We need to compute the definite integral:

$\int_{0}^{6} \left( 7e^{-0.5x} + 6 \right) dx$

Step 1: Break the integral into two parts

We can split the integral as: $\int_{0}^{6} \left( 7e^{-0.5x} + 6 \right) dx = \int_{0}^{6} 7e^{-0.5x} dx + \int_{0}^{6} 6 dx$

Step 2: Evaluate each integral separately

Part 1: $\int_{0}^{6} 7e^{-0.5x} dx$

Let’s first find the indefinite integral of $7e^{-0.5x}$: $\int 7e^{-0.5x} dx = 7 \cdot \int e^{-0.5x} dx$

The integral of $e^{-0.5x}$ is: $\int e^{-0.5x} dx = \frac{e^{-0.5x}}{-0.5} = -2e^{-0.5x}$

Thus: $\int 7e^{-0.5x} dx = 7 \cdot (-2e^{-0.5x}) = -14e^{-0.5x}$

Now evaluate the definite integral from $0$ to $6$: $\int_{0}^{6} 7e^{-0.5x} dx = \left[ -14e^{-0.5x} \right]_{0}^{6}$

Substitute the limits: $= -14e^{-0.5(6)} + 14e^{-0.5(0)}$ $= -14e^{-3} + 14$

Part 2: $\int_{0}^{6} 6 dx$

The integral of a constant $6$ is: $\int 6 dx = 6x$

Evaluate the definite integral from $0$ to $6$: $\int_{0}^{6} 6 dx = \left[ 6x \right]_{0}^{6} = 6(6) - 6(0) = 36$

Step 3: Combine the results

Now, add the results of the two parts: $\int_{0}^{6} \left( 7e^{-0.5x} + 6 \right) dx = \left( -14e^{-3} + 14 \right) + 36$ $= 50 - 14e^{-3}$

Final Answer:

$\boxed{50 - 14e^{-3}}$

Let me know if you want details on any step or have additional questions. Here are 5 related questions and a tip for further exploration:

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1. How is the exponential function $e^{-x}$ commonly used in integral computations?
2. Can you explain why we split the integral into two parts?
3. What happens if we change the limits to $[1, 4]$ in the integral?
4. What is the importance of using substitution methods in integrals involving exponential terms?
5. How does the value of $e^{-3}$ approximately affect the final result?

Tip: Always simplify expressions during integration to identify constants or terms that don’t depend on the variable of integration, making calculations easier.