Find the integral S vVy+8 dy = .

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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### Problem Statement:
**Find the integral**

\[ \int y \sqrt[5]{y} + 8 \, dy = \]

In this problem, you are required to calculate the indefinite integral of the function \( y \sqrt[5]{y} + 8 \) with respect to \( y \).

### Explanation:

- The integral symbol \(\int\) indicates that you need to integrate the given function.
- \( y \sqrt[5]{y} \) is a term where \( y \) is multiplied by the fifth root of \( y \).
- \( 8 \) is a constant term.

This integral combines polynomial terms and involves applying the power rule for integration. Write your answer by integrating each term separately and then combining the results.

### Steps to Solve:

1. Express \( y \sqrt[5]{y} \) as \( y^{1 + \frac{1}{5}} = y^{\frac{6}{5}} \).
2. Integrate \( y^{\frac{6}{5}} \) using the power rule.
3. Integrate the constant \( 8 \).

After integrating, combine the results to get the final solution.
Transcribed Image Text:### Problem Statement: **Find the integral** \[ \int y \sqrt[5]{y} + 8 \, dy = \] In this problem, you are required to calculate the indefinite integral of the function \( y \sqrt[5]{y} + 8 \) with respect to \( y \). ### Explanation: - The integral symbol \(\int\) indicates that you need to integrate the given function. - \( y \sqrt[5]{y} \) is a term where \( y \) is multiplied by the fifth root of \( y \). - \( 8 \) is a constant term. This integral combines polynomial terms and involves applying the power rule for integration. Write your answer by integrating each term separately and then combining the results. ### Steps to Solve: 1. Express \( y \sqrt[5]{y} \) as \( y^{1 + \frac{1}{5}} = y^{\frac{6}{5}} \). 2. Integrate \( y^{\frac{6}{5}} \) using the power rule. 3. Integrate the constant \( 8 \). After integrating, combine the results to get the final solution.
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