To find the area between the graphs of $f(x) = 9x + 8$ and $g(x) = x^2 + 4x + 2$ over the interval $[0, 2]$, we can follow these steps:

1. Determine the difference between the functions:
   The area between the graphs is given by the integral of the difference between the two functions, where the upper function is $f(x)$ and the lower function is $g(x)$.
   So, we need to compute the integral of $f(x) - g(x)$ over the interval $[0, 2]$.

   $f(x) - g(x) = (9x + 8) - (x^2 + 4x + 2) = -x^2 + 5x + 6$

2. Set up the integral:
   We now integrate the expression $-x^2 + 5x + 6$ from 0 to 2.

   $\text{Area} = \int_0^2 (-x^2 + 5x + 6) \, dx$

3. Calculate the integral:
   The integral of $-x^2 + 5x + 6$ is:

   $\int (-x^2 + 5x + 6) \, dx = -\frac{x^3}{3} + \frac{5x^2}{2} + 6x$

   Now, evaluate this from 0 to 2.

4. Evaluate at the bounds: Substitute $x = 2$ and $x = 0$ into the antiderivative:

   At $x = 2$: $-\frac{(2)^3}{3} + \frac{5(2)^2}{2} + 6(2) = -\frac{8}{3} + \frac{20}{2} + 12 = -\frac{8}{3} + 10 + 12 = 22 - \frac{8}{3} = \frac{66}{3} - \frac{8}{3} = \frac{58}{3}$

   At $x = 0$: $-\frac{(0)^3}{3} + \frac{5(0)^2}{2} + 6(0) = 0$

5. Subtract and simplify: Now, subtract the result at $x = 0$ from the result at $x = 2$:

   $\text{Area} = \frac{58}{3} - 0 = \frac{58}{3}$

Thus, the area between the graphs of $f(x) = 9x + 8$ and $g(x) = x^2 + 4x + 2$ over the interval $[0, 2]$ is $\frac{58}{3}$.

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Would you like to see a more detailed breakdown of any steps?

Relative questions:

1. How do you determine which function is the upper or lower one in a given interval?
2. What happens if the functions intersect within the interval?
3. Can you explain the role of definite integrals in finding the area between curves?
4. How does the integral change if the limits of integration are reversed?
5. How would the problem change if the functions were not continuous over the interval?

Tip:

When finding the area between curves, always check if the functions intersect within the given interval. If they do, you need to split the integral at those points of intersection.