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
![**Problem 2: Analysis of Integrals Without Direct Calculation**
**Objective:**
Prove, without calculating the integrals explicitly, that:
\[
\int_{0}^{1} \sqrt{1 + x^3} \, dx \leq \int_{0}^{1} \sqrt{1 + x^2} \, dx
\]
**Explanation:**
This problem requires understanding the properties of the functions under the integral signs. Here, \(\sqrt{1 + x^3}\) and \(\sqrt{1 + x^2}\) are compared over the interval from 0 to 1.
**Approach:**
1. **Function Analysis:** Compare \( \sqrt{1 + x^3} \) and \( \sqrt{1 + x^2} \).
- Note that \( x^3 \leq x^2 \) for \( x \) in the interval \([0, 1]\).
- Consequentially, \( 1 + x^3 \leq 1 + x^2 \).
- This implies \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \) since the square root function is monotonically increasing for non-negative values.
2. **Integral Comparison:**
- Since \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \) for all \( x \) in the interval \([0, 1]\), the integral of the lesser function is bounded by the integral of the greater function over the same interval.
Conclusion:
Thus, the inequality \(\int_{0}^{1} \sqrt{1 + x^3} \, dx \leq \int_{0}^{1} \sqrt{1 + x^2} \, dx\) holds true without explicitly computing these integrals.
**Educational Insight:**
This problem exemplifies using function properties and comparisons to infer relations between integrals, demonstrating an important analytical technique in calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faea7a391-77a2-4be9-b3a3-98e0f70bf810%2F0ba221b1-18df-4c28-8502-73b98d3613a4%2Fervgi8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Analysis of Integrals Without Direct Calculation**
**Objective:**
Prove, without calculating the integrals explicitly, that:
\[
\int_{0}^{1} \sqrt{1 + x^3} \, dx \leq \int_{0}^{1} \sqrt{1 + x^2} \, dx
\]
**Explanation:**
This problem requires understanding the properties of the functions under the integral signs. Here, \(\sqrt{1 + x^3}\) and \(\sqrt{1 + x^2}\) are compared over the interval from 0 to 1.
**Approach:**
1. **Function Analysis:** Compare \( \sqrt{1 + x^3} \) and \( \sqrt{1 + x^2} \).
- Note that \( x^3 \leq x^2 \) for \( x \) in the interval \([0, 1]\).
- Consequentially, \( 1 + x^3 \leq 1 + x^2 \).
- This implies \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \) since the square root function is monotonically increasing for non-negative values.
2. **Integral Comparison:**
- Since \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \) for all \( x \) in the interval \([0, 1]\), the integral of the lesser function is bounded by the integral of the greater function over the same interval.
Conclusion:
Thus, the inequality \(\int_{0}^{1} \sqrt{1 + x^3} \, dx \leq \int_{0}^{1} \sqrt{1 + x^2} \, dx\) holds true without explicitly computing these integrals.
**Educational Insight:**
This problem exemplifies using function properties and comparisons to infer relations between integrals, demonstrating an important analytical technique in calculus.
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