Question

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

• A. 1021
• B. 1120
• C. 2012
• D. 2120

Answer 1

The Correct Option is D

Given:

\[ \int_0^1 (1 - x^{10})^{20} dx = a \times \beta(b, c). \]

Step 1: Comparison with the Beta Function

Recall the beta function definition:

\[ \beta(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx. \]

To match this form, let \( u = x^{10} \). Then, \( du = 10x^9 dx \), or:

\[ dx = \frac{du}{10x^9}. \]

Changing the limits of integration, when \( x \) varies from 0 to 1, \( u \) also varies from 0 to 1. Thus, the integral becomes:

\[ \int_0^1 (1 - u)^{20} \cdot \frac{du}{10x^9}. \]

Since \( x = u^{1/10} \), we have:

\[ x^9 = u^{9/10}, \quad \text{so} \quad \frac{1}{10x^9} = \frac{1}{10}u^{-9/10}. \]

The integral becomes:

\[ \int_0^1 (1 - u)^{20} \cdot \frac{1}{10}u^{-9/10} du = \frac{1}{10} \int_0^1 u^{-9/10}(1 - u)^{20} du. \]

Comparing this with the beta function form:

\[ \beta\left(\frac{1}{10}, 21\right) = \int_0^1 u^{\frac{1}{10}-1}(1-u)^{20} du. \]

Thus, we can set:

\[ a = \frac{1}{10}, \quad b = \frac{1}{10}, \quad c = 21. \]

Step 2: Calculate the Expression

Now, we need to evaluate:

\[ 100(a + b + c) = 100 \left(\frac{1}{10} + \frac{1}{10} + 21\right) = 100 \times \left(0.2 + 21\right) = 100 \times 21.2 = 2120. \]

Therefore, the correct answer is Option (4).