Which of the following are equivalent to T 1. II. III. A (В (С) D (3x+11) d'a (x+2)dx √₁² (³ 5 x + 2 [² (3 + 5 ) du U 3+ I only Il only Ill only II and III only dx 5 3x + 11 x + 2 -dx?

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...
icon
Related questions
Question
100%
**Topic: Evaluating Equivalent Integrals**

**Question:**

Which of the following are equivalent to 

\[
\int_0^5 \frac{3x + 11}{x + 2} \, dx \, ?
\]

**Options:**

I. \(\int_0^5 (3x + 11) \, dx - \int_0^5 (x + 2) \, dx\)

II. \(\int_0^5 \left( 3 + \frac{5}{x + 2} \right) \, dx\)

III. \(\int_2^7 \left( 3 + \frac{5}{u} \right) \, du\)

**Choices:**

A) I only

B) II only

C) III only

D) II and III only

---

**Solution Explanation:**

To find equivalency, we transform the original integral. The given expression \(\int_0^5 \frac{3x + 11}{x + 2} \, dx\) can be broken down and compared with the options:

- **Option I** states that the given integral can be split into two separate integrals. This is incorrect because splitting the integrand in such a manner does not preserve the original structure of the rational function.
  
- **Option II** correctly simplifies the integrand using polynomial division: \(3x+11 = (3)(x + 2) + 5\). Thus, \(\frac{3x + 11}{x + 2} = 3 + \frac{5}{x + 2}\). This matches Option II.
  
- **Option III** involves changing the variable of integration from \(x\) to \(u\), with limits adjusted appropriately if \(u = x + 2\), making \(u\) range from 2 to 7. This means Option III is representing the same integral as Option II under a substitution.

Thus, both Option II and Option III are equivalent to the original integral.

**Correct Answer:** D) II and III only
Transcribed Image Text:**Topic: Evaluating Equivalent Integrals** **Question:** Which of the following are equivalent to \[ \int_0^5 \frac{3x + 11}{x + 2} \, dx \, ? \] **Options:** I. \(\int_0^5 (3x + 11) \, dx - \int_0^5 (x + 2) \, dx\) II. \(\int_0^5 \left( 3 + \frac{5}{x + 2} \right) \, dx\) III. \(\int_2^7 \left( 3 + \frac{5}{u} \right) \, du\) **Choices:** A) I only B) II only C) III only D) II and III only --- **Solution Explanation:** To find equivalency, we transform the original integral. The given expression \(\int_0^5 \frac{3x + 11}{x + 2} \, dx\) can be broken down and compared with the options: - **Option I** states that the given integral can be split into two separate integrals. This is incorrect because splitting the integrand in such a manner does not preserve the original structure of the rational function. - **Option II** correctly simplifies the integrand using polynomial division: \(3x+11 = (3)(x + 2) + 5\). Thus, \(\frac{3x + 11}{x + 2} = 3 + \frac{5}{x + 2}\). This matches Option II. - **Option III** involves changing the variable of integration from \(x\) to \(u\), with limits adjusted appropriately if \(u = x + 2\), making \(u\) range from 2 to 7. This means Option III is representing the same integral as Option II under a substitution. Thus, both Option II and Option III are equivalent to the original integral. **Correct Answer:** D) II and III only
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning