If \(a{\lt}b\) then \(\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx\) is equal to
Step-by-step Solution:
\[ \begin{aligned} & \text{Given integration is } \int_{a}^{b} |x-a| \, dx + \int_{a}^{b} |x-b| \, dx, \\ & = \int_{a}^{b} (x-a) \, dx + \int_{a}^{b} -(x-b) \, dx, \\ & = \left[\frac{x^2}{2} - ax \right]_a^b + \left[bx - \frac{x^2}{2} \right]_a^b, \\ & = \left(\frac{b^2}{2} - ab\right) - \left(\frac{a^2}{2} - a^2\right) + \left(b^2 - \frac{b^2}{2}\right) - \left(ab - \frac{a^2}{2}\right), \\ & = \frac{b^2}{2} - ab - \frac{a^2}{2} + a^2 + b^2 - \frac{b^2}{2} - ab + \frac{a^2}{2}, \\ & = a^2 + b^2 - 2ab, \\ & = (b-a)^2. \end{aligned} \]