Question 74

Let $$\beta(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$$, $$m, n > 0$$. If $$\int_0^1 (1 - x^{10})^{20} dx = a \times \beta(b, c)$$, then $$100(a + b + c)$$ equals

Solution

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We need to express $$\int_0^1 (1 - x^{10})^{20} dx$$ in terms of the Beta function $$\beta(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}dx$$.

Substitution: Let $$x^{10} = t$$, so $$dx = \frac{dt}{10t^{9/10}}$$.

$$ \int_0^1 (1-t)^{20} \cdot \frac{dt}{10t^{9/10}} = \frac{1}{10}\int_0^1 t^{1/10 - 1}(1-t)^{21-1} dt = \frac{1}{10}\beta\left(\frac{1}{10}, 21\right) $$

So $$a = \frac{1}{10}$$, $$b = \frac{1}{10}$$, $$c = 21$$.

$$ 100(a + b + c) = 100\left(\frac{1}{10} + \frac{1}{10} + 21\right) = 100 \times \frac{212}{10} = 2120 $$

The correct answer is Option 2: 2120.

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