Question:

The integral \( \int_{0}^{1} x^{5/2} (1-x)^{3/2} dx \) is equal to:

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The Beta function is a powerful tool for solving definite integrals ranging from 0 to 1 involving powers of \( x \) and \( (1-x) \).
Updated On: Jun 9, 2026
  • \( \frac{3\pi}{128} \)
  • \( \frac{5\pi}{128} \)
  • \( \frac{3\pi}{256} \)
  • \( \frac{5\pi}{256} \)
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The Correct Option is C

Solution and Explanation

Concept: This integral is in the form of the Beta function, which is defined as \( \beta(m, n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)} \).

Step 1: Identify parameters m and n.
Comparing \( \int_{0}^{1} x^{5/2} (1-x)^{3/2} dx \) with the Beta function: \( m-1 = 5/2 \implies m = 7/2 \) \( n-1 = 3/2 \implies n = 5/2 \)

Step 2: Apply the Beta function formula.
$$I = \frac{\Gamma(7/2) \Gamma(5/2)}{\Gamma(7/2 + 5/2)} = \frac{\Gamma(7/2) \Gamma(5/2)}{\Gamma(6)} = \frac{\Gamma(7/2) \Gamma(5/2)}{5!}$$

Step 3: Evaluate Gamma functions.
Using \( \Gamma(n+1) = n \Gamma(n) \) and \( \Gamma(1/2) = \sqrt{\pi} \): \( \Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15}{8} \sqrt{\pi} \) \( \Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3}{4} \sqrt{\pi} \) \( 5! = 120 \)

Step 4: Calculate the final value.
$$I = \frac{\left( \frac{15}{8} \sqrt{\pi} \right) \left( \frac{3}{4} \sqrt{\pi} \right)}{120} = \frac{\frac{45}{32} \pi}{120} = \frac{45\pi}{3840} = \frac{3\pi}{256}$$ $$\boxed{\frac{3\pi}{256}}$$
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