Questions: Quiz 4 6 points possible Answered: 1 / 6 Question 2 The average value of f(x, y)=(3x+5y)^5 on the region R=(x, y) 2 ≤ x ≤ 8 ; 1 ≤ y ≤ 6 can be calculated with which of the following (choose all that apply): 1/(8-2)(6-1) ∫ from 2 to 8 ∫ from 1 to 6 (3x+5y)^5 dx dy 1/(2-1)(8-6) ∫ from 1 to 2 ∫ from 6 to 8 (3x+5y)^5 dx dy 1/(6-1)(8-2) ∫ from 1 to 6 ∫ from 2 to 8 (3x+5y)^5 dy dx 1/(1-2)(6-8) ∫ from 2 to 1 ∫ from 8 to 6 (3x+5y)^5 dy dx 1/(8-2)(6-1) ∫ from 2 to 8 ∫ from 1 to 6 (3x+5y)^5 dy dx

Quiz 4
6 points possible Answered: 1 / 6
Question 2

The average value of f(x, y)=(3x+5y)^5 on the region R=(x, y) 2 ≤ x ≤ 8 ; 1 ≤ y ≤ 6 can be calculated with which of the following (choose all that apply):
1/(8-2)(6-1) ∫ from 2 to 8 ∫ from 1 to 6 (3x+5y)^5 dx dy
1/(2-1)(8-6) ∫ from 1 to 2 ∫ from 6 to 8 (3x+5y)^5 dx dy
1/(6-1)(8-2) ∫ from 1 to 6 ∫ from 2 to 8 (3x+5y)^5 dy dx
1/(1-2)(6-8) ∫ from 2 to 1 ∫ from 8 to 6 (3x+5y)^5 dy dx
1/(8-2)(6-1) ∫ from 2 to 8 ∫ from 1 to 6 (3x+5y)^5 dy dx

Transcript text: Quiz 4 6 points possible Answered: $1 / 6$ Question 2 The average value of $f(x, y)=(3 x+5 y)^{5}$ on the region $R=\{(x, y) \mid 2 \leq x \leq 8 ; 1 \leq y \leq 6\}$ can be calculated with which of the following (choose all that apply): $\frac{1}{(8-2)(6-1)} \int_{2}^{8} \int_{1}^{6}(3 x+5 y)^{5} d x d y$ $\frac{1}{(2-1)(8-6)} \int_{1}^{2} \int_{6}^{8}(3 x+5 y)^{5} d x d y$ $\frac{1}{(6-1)(8-2)} \int_{1}^{6} \int_{2}^{8}(3 x+5 y)^{5} d y d x$ $\frac{1}{(1-2)(6-8)} \int_{2}^{1} \int_{8}^{6}(3 x+5 y)^{5} d y d x$ $\frac{1}{(8-2)(6-1)} \int_{2}^{8} \int_{1}^{6}(3 x+5 y)^{5} d y d x$

Solution

Solution Steps

To find the average value of a function $ f(x, y) $ over a region $ R $, we use the formula for the average value of a function over a region:

\[ \text{Average value} = \frac{1}{\text{Area of } R} \int \int_{R} f(x, y) \, dA \]

For the given problem, the region $ R $ is defined by $ 2 \leq x \leq 8 $ and $ 1 \leq y \leq 6 $. The area of $ R $ is $(8-2) \times (6-1)$. We need to evaluate the double integral of $ f(x, y) = (3x + 5y)^5 $ over this region. The correct setup for the integral is:

The limits of integration for $ x $ are from 2 to 8, and for $ y $ are from 1 to 6.
The order of integration can be either $ dx \, dy $ or $ dy \, dx $, but the limits must match the region $ R $.

Step 1: Define the Function and Region

We are given the function $ f(x, y) = (3x + 5y)^5 $ and the region $ R = \{(x, y) \mid 2 \leq x \leq 8, 1 \leq y \leq 6\} $.

Step 2: Calculate the Area of the Region

The area $ A $ of the region $ R $ is calculated as follows: \[ A = (8 - 2)(6 - 1) = 6 \times 5 = 30 \]

Step 3: Evaluate the Double Integral

We compute the double integral of $ f(x, y) $ over the region $ R $: \[ \int_{2}^{8} \int_{1}^{6} (3x + 5y)^5 \, dy \, dx = 1973540725 \]

Step 4: Calculate the Average Value

The average value $ \text{Avg} $ of the function over the region $ R $ is given by: \[ \text{Avg} = \frac{1}{A} \int \int_{R} f(x, y) \, dA = \frac{1973540725}{30} = 394708145.8333 \]