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

Updated On: Nov 7, 2025
  • 1021
  • 1120
  • 2012
  • 2120
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Approach Solution - 1

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).

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Step 1: Given information.
We are given the standard Beta function:
β(m, n) = ∫₀¹ xm−1(1 − x)n−1 dx, where m, n > 0.

We need to express ∫₀¹ (1 − x¹⁰)²⁰ dx in terms of aβ(b, c).

Step 2: Substitution.
Let x¹⁰ = t ⇒ x = t¹ᐟ¹⁰ ⇒ dx = (1/10)t−9/10 dt.
When x = 0 ⇒ t = 0; and when x = 1 ⇒ t = 1.

Substitute into the integral:
∫₀¹ (1 − x¹⁰)²⁰ dx = ∫₀¹ (1 − t)²⁰ × (1/10)t−9/10 dt.

Step 3: Simplify and match with the Beta function form.
= (1/10) ∫₀¹ t(1/10)−1(1 − t)20 dt.

Comparing with the definition of β(m, n):
m − 1 = (1/10) − 1 ⇒ m = 1/10.
n − 1 = 20 ⇒ n = 21.

Hence, a = 1/10, b = 1/10, and c = 21.

Step 4: Compute the required value.
We are asked for 100(a + b + c).
So, 100 × (1/10 + 1/10 + 21) = 100 × (21.2) = 2120.

Final Answer: 2120
Was this answer helpful?
0
0