To find the average value of the function $f(x) = 8x + 2$ over the interval $[4, 8]$, we use the formula for the average value of a function:

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

Step-by-step Solution:

1. The interval $[a, b]$ is $[4, 8]$, so $a = 4$ and $b = 8$.
2. $f(x) = 8x + 2$.

Using the formula:

$\text{Average Value} = \frac{1}{8-4} \int_4^8 (8x + 2) \, dx$

$\text{Average Value} = \frac{1}{4} \int_4^8 (8x + 2) \, dx$

3. Compute the integral: $\int (8x + 2) \, dx = 4x^2 + 2x$

Now evaluate this from 4 to 8: $\int_4^8 (8x + 2) \, dx = \left[ 4x^2 + 2x \right]_4^8$

Substitute the limits: [ \left[ 4(8)^2 + 2(8) \right] - \left[ 4(4)^2 + 2(4) \right] ]

[ = \left[ 4(64) + 16 \right] - \left[ 4(16) + 8 \right] ]

$= (256 + 16) - (64 + 8)$

$= 272 - 72 = 200$

4. Divide by 4 to find the average: $\text{Average Value} = \frac{1}{4} \cdot 200 = 50$

Final Answer:

The average value is 50.

Would you like further explanation or related examples?

Related Questions:

1. How do you interpret the average value of a function geometrically?
2. What would change if the interval was $[2, 6]$ instead of $[4, 8]$?
3. What is the significance of the average value in real-world applications?
4. How do you compute average value for more complex functions like $f(x) = x^2 + x + 1$?
5. How does the average value formula relate to the Mean Value Theorem for integrals?

Tip:

Always simplify the integral limits and ensure you correctly evaluate both upper and lower bounds when finding definite integrals.