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
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