Question

Evaluate $ \int_0^1 x^5 \sqrt{1 - x^3} \, dx $

• A. \(\frac{1}{15}\)
• B. \(\frac{2}{45}\)
• C. \(\frac{2}{15}\)
• D. \(\frac{4}{45}\)

Answer 1

The Correct Option is D

Let \(x^3 = t \Rightarrow 3x^2 dx = dt\), When \(x = 0, t = 0;\ x = 1, t = 1\) Then: \[ x^5 \sqrt{1 - x^3} dx = x^5 \sqrt{1 - t} \cdot \frac{dt}{3x^2} = \frac{1}{3} x^3 \sqrt{1 - t} dt = \frac{1}{3} t \sqrt{1 - t} dt \] So: \[ I = \frac{1}{3} \int_0^1 t(1 - t)^{1/2} dt \] Use Beta function: \[ \int_0^1 t^m (1 - t)^n dt = \frac{\Gamma(m + 1)\Gamma(n + 1)}{\Gamma(m + n + 2)} \] Here: \(m = 1, n = \frac{1}{2}\) \[ I = \frac{1}{3} \cdot \frac{\Gamma(2)\Gamma(3/2)}{\Gamma(7/2)} = \frac{1}{3} \cdot \frac{1 \cdot \frac{\sqrt{\pi}}{2}}{\frac{15\sqrt{\pi}}{8}} = \frac{1}{3} \cdot \frac{4}{15} = \frac{4}{45} \]