To solve the integral

$\int_0^2 |2x - 3| \, dx,$

we need to analyze the behavior of the absolute value expression $|2x - 3|$ within the given interval. This can be broken into two parts:

1. Find where $2x - 3 = 0$ within the interval $[0, 2]$ to determine where the absolute value expression changes sign.
2. Split the integral at this point of change and compute each segment separately.

Step 1: Determine the critical point

Solve $2x - 3 = 0$: $2x = 3 \Rightarrow x = \frac{3}{2}.$

This means the expression $2x - 3$ changes sign at $x = \frac{3}{2}$. Therefore, we split the integral at $x = \frac{3}{2}$ and rewrite it as two separate integrals:

$\int_0^2 |2x - 3| \, dx = \int_0^{\frac{3}{2}} (3 - 2x) \, dx + \int_{\frac{3}{2}}^2 (2x - 3) \, dx.$

Step 2: Evaluate each integral separately

1. First integral from $0$ to $\frac{3}{2}$: $\int_0^{\frac{3}{2}} (3 - 2x) \, dx.$

   Integrate $3 - 2x$: $\int (3 - 2x) \, dx = 3x - x^2.$

   Now evaluate from $0$ to $\frac{3}{2}$: $\left[ 3x - x^2 \right]_0^{\frac{3}{2}} = \left( 3 \cdot \frac{3}{2} - \left( \frac{3}{2} \right)^2 \right) - (3 \cdot 0 - 0^2).$ Simplify: $= \left( \frac{9}{2} - \frac{9}{4} \right) = \frac{18}{4} - \frac{9}{4} = \frac{9}{4}.$

2. Second integral from $\frac{3}{2}$ to $2$: $\int_{\frac{3}{2}}^2 (2x - 3) \, dx.$

   Integrate $2x - 3$: $\int (2x - 3) \, dx = x^2 - 3x.$

   Now evaluate from $\frac{3}{2}$ to $2$: $\left[ x^2 - 3x \right]_{\frac{3}{2}}^2 = \left( 2^2 - 3 \cdot 2 \right) - \left( \left( \frac{3}{2} \right)^2 - 3 \cdot \frac{3}{2} \right).$ Simplify: $= (4 - 6) - \left( \frac{9}{4} - \frac{9}{2} \right) = -2 - \left( \frac{9}{4} - \frac{18}{4} \right) = -2 + \frac{9}{4} = -\frac{8}{4} + \frac{9}{4} = \frac{1}{4}.$

Step 3: Combine the results

Add the results of both integrals: $\frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}.$

Final Answer

$\int_0^2 |2x - 3| \, dx = \frac{5}{2}.$

Would you like further explanation on any steps or concepts used here?

Here are five questions related to this integral:

1. How does the absolute value affect the integration process?
2. What would happen if the limits were different, say $\int_0^3 |2x - 3| \, dx$?
3. Can you graph the function $|2x - 3|$ to visually see the break point at $x = \frac{3}{2}$?
4. How can this method be applied to other functions with absolute values?
5. What are other ways to handle integrals involving piecewise functions?

Tip: When integrating absolute values, identifying where the expression inside changes sign is crucial, as it allows you to split the integral and remove the absolute value by treating each interval accordingly.