We are tasked with evaluating the definite integral:
$\int_0^6 \left( 2e^{-0.5x} - 2x \right) dx$

Step 1: Split the Integral

The integral can be split into two separate terms: $\int_0^6 \left( 2e^{-0.5x} \right) dx - \int_0^6 \left( 2x \right) dx$

Step 2: Evaluate Each Integral Separately

Part 1: $\int_0^6 2e^{-0.5x} dx$

Let $u = -0.5x$, so $du = -0.5 dx$ or $dx = -2du$. The limits transform as follows:

• When $x = 0$, $u = 0$
• When $x = 6$, $u = -3$

The integral becomes: $\int_0^6 2e^{-0.5x} dx = \int_0^{-3} 2e^u \cdot (-2) du = -4 \int_0^{-3} e^u du$

Evaluate $\int e^u du = e^u$: $-4 \left[ e^u \right]_0^{-3} = -4 \left( e^{-3} - e^0 \right) = -4 \left( e^{-3} - 1 \right)$

Simplify: $-4 \left( e^{-3} - 1 \right) = 4 - 4e^{-3}$

Part 2: $\int_0^6 2x dx$

Use the power rule for integration: [ \int_0^6 2x dx = 2 \int_0^6 x dx = 2 \left[ \frac{x^2}{2} \right]_0^6 = \left[ x^2 \right]_0^6 ]

Evaluate: $\left[ x^2 \right]_0^6 = 6^2 - 0^2 = 36$

Multiply by 2: $2 \times 36 = 72$

Step 3: Combine Results

The integral is: $\int_0^6 \left( 2e^{-0.5x} - 2x \right) dx = \left( 4 - 4e^{-3} \right) - 72$

Simplify: $4 - 4e^{-3} - 72 = -68 - 4e^{-3}$

Final Answer:

$\boxed{-68 - 4e^{-3}}$

Would you like a detailed explanation of any step?

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Related Questions:

1. How do you integrate exponential functions with a linear argument?
2. What are common techniques for definite integrals?
3. How does substitution work in integral transformations?
4. How do you simplify expressions involving exponential terms?
5. What are the properties of definite integrals that make them useful?

Tip:

When solving integrals involving exponentials, always check if substitution simplifies the process.