To solve the problem, we need to calculate the average value of the function $f(x) = 4x + 3$ on the interval $[0, 2]$.

The formula for the average value of a function $f(x)$ over the interval $[a, b]$ is:

$\text{Average Value} = \frac{1}{b - a} \int_a^b f(x) \, dx$

In this case, $a = 0$, $b = 2$, and $f(x) = 4x + 3$. Plugging in the values:

$\text{Average Value} = \frac{1}{2 - 0} \int_0^2 (4x + 3) \, dx = \frac{1}{2} \int_0^2 (4x + 3) \, dx$

Now, let's compute the integral of $4x + 3$:

$\int (4x + 3) \, dx = 2x^2 + 3x$

Evaluating this from 0 to 2:

$\left[ 2x^2 + 3x \right]_0^2 = (2(2)^2 + 3(2)) - (2(0)^2 + 3(0)) = (8 + 6) - 0 = 14$

Now, dividing by 2:

$\text{Average Value} = \frac{14}{2} = 7$

Thus, the average value of the function is $\boxed{7}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you calculate the average value of a function over a different interval?
2. What does the average value of a linear function represent geometrically?
3. How does the average value formula apply to non-linear functions?
4. Can you find the average value of a function using numerical integration methods?
5. What happens to the average value if the function is shifted vertically?

Tip: For linear functions, the average value is always the midpoint between the function's values at the endpoints.