To find the average value of the function $f(x) = x \ln(x^2) + 5$ on the interval $[e, e^2]$, we use the formula for the average value of a function over an interval $[a, b]$:

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

Step 1: Simplify the function

First, simplify the function $f(x) = x \ln(x^2) + 5$.

Notice that:

$\ln(x^2) = 2\ln(x)$

So, the function simplifies to:

$f(x) = x \cdot 2\ln(x) + 5 = 2x \ln(x) + 5$

Step 2: Set up the integral

Now, we need to find the integral of $f(x) = 2x \ln(x) + 5$ over the interval $[e, e^2]$:

$\int_e^{e^2} f(x) \, dx = \int_e^{e^2} \left(2x \ln(x) + 5\right) \, dx$

Step 3: Integrate the function

We will integrate $2x \ln(x)$ and $5$ separately.

1. Integrate $2x \ln(x)$:

   To integrate $2x \ln(x)$, use integration by parts: $\int u \, dv = uv - \int v \, du$ Let: $u = \ln(x), \quad dv = 2x \, dx$ Then: $du = \frac{1}{x} \, dx, \quad v = x^2$ Applying integration by parts: $\int 2x \ln(x) \, dx = x^2 \ln(x) - \int x^2 \cdot \frac{1}{x} \, dx = x^2 \ln(x) - \int x \, dx = x^2 \ln(x) - \frac{x^2}{2} + C$

2. Integrate $5$:

   $\int 5 \, dx = 5x$

Thus, the integral of the function over $[e, e^2]$ is:

$\int_e^{e^2} \left(2x \ln(x) + 5\right) \, dx = \left[x^2 \ln(x) - \frac{x^2}{2} + 5x\right]_e^{e^2}$

Step 4: Evaluate the integral

Now, we evaluate the above expression at $x = e^2$ and $x = e$:

1. At $x = e^2$: $(e^2)^2 \ln(e^2) - \frac{(e^2)^2}{2} + 5e^2 = e^4 \cdot 2 - \frac{e^4}{2} + 5e^2 = 2e^4 - \frac{e^4}{2} + 5e^2 = \frac{4e^4 - e^4}{2} + 5e^2 = \frac{3e^4}{2} + 5e^2$

2. At $x = e$: $e^2 \ln(e) - \frac{e^2}{2} + 5e = e^2 \cdot 1 - \frac{e^2}{2} + 5e = \frac{2e^2 - e^2}{2} + 5e = \frac{e^2}{2} + 5e$

The definite integral from $e$ to $e^2$ is:

[ \left[\frac{3e^4}{2} + 5e^2\right] - \left[\frac{e^2}{2} + 5e\right] = \frac{3e^4}{2} + 5e^2 - \frac{e^2}{2} - 5e = \frac{3e^4}{2} + \frac{4e^2}{2} - 5e = \frac{3e^4 + 4e^2}{2} - 5e ]

Step 5: Find the average value

Finally, divide this by the length of the interval, $e^2 - e$:

$\text{Average value} = \frac{1}{e^2 - e} \left(\frac{3e^4 + 4e^2}{2} - 5e\right)$

This simplifies further to the final expression, giving us the average value of the function over the interval $[e, e^2]$.

Would you like the final simplified result or have any questions?

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Relative Questions:

1. How do you apply integration by parts in more complex integrals?
2. What are some real-life applications of finding the average value of a function?
3. How does the logarithm property $\ln(x^2) = 2\ln(x)$ simplify integration?
4. What is the significance of the interval $[e, e^2]$ in this problem?
5. How can the definite integral be used to find the area under a curve?

Tip: When applying integration by parts, always choose $u$ and $dv$ wisely to simplify the resulting integral.