This integral involves an absolute value function, which needs to be handled by splitting the integral at the point where the expression inside the absolute value equals zero.

Given:

$\int_{-1}^1 (|4x| + 2) \, dx$

1. The expression $|4x|$ means that we must account for the sign of $4x$:

   • When $x \geq 0$, $|4x| = 4x$,
   • When $x < 0$, $|4x| = -4x$.

2. Split the integral at $x = 0$: $\int_{-1}^1 (|4x| + 2) \, dx = \int_{-1}^0 ((-4x) + 2) \, dx + \int_{0}^1 (4x + 2) \, dx.$

Step 1: Compute $\int_{-1}^0 (-4x + 2) \, dx$

$\int_{-1}^0 (-4x + 2) \, dx = \int_{-1}^0 (-4x) \, dx + \int_{-1}^0 2 \, dx.$

• For $\int_{-1}^0 (-4x) \, dx$: $\int_{-1}^0 (-4x) \, dx = \left[-2x^2\right]_{-1}^0 = \left(-2(0)^2\right) - \left(-2(-1)^2\right) = 0 - (-2) = 2.$
• For $\int_{-1}^0 2 \, dx$: $\int_{-1}^0 2 \, dx = \left[2x\right]_{-1}^0 = (2(0)) - (2(-1)) = 0 - (-2) = 2.$ Thus: $\int_{-1}^0 (-4x + 2) \, dx = 2 + 2 = 4.$

Step 2: Compute $\int_{0}^1 (4x + 2) \, dx$

$\int_{0}^1 (4x + 2) \, dx = \int_{0}^1 (4x) \, dx + \int_{0}^1 2 \, dx.$

• For $\int_{0}^1 (4x) \, dx$: $\int_{0}^1 (4x) \, dx = \left[2x^2\right]_{0}^1 = \left(2(1)^2\right) - \left(2(0)^2\right) = 2 - 0 = 2.$
• For $\int_{0}^1 2 \, dx$: $\int_{0}^1 2 \, dx = \left[2x\right]_{0}^1 = (2(1)) - (2(0)) = 2 - 0 = 2.$ Thus: $\int_{0}^1 (4x + 2) \, dx = 2 + 2 = 4.$

Step 3: Combine the results

$\int_{-1}^1 (|4x| + 2) \, dx = \int_{-1}^0 (-4x + 2) \, dx + \int_{0}^1 (4x + 2) \, dx = 4 + 4 = 8.$

Final Answer:

$\boxed{8}$

Let me know if you'd like further clarification or have additional questions! Here are five related questions for further exploration:

1. How do you integrate absolute value functions in general?
2. What are the steps for determining where the expression inside an absolute value changes sign?
3. How would the integral change if the limits were different (e.g., $[-2, 2]$)?
4. How can you interpret this integral geometrically?
5. Can this problem be solved using symmetry?

Tip: Always visualize the absolute value function to identify the points where it transitions between positive and negative values.