To evaluate the integral \( \int_0^2 |2x - 3| \, dx \), the absolute value function requires splitting the integration interval.
Step 1: Find the root of the expression within the absolute value.
\[
2x - 3 = 0 \implies x = \frac{3}{2}
\]
This point partitions the interval \([0, 2]\) into:
\[
[0, \frac{3}{2}] \quad \text{and} \quad [\frac{3}{2}, 2]
\]
Step 2: Redefine the integrand without the absolute value for each sub-interval.
1. For \(0 \leq x \leq \frac{3}{2}\), \(2x - 3 \leq 0\), so \(|2x - 3| = -(2x - 3) = 3 - 2x\).
2. For \(\frac{3}{2} \leq x \leq 2\), \(2x - 3 \geq 0\), so \(|2x - 3| = 2x - 3\).
Step 3: Decompose the original integral into a sum of integrals over the sub-intervals.
\[
\int_0^2 |2x - 3| \, dx = \int_0^{\frac{3}{2}} (3 - 2x) \, dx + \int_{\frac{3}{2}}^2 (2x - 3) \, dx
\]
Step 4: Compute each integral.
1. First integral:
\[
\int_0^{\frac{3}{2}} (3 - 2x) \, dx = \left[ 3x - x^2 \right]_0^{\frac{3}{2}} = \left( \frac{9}{2} - \frac{9}{4} \right) - 0 = \frac{9}{4}
\]
2. Second integral:
\[
\int_{\frac{3}{2}}^2 (2x - 3) \, dx = \left[ x^2 - 3x \right]_{\frac{3}{2}}^2 = (4 - 6) - \left( \frac{9}{4} - \frac{9}{2} \right) = -2 - (-\frac{9}{4}) = -2 + \frac{9}{4} = \frac{1}{4}
\]
Step 5: Combine the results.
\[
\int_0^2 |2x - 3| \, dx = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}
\]
Final Answer:
\[
\boxed{\frac{5}{2}}
\]