Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Transcription of the Image:
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**Problem 4:**
Find the average value of the function \( f(x, y) = e^{x + 2y} \) over the region
\[ R = \{ (x, y) \mid 0 \leq x \leq \ln(2), \ 0 \leq y \leq \ln(3) \} . \]
\[ \]
The average value of the function is calculated as:
\[
\frac{1}{b-A} \iint_R e^{x+2y} \, dA
\]
This is specified as:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} \int_0^{\ln(3)} e^{x + 2y} \, dy \, dx
\]
\[ \]
The next steps are:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} \left[ e^x \cdot \frac{1}{2}e^{2y} \right]_0^{\ln(3)} \, dx
\]
\[ \]
Further simplifying:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} e^x \left( \frac{e^{2\ln(3)}}{2} \right) \, dx
\]
Please note: There are no graphs or diagrams present in the image, only mathematical expressions.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65e8876c-9cfb-4d00-b5f3-6286bfdc9996%2F3fe86869-7a9e-4b89-9560-f822e0ebc00d%2Fwlfaxmf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Transcription of the Image:
---
**Problem 4:**
Find the average value of the function \( f(x, y) = e^{x + 2y} \) over the region
\[ R = \{ (x, y) \mid 0 \leq x \leq \ln(2), \ 0 \leq y \leq \ln(3) \} . \]
\[ \]
The average value of the function is calculated as:
\[
\frac{1}{b-A} \iint_R e^{x+2y} \, dA
\]
This is specified as:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} \int_0^{\ln(3)} e^{x + 2y} \, dy \, dx
\]
\[ \]
The next steps are:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} \left[ e^x \cdot \frac{1}{2}e^{2y} \right]_0^{\ln(3)} \, dx
\]
\[ \]
Further simplifying:
\[
= \frac{1}{\ln(3) - \ln(2)} \int_0^{\ln(2)} e^x \left( \frac{e^{2\ln(3)}}{2} \right) \, dx
\]
Please note: There are no graphs or diagrams present in the image, only mathematical expressions.
---
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