1. **State the problem:** Calculate the definite integral $$\int_0^1 x^{\frac{1}{2}} (1-x)^{\frac{1}{3}} \, dx$$. 2. **Use the Beta function formula:** The Beta function is defined as $$B(p,q) = \int_0^1 x^{p-1} (1-x)^{q-1} \, dx = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$. 3. **Identify parameters:** Here, $$p = \frac{3}{2}$$ and $$q = \frac{4}{3}$$ because the integral matches the form with exponents $$p-1 = \frac{1}{2}$$ and $$q-1 = \frac{1}{3}$$. 4. **Apply the Beta function:** $$\int_0^1 x^{\frac{1}{2}} (1-x)^{\frac{1}{3}} \, dx = B\left(\frac{3}{2}, \frac{4}{3}\right) = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{3}{2} + \frac{4}{3}\right)}$$. 5. **Evaluate Gamma functions:** - $$\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$$. - Use the property $$\Gamma(z+1) = z \Gamma(z)$$ to write $$\Gamma\left(\frac{4}{3}\right) = \frac{1}{3} \Gamma\left(\frac{1}{3}\right)$$. - Sum in denominator: $$\frac{3}{2} + \frac{4}{3} = \frac{9}{6} + \frac{8}{6} = \frac{17}{6}$$. 6. **Final expression:** $$\int_0^1 x^{\frac{1}{2}} (1-x)^{\frac{1}{3}} \, dx = \frac{\frac{\sqrt{\pi}}{2} \times \frac{1}{3} \Gamma\left(\frac{1}{3}\right)}{\Gamma\left(\frac{17}{6}\right)} = \frac{\sqrt{\pi}}{6} \frac{\Gamma\left(\frac{1}{3}\right)}{\Gamma\left(\frac{17}{6}\right)}$$. This completes the evaluation of the integral using Beta and Gamma functions.