To find the average value of the function $f(x) = 5x + \frac{1}{4x + 3}$ on the interval $[6, 10]$, we use the formula for the average value of a function on a closed interval $[a, b]$:

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

Here, $a = 6$, $b = 10$, and $f(x) = 5x + \frac{1}{4x + 3}$.

So, the average value will be:

$\text{Average value} = \frac{1}{10 - 6} \int_6^{10} \left( 5x + \frac{1}{4x + 3} \right) dx$

Now, let's compute the integral.

1. The integral of $5x$ is straightforward: $\int 5x \, dx = \frac{5x^2}{2}$

2. The integral of $\frac{1}{4x + 3}$ is computed using substitution. Let $u = 4x + 3$, then $du = 4 dx$, or $dx = \frac{du}{4}$. This gives: $\int \frac{1}{4x + 3} dx = \frac{1}{4} \ln|4x + 3|$

Thus, the integral becomes:

$\int_6^{10} \left( 5x + \frac{1}{4x + 3} \right) dx = \left[ \frac{5x^2}{2} + \frac{1}{4} \ln|4x + 3| \right]_6^{10}$

Let me calculate this expression.It seems there was an issue with the calculation. Let me break the steps down and manually handle the evaluation for the integral:

We have the integral to compute:

$\int_6^{10} \left( 5x + \frac{1}{4x + 3} \right) dx = \left[ \frac{5x^2}{2} + \frac{1}{4} \ln|4x + 3| \right]_6^{10}$

First, compute each part:

1. For $\frac{5x^2}{2}$: $\frac{5(10)^2}{2} - \frac{5(6)^2}{2} = \frac{5 \times 100}{2} - \frac{5 \times 36}{2} = 250 - 90 = 160$

2. For $\frac{1}{4} \ln|4x + 3|$: $\frac{1}{4} \ln|4(10) + 3| - \frac{1}{4} \ln|4(6) + 3| = \frac{1}{4} \ln(43) - \frac{1}{4} \ln(27)$ Using logarithmic properties: $\frac{1}{4} \ln \left( \frac{43}{27} \right)$

Now, let's add these together and then divide by $10 - 6 = 4$ to get the average value. Let me calculate this manually.The average value of the function $f(x) = 5x + \frac{1}{4x + 3}$ on the interval $[6, 10]$ is approximately 40.03.

Would you like a further breakdown of the steps or have any other questions?

Here are some related questions you might explore:

1. How does changing the interval affect the average value of the function?
2. How can we compute the average value if $f(x)$ involves more complex functions like trigonometric or exponential terms?
3. What is the geometric interpretation of the average value of a function?
4. Can the average value of a function be negative, and under what conditions?
5. How does substitution help in simplifying certain integrals?

Tip: When solving definite integrals that involve logarithms, always simplify using logarithmic properties, like $\ln \frac{a}{b} = \ln(a) - \ln(b)$.