∫limits_2^5|x - 3|dx equals

The correct answer is Option (2) → $\frac{5}{2}$

To evaluate $\int_2^5 |x - 3| \, dx$ directly, observe the behavior of the absolute value:

For $x \in [2, 3]$, $|x - 3| = 3 - x$, and for $x \in [3, 5]$, $|x - 3| = x - 3$.

So directly applying definition over each subinterval:

$\int_2^5 |x - 3| \, dx = \int_2^3 (3 - x) \, dx + \int_3^5 (x - 3) \, dx$

$= \left[3x - \frac{x^2}{2}\right]_2^3 + \left[\frac{x^2}{2} - 3x\right]_3^5$

$= \left(9 - \frac{9}{2}\right) - \left(6 - \frac{4}{2}\right) + \left(\frac{25}{2} - 15\right) - \left(\frac{9}{2} - 9\right)$

$= (4.5) - (4) + (12.5 - 15) - (4.5 - 9)$

$= 0.5 + (-2.5 + 4.5) = 0.5 + 2 = {\frac{5}{2}}$