To find the area between the curves $g(x) = 4x + 8$ and $f(x) = x^2 + 19x + 62$ over the interval $[-9, -6]$, we need to compute the integral of the absolute difference between these two functions over this interval. Here are the steps:

1. Set up the integral: The area between the two curves from $x = -9$ to $x = -6$ is given by: $\text{Area} = \int_{-9}^{-6} |f(x) - g(x)| \, dx$

2. Determine $f(x) - g(x)$: $f(x) - g(x) = (x^2 + 19x + 62) - (4x + 8)$ Simplify this expression: $= x^2 + 19x + 62 - 4x - 8$ $= x^2 + 15x + 54$

3. Set up the integral with the simplified expression: Since $f(x) \geq g(x)$ over this interval, we don't need the absolute value and can directly integrate $f(x) - g(x)$ from $-9$ to $-6$: $\text{Area} = \int_{-9}^{-6} (x^2 + 15x + 54) \, dx$

4. Integrate: Integrate each term separately: $\int (x^2 + 15x + 54) \, dx = \frac{x^3}{3} + \frac{15x^2}{2} + 54x$

5. Evaluate the integral from $-9$ to $-6$: Substitute the limits: $= \left[ \frac{x^3}{3} + \frac{15x^2}{2} + 54x \right]_{-9}^{-6}$

   Calculating for $x = -6$: $\frac{(-6)^3}{3} + \frac{15(-6)^2}{2} + 54(-6) = -72 + 270 - 324 = -126$

   Calculating for $x = -9$: $\frac{(-9)^3}{3} + \frac{15(-9)^2}{2} + 54(-9) = -243 + 607.5 - 486 = -121.5$

   Now, find the difference: $\text{Area} = -126 - (-121.5) = -126 + 121.5 = 4.5$

So, the area bounded by $g(x)$ and $f(x)$ over $[-9, -6]$ is $4.5$ square units.

Would you like a detailed breakdown of any step or have additional questions?

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Here are 5 related questions:

1. How do you find the area between two curves over different intervals?
2. What happens if $f(x)$ and $g(x)$ intersect within the interval?
3. Can you explain how to determine which function is above the other over an interval?
4. How does the absolute value affect the integration of differences between functions?
5. What are the applications of calculating the area between curves?

Tip: For integration problems, simplify the integrand as much as possible before starting to integrate; it can make both integration and evaluation easier.