To find the average value of $f(x, y) = \sqrt{e^{y} 4x + e^y}$ over the rectangle $R = [0, 3] \times [0, 2]$, we use the formula for the average value of a function over a region $R$ in two dimensions:

$\text{Average value of } f(x, y) = \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA$

Step 1: Calculate the Area of $R$

The area of $R$ is: $\text{Area of } R = (3 - 0)(2 - 0) = 6$

Step 2: Set up the Integral

We want to compute: $\frac{1}{6} \int_0^3 \int_0^2 \sqrt{e^{y} 4x + e^y} \, dy \, dx$

Step 3: Evaluate the Integral

Rewrite the function inside the integral: $\sqrt{e^{y} 4x + e^y} = \sqrt{e^y (4x + 1)} = e^{y/2} \sqrt{4x + 1}$

Thus, the integral becomes: $\frac{1}{6} \int_0^3 \int_0^2 e^{y/2} \sqrt{4x + 1} \, dy \, dx$

Step 4: Integrate with respect to $y$

Since $e^{y/2}$ depends only on $y$ and $\sqrt{4x + 1}$ depends only on $x$, we can separate the integration: $\frac{1}{6} \int_0^3 \sqrt{4x + 1} \left( \int_0^2 e^{y/2} \, dy \right) dx$

Inner Integral:

$\int_0^2 e^{y/2} \, dy = 2 \left( e^{y/2} \right) \Big|_0^2 = 2 \left( e - 1 \right)$

So we get: $\frac{1}{6} \int_0^3 \sqrt{4x + 1} \cdot 2(e - 1) \, dx = \frac{e - 1}{3} \int_0^3 \sqrt{4x + 1} \, dx$

Step 5: Integrate with respect to $x$

Let $u = 4x + 1$, then $du = 4 \, dx$ or $dx = \frac{du}{4}$. When $x = 0$, $u = 1$, and when $x = 3$, $u = 13$.

The integral becomes: $\frac{e - 1}{3} \int_1^{13} \sqrt{u} \, \frac{du}{4} = \frac{e - 1}{12} \int_1^{13} u^{1/2} \, du$

Evaluate:

$\frac{e - 1}{12} \cdot \frac{2}{3} u^{3/2} \Big|_1^{13} = \frac{e - 1}{18} \left( 13^{3/2} - 1^{3/2} \right)$

So, the average value is: $\frac{e - 1}{18} \left( 13^{3/2} - 1 \right)$

Would you like more details or have any questions?

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Related Questions

1. How do we find the average value over regions other than rectangles?
2. How does the presence of $e^y$ in the function affect the integration?
3. Can the average value formula apply to higher dimensions?
4. How would the answer change if $R$ had different limits?
5. Why did we separate the integral in Step 4?

Tip

For average values, simplify the integrand wherever possible before integrating.