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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![**Evaluate the Iterated Integral**
To evaluate the iterated integral given by:
\[
\int_{0}^{\ln 3} \int_{1}^{\ln 5} e^{2x + 4y} \, dy \, dx
\]
We first need to integrate with respect to \( y \), and then with respect to \( x \).
To begin, we integrate the inner integral:
\[
\int_{1}^{\ln 5} e^{2x + 4y} \, dy
\]
Treating \( e^{2x} \) as a constant with respect to \( y \).
The integrand \( e^{2x + 4y} \) can be rewritten as:
\[
e^{2x} \cdot e^{4y}
\]
Thus, we integrate \( e^{4y} \) with respect to \( y \):
\[
\int e^{4y} \, dy = \frac{1}{4} e^{4y}
\]
Evaluating this from \( y = 1 \) to \( y = \ln 5 \), we get:
\[
\left. \frac{1}{4} e^{4y} \right|_{1}^{\ln 5} = \frac{1}{4} \left( e^{4 \ln 5} - e^4 \right) = \frac{1}{4} \left( 5^4 - e^4 \right)
\]
Since \( 5^4 = 625 \), we have:
\[
\frac{1}{4} (625 - e^4)
\]
Now we multiply this result by \( e^{2x} \) and perform the outer integral with respect to \( x \):
\[
\int_{0}^{\ln 3} e^{2x} \cdot \frac{1}{4} (625 - e^4) \, dx
\]
Simplifying constants:
\[
\frac{625 - e^4}{4} \int_{0}^{\ln 3} e^{2x} \, dx
\]
Integrate \( e^{2x} \) with respect to \( x \):
\[
\int e^{2x} \, dx = \frac{1}{2} e^{2x}
\]
Evaluating](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1840ebe-b0df-4fe8-9210-d8e3dcfa32cc%2F9a265fe2-a467-4803-b68d-dd7898569b34%2Faq3l7k5_processed.png&w=3840&q=75)
Transcribed Image Text:**Evaluate the Iterated Integral**
To evaluate the iterated integral given by:
\[
\int_{0}^{\ln 3} \int_{1}^{\ln 5} e^{2x + 4y} \, dy \, dx
\]
We first need to integrate with respect to \( y \), and then with respect to \( x \).
To begin, we integrate the inner integral:
\[
\int_{1}^{\ln 5} e^{2x + 4y} \, dy
\]
Treating \( e^{2x} \) as a constant with respect to \( y \).
The integrand \( e^{2x + 4y} \) can be rewritten as:
\[
e^{2x} \cdot e^{4y}
\]
Thus, we integrate \( e^{4y} \) with respect to \( y \):
\[
\int e^{4y} \, dy = \frac{1}{4} e^{4y}
\]
Evaluating this from \( y = 1 \) to \( y = \ln 5 \), we get:
\[
\left. \frac{1}{4} e^{4y} \right|_{1}^{\ln 5} = \frac{1}{4} \left( e^{4 \ln 5} - e^4 \right) = \frac{1}{4} \left( 5^4 - e^4 \right)
\]
Since \( 5^4 = 625 \), we have:
\[
\frac{1}{4} (625 - e^4)
\]
Now we multiply this result by \( e^{2x} \) and perform the outer integral with respect to \( x \):
\[
\int_{0}^{\ln 3} e^{2x} \cdot \frac{1}{4} (625 - e^4) \, dx
\]
Simplifying constants:
\[
\frac{625 - e^4}{4} \int_{0}^{\ln 3} e^{2x} \, dx
\]
Integrate \( e^{2x} \) with respect to \( x \):
\[
\int e^{2x} \, dx = \frac{1}{2} e^{2x}
\]
Evaluating
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