$\int_1^4 |x - 2|dx$ is equal to

The absolute value $|x - 2|$ changes behavior at $x = 2$.

When $x < 2$: $(x - 2)$ is negative, so $|x - 2| = -(x - 2) = 2 - x$

When $x \geq 2$: $(x - 2)$ is positive, so $|x - 2| = x - 2$

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Since the limits go from 1 to 4, and the behavior changes at $x = 2$:

$\int_1^4 |x - 2|dx = \int_1^2 |x - 2|dx + \int_2^4 |x - 2|dx$

$= \int_1^2 (2 - x)dx + \int_2^4 (x - 2)dx$

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For the first integral:

$\int_1^2 (2 - x)dx = \left[2x - \frac{x^2}{2}\right]_1^2$

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

$= (4 - 2) - (2 - 0.5)$

$= 2 - 1.5$

$= \frac{1}{2}$

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For the second integral:

$\int_2^4 (x - 2)dx = \left[\frac{x^2}{2} - 2x\right]_2^4$

$= \left(\frac{16}{2} - 8\right) - \left(\frac{4}{2} - 4\right)$

$= (8 - 8) - (2 - 4)$

$= 0 - (-2)$

$= 2$

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Adding both parts:

$\int_1^4 |x - 2|dx = \frac{1}{2} + 2$

$= \frac{1}{2} + \frac{4}{2}$

$= \frac{5}{2}$