Step-by-step Solution:

\[ I = \int_0^1 x(1 - x)^n \,dx \] Let \( t = 1 - x \), so that \( dt = -dx \). Changing Limits: - When \( x = 0 \Rightarrow t = 1 \) - When \( x = 1 \Rightarrow t = 0 \) Rewriting the integral in terms of \( t \): \[ I = - \int_1^0 (1 - t)t^n \, dt \] \[ = - \int_1^0 (t^n - t^{n+1}) \, dt \] \[ = - \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_1^0 \] Evaluating at limits: \[ = - \left[ \left( \frac{0^{n+1}}{n+1} - \frac{0^{n+2}}{n+2} \right) - \left( \frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2} \right) \right] \] \[ = - \left[ (0 - 0) - \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \right] \] \[ = \frac{1}{n+1} - \frac{1}{n+2} \] \[ = \frac{(n+2) - (n+1)}{(n+1)(n+2)} \] \[ = \frac{1}{(n+1)(n+2)} \] Thus, the final result is: \[ I = \frac{1}{(n+1)(n+2)} \]