Let (m,n)= _0^1x^m-1(1-x)^n-1dx,m,n0.

Question

Let $\beta (m,n)=\int _{0}^{1}{x}^{m-1}(1-x{)}^{n-1}dx,m,n>0$. If $\int _{0}^{1}{(1-{x}^{10})}^{20}dx=a	imes \beta (b,c)$, then $100(a+b+c)$ equals \_\_\_\_\_\_\_\_\_.

Accepted Answer

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

$$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$$

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:

$$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$$

which simplifies to:

$$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$$

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:

$$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$$

Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.

Now we calculate $ 100(a + b + c) $:

$$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$$

So, the answer is Option D, 2120.

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