Question

Evaluate the integral: \[ \int_0^1 x^{5/2} (1 - x)^{3/2} \, dx = \]

• A. \( \dfrac{5\pi}{256} \)
• B. \( \dfrac{3\pi}{256} \)
• C. \( \dfrac{3\pi}{128} \)
• D. \( \dfrac{5\pi}{128} \)

Hint:

For integrals of the form \( \int_0^1 x^m (1 - x)^n \, dx \), use the Beta function formula: \( B(m + 1, n + 1) = \dfrac{\Gamma(m + 1)\Gamma(n + 1)}{\Gamma(m + n + 2)} \).

Answer 1

The Correct Option is B

We are given: \[ \int_0^1 x^{5/2} (1 - x)^{3/2} \, dx \] This integral is in the form of a Beta function: \[ \int_0^1 x^{m - 1}(1 - x)^{n - 1} dx = B(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)} \] Step 1: Match the form Let us write: \[ x^{5/2} = x^{(7/2) - 1},
(1 - x)^{3/2} = (1 - x)^{(5/2) - 1} \] So we can set: \[ m = \frac{7}{2},
n = \frac{5}{2} \] Step 2: Use the Beta function \[ \int_0^1 x^{5/2} (1 - x)^{3/2} dx = B\left(\frac{7}{2}, \frac{5}{2}\right) = \frac{\Gamma\left(\frac{7}{2}\right) \Gamma\left(\frac{5}{2}\right)}{\Gamma\left(6\right)} \] Step 3: Use values of Gamma functions \[ \Gamma\left(\frac{7}{2}\right) = \frac{5}{2} . \frac{3}{2} . \frac{1}{2} . \sqrt{\pi} = \frac{15}{8} \sqrt{\pi} \] \[ \Gamma\left(\frac{5}{2}\right) = \frac{3}{2} . \frac{1}{2} . \sqrt{\pi} = \frac{3}{4} \sqrt{\pi} \] \[ \Gamma(6) = (6 - 1)! = 5! = 120 \] Step 4: Substitute back \[ \frac{\Gamma\left(\frac{7}{2}\right) \Gamma\left(\frac{5}{2}\right)}{\Gamma(6)} = \frac{\left(\frac{15}{8} \sqrt{\pi}\right) \left(\frac{3}{4} \sqrt{\pi}\right)}{120} = \frac{45 \pi}{32 . 120} = \frac{45 \pi}{3840} = \frac{3 \pi}{256} \] \[ \boxed{ \int_0^1 x^{5/2} (1 - x)^{3/2} \, dx = \frac{3 \pi}{256} } \]