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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Question
q5
![**Problem Statement:**
Evaluate the integral
\[ \int_{0}^{2e} \frac{x^2}{x^3 + e} \, dx \]
**Explanation:**
In this problem, you are tasked with evaluating the given definite integral. Below is a step-by-step solution to solve the integral.
1. **Integrate using the given bounds:**
Define the integral \( I \) as:
\[
I = \int_{0}^{2e} \frac{x^2}{x^3 + e} \, dx
\]
2. **Substitution Method:**
To simplify the integral, let \( u = x^3 + e \). Consequently, the differential \( du \) is:
\[
du = 3x^2 \, dx \implies x^2 \, dx = \frac{1}{3} \, du
\]
Now adjust the limits of integration according to the substitution:
- When \( x = 0 \),
\[
u = 0^3 + e = e
\]
- When \( x = 2e \),
\[
u = (2e)^3 + e = 8e^3 + e
\]
3. **Rewrite the integral:**
Substitute the variables and limits into the integral:
\[
I = \int_{e}^{8e^3 + e} \frac{1}{3} \frac{1}{u} \, du = \frac{1}{3} \int_{e}^{8e^3 + e} \frac{1}{u} \, du
\]
4. **Integrate with respect to u:**
The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln |u| \):
\[
I = \frac{1}{3} \left[ \ln |u| \right]_{e}^{8e^3 + e}
\]
Evaluate the limits:
\[
I = \frac{1}{3} \left[ \ln(8e^3 + e) - \ln(e) \right]
\]
5. **Simplify using logarithm properties:**
Use the property \( \ln(a) - \ln(b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddf14729-dbf2-4574-b793-bebd8df9c378%2F519f7947-3652-411a-9b4b-a0cce6e4db8b%2Fg29a71h_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Evaluate the integral
\[ \int_{0}^{2e} \frac{x^2}{x^3 + e} \, dx \]
**Explanation:**
In this problem, you are tasked with evaluating the given definite integral. Below is a step-by-step solution to solve the integral.
1. **Integrate using the given bounds:**
Define the integral \( I \) as:
\[
I = \int_{0}^{2e} \frac{x^2}{x^3 + e} \, dx
\]
2. **Substitution Method:**
To simplify the integral, let \( u = x^3 + e \). Consequently, the differential \( du \) is:
\[
du = 3x^2 \, dx \implies x^2 \, dx = \frac{1}{3} \, du
\]
Now adjust the limits of integration according to the substitution:
- When \( x = 0 \),
\[
u = 0^3 + e = e
\]
- When \( x = 2e \),
\[
u = (2e)^3 + e = 8e^3 + e
\]
3. **Rewrite the integral:**
Substitute the variables and limits into the integral:
\[
I = \int_{e}^{8e^3 + e} \frac{1}{3} \frac{1}{u} \, du = \frac{1}{3} \int_{e}^{8e^3 + e} \frac{1}{u} \, du
\]
4. **Integrate with respect to u:**
The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln |u| \):
\[
I = \frac{1}{3} \left[ \ln |u| \right]_{e}^{8e^3 + e}
\]
Evaluate the limits:
\[
I = \frac{1}{3} \left[ \ln(8e^3 + e) - \ln(e) \right]
\]
5. **Simplify using logarithm properties:**
Use the property \( \ln(a) - \ln(b
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