}{5} + \frac{105}{5} ight) = 100 \left( \frac{106}{5} ight) = 100 	imes 21.2 = 2120 So, the answer is Option D, 2120.

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