∫_1^5 |x-2| dx =

The correct answer is Option (1) → 5

The expression x − 2 changes sign at x = 2, so split the integral at 2:

    \( \displaystyle \int_{1}^{5}|x-2|\,dx = \int_{1}^{2}(2-x)\,dx + \int_{2}^{5}(x-2)\,dx \)

First part:

    \( \displaystyle \int_{1}^{2}(2-x)\,dx = \left[2x - \frac{x^2}{2}\right]_{1}^{2} \)

    \( \displaystyle = 2 - \frac{3}{2} = \frac{1}{2} \)

Second part:

    \( \displaystyle \int_{2}^{5}(x-2)\,dx = \left[\frac{(x-2)^2}{2}\right]_{2}^{5} \)

    \( \displaystyle = \frac{9}{2} \)

So the integral is

    \( \displaystyle \frac{1}{2} + \frac{9}{2} = 5 \)