To find the average value of a function $f(x)$ over an interval $[a, b]$, we use the formula:

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

You haven't provided an interval, so let's first calculate the general form for the average value over $[a, b]$ for the function $f(x) = 2x^2 + \frac{1}{e^{x/3}}$.

Step 1: Set up the integral

We want to compute:

$\int_a^b \left( 2x^2 + \frac{1}{e^{x/3}} \right) dx$

This integral is split into two parts:

$\int_a^b 2x^2 \, dx + \int_a^b \frac{1}{e^{x/3}} \, dx$

Step 2: Compute the integrals

1. First integral: $\int 2x^2 \, dx$

$\int 2x^2 \, dx = \frac{2x^3}{3} + C$

2. Second integral: $\int \frac{1}{e^{x/3}} \, dx$

Let’s use substitution: $u = \frac{x}{3}$, so $du = \frac{1}{3} dx$, and the integral becomes:

$3 \int e^{-u} \, du = -3e^{-x/3} + C$

Step 3: Combine results

The integral of $f(x)$ over $[a, b]$ is:

[ \int_a^b f(x) , dx = \left[ \frac{2x^3}{3} \right]_a^b - 3 \left[ e^{-x/3} \right]_a^b ]

Thus,

$\int_a^b f(x) \, dx = \left( \frac{2b^3}{3} - \frac{2a^3}{3} \right) - 3 \left( e^{-b/3} - e^{-a/3} \right)$

Step 4: Find the average value

Now, divide the result of the integral by $b - a$:

$\text{Average value} = \frac{1}{b - a} \left( \frac{2b^3}{3} - \frac{2a^3}{3} - 3\left( e^{-b/3} - e^{-a/3} \right) \right)$

This gives the average value of the function over any interval $[a, b]$.

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Would you like me to calculate the average for a specific interval or give more details on any part?

5 Related Questions:

1. How is the average value of a function used in physics?
2. What is the significance of finding average values over an interval?
3. Can we approximate this average value numerically for large or difficult integrals?
4. What happens if $a$ and $b$ are negative in this function?
5. How does substitution simplify more complex integrals like this one?

Tip: Always double-check integral bounds and substitution steps for accuracy in complex problems.