$\int_2^5 |x - 3|dx$ equals

The absolute value $|x - 3|$ represents the distance between $x$ and $3$. To integrate, the absolute value must be removed:

• When $x \geq 3$: $|x - 3| = x - 3$
• When $x < 3$: $|x - 3| = -(x - 3) = 3 - x$

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Since the integration is from $2$ to $5$ and the expression changes at $x = 3$, split the integral at $x = 3$:

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

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For $x \in [2, 3]$: $|x - 3| = 3 - x$

For $x \in [3, 5]$: $|x - 3| = x - 3$

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

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$\int_2^3 (3-x)dx = \left[3x - \frac{x^2}{2}\right]_2^3$

At $x = 3$: $3(3) - \frac{3^2}{2} = 9 - \frac{9}{2} = \frac{9}{2}$

At $x = 2$: $3(2) - \frac{2^2}{2} = 6 - 2 = 4$

$\int_2^3 (3-x)dx = \frac{9}{2} - 4 = \frac{1}{2}$

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$\int_3^5 (x-3)dx = \left[\frac{x^2}{2} - 3x\right]_3^5$

At $x = 5$: $\frac{5^2}{2} - 3(5) = \frac{25}{2} - 15 = -\frac{5}{2}$

At $x = 3$: $\frac{3^2}{2} - 3(3) = \frac{9}{2} - 9 = -\frac{9}{2}$

$\int_3^5 (x-3)dx = -\frac{5}{2} - \left(-\frac{9}{2}\right) = 2$

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$\int_2^5 |x - 3|dx = \frac{1}{2} + 2$

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

$= \frac{5}{2}$