The problem involves evaluating definite integrals of the function $8x - 2x^2$, followed by subtraction. Let's solve step by step.

The expression is: $\int_0^{1/2} (8x - 2x^2) \, dx - \int_4^{1/2} (8x - 2x^2) \, dx$

Step 1: Evaluate the integral formula

The antiderivative of $8x - 2x^2$ is: $\int (8x - 2x^2) \, dx = 4x^2 - \frac{2x^3}{3} + C$

Step 2: Evaluate $\int_0^{1/2} (8x - 2x^2) \, dx$

Using the limits $0$ and $1/2$, the integral becomes: $\left[ 4x^2 - \frac{2x^3}{3} \right]_0^{1/2}$ At $x = 1/2$: $4 \left(\frac{1}{2}\right)^2 - \frac{2\left(\frac{1}{2}\right)^3}{3} = 4 \cdot \frac{1}{4} - \frac{2 \cdot \frac{1}{8}}{3} = 1 - \frac{1}{12} = \frac{12}{12} - \frac{1}{12} = \frac{11}{12}$ At $x = 0$: $4(0)^2 - \frac{2(0)^3}{3} = 0$ Thus: $\int_0^{1/2} (8x - 2x^2) \, dx = \frac{11}{12}$

Step 3: Evaluate $\int_4^{1/2} (8x - 2x^2) \, dx$

This integral is calculated as: $\left[ 4x^2 - \frac{2x^3}{3} \right]_4^{1/2}$ At $x = 1/2$: $4 \left(\frac{1}{2}\right)^2 - \frac{2\left(\frac{1}{2}\right)^3}{3} = \frac{11}{12} \quad \text{(already computed)}.$ At $x = 4$: $4(4)^2 - \frac{2(4)^3}{3} = 4 \cdot 16 - \frac{2 \cdot 64}{3} = 64 - \frac{128}{3} = \frac{192}{3} - \frac{128}{3} = \frac{64}{3}$ Thus: $\int_4^{1/2} (8x - 2x^2) \, dx = \frac{11}{12} - \frac{64}{3} = \frac{11}{12} - \frac{256}{12} = -\frac{245}{12}$

Step 4: Subtract the results

$\frac{11}{12} - \left(-\frac{245}{12}\right) = \frac{11}{12} + \frac{245}{12} = \frac{256}{12} = \frac{64}{3}$

Final Answer:

$\boxed{\frac{64}{3}}$

Let me know if you'd like more details or further clarification!

Relative Questions:

1. How can definite integrals be used to calculate areas under curves?
2. What are some common applications of definite integrals in physics?
3. How do you handle integrals where the limits are reversed (e.g., $\int_b^a f(x) dx$)?
4. Why is the antiderivative important for solving definite integrals?
5. What happens if the function being integrated contains discontinuities?

Tip: Always simplify constants carefully when working with fractions in definite integrals to avoid errors!