∫limits_1^4|x - 2|dx is equal to

The correct answer is Option (4) → $\frac{5}{2}$

$\displaystyle \int_{1}^{4} |x-2|\,dx$

Break at $x=2$:

$|x-2|=2-x$ for $1\le x\le 2$

$|x-2|=x-2$ for $2\le x\le 4$

$\int_{1}^{2} (2-x)\,dx + \int_{2}^{4} (x-2)\,dx$

$\left[2x-\frac{x^{2}}{2}\right]_{1}^{2} + \left[\frac{x^{2}}{2}-2x\right]_{2}^{4}$

$\left(4-2\right)-\left(2-\frac{1}{2}\right) + \left(8-8\right)-\left(2-4\right)$

$2 - \frac{3}{2} + 0 + 2$

$\frac{1}{2} + 2$

$\frac{5}{2}$

Final answer: $\frac{5}{2}$