Let \(I\) =\(\int_{2}^{8} |x-5| \,dx\)
It can be seen that (x−5)≤0 on [2,5] and (x−5)≥0 on [5,8].
\(I\) =\(\int_{2}^{8} -(x-5) \,dx\)+\(\int_{2}^{8} (x-5) \,dx\) \(\bigg(\int_{a}^{b} f(x) \,dx\) =\(\int_{a}^{c} f(x)\)+\(\int_{c}^{b} f(x)\bigg)\)
=\(-\bigg[\frac{x^2}{2}-5x\bigg]^5_2+\bigg[\frac{x^2}{2}-5x\bigg]^8_5\)
=-\(\bigg[\frac {25}{2}\)-25-2+10\(\bigg]\)+\(\bigg[\)32-40-\(\frac {25}{2}\)+25\(\bigg]\)
=9
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]