Let I=\(\int_{0}^{1} x(1-x)^n \,dx\)
∴I=\(\int_{0}^{1} (1-x)(1-(1-x))^n \,dx\)
=\(\int_{0}^{1} x(1-x) (x)^n \,dx\)
\(=∫^1_0(x^n-x^{n+1})dx\)
\(=[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]\,\,\,\,\, (∫^a_0ƒ(x)dx=∫^a_0ƒ(a-x)dx)\)
=\([\frac{1}{n+1}-\frac{1}{n+2}]\)
=\(\frac{(n+2)-(n+1)}{(n+1)(n+2)}\)
=\(\frac{1}{(n+1)(n+2)}\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
