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

Answer: (D)

2120

Answer 2

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).