To evaluate the integral:

$\int |2x - 4| \, dx$

Step 1: Analyze the absolute value function

The absolute value function $|2x - 4|$ changes sign at the point where $2x - 4 = 0$. Solve for $x$:

$2x - 4 = 0 \implies x = 2$

Thus, $|2x - 4|$ is defined as:

• $2x - 4$ when $x \geq 2$,
• $-(2x - 4)$ when $x < 2$.

Step 2: Split the integral at $x = 2$

Given the absolute value function, the integral can be split as: $\int |2x - 4| \, dx = \int_{1}^{2} -(2x - 4) \, dx + \int_{2}^{4} (2x - 4) \, dx$

Step 3: Evaluate each integral

First integral: $\int_{1}^{2} -(2x - 4) \, dx$

$\int_{1}^{2} -(2x - 4) \, dx = \int_{1}^{2} (-2x + 4) \, dx$ Find the antiderivative: $\int (-2x + 4) \, dx = -x^2 + 4x + C$ Evaluate at bounds $x = 1$ and $x = 2$: $\left[ -x^2 + 4x \right]_{1}^{2} = \left( -(2)^2 + 4(2) \right) - \left( -(1)^2 + 4(1) \right)$ $= \left( -4 + 8 \right) - \left( -1 + 4 \right) = 4 - 3 = 1$

Second integral: $\int_{2}^{4} (2x - 4) \, dx$

$\int_{2}^{4} (2x - 4) \, dx$ Find the antiderivative: $\int (2x - 4) \, dx = x^2 - 4x + C$ Evaluate at bounds $x = 2$ and $x = 4$: $\left[ x^2 - 4x \right]_{2}^{4} = \left( (4)^2 - 4(4) \right) - \left( (2)^2 - 4(2) \right)$ $= \left( 16 - 16 \right) - \left( 4 - 8 \right) = 0 - (-4) = 4$

Step 4: Add the results

$\int_{1}^{4} |2x - 4| \, dx = 1 + 4 = 5$

Final Answer:

$\boxed{5}$

Let me know if you'd like further clarification or additional examples! 😊

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Related Questions:

1. How do you handle integrals involving other piecewise functions?
2. What happens if the limits of integration include a critical point of the absolute value function?
3. Can the integral be computed geometrically for $|2x - 4|$?
4. How would this problem change if the limits were different?
5. How do you generalize the approach for $|ax + b|$?

Tip:

Always identify the critical points of the absolute value function to properly split the integral into manageable parts.