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 Statement:**
Find the general anti-derivative of \( f(x) = \frac{1}{\sqrt[3]{x}} + \sqrt[3]{x} \) on a continuous interval.
**Explanation:**
To solve this problem, you will need to integrate the function \( f(x) \). The function is composed of two terms:
1. \( \frac{1}{\sqrt[3]{x}} \) can be rewritten as \( x^{-\frac{1}{3}} \).
2. \( \sqrt[3]{x} \) can be rewritten as \( x^{\frac{1}{3}} \).
**Integration Steps:**
- For \( x^{-\frac{1}{3}} \), the integral is \( \frac{x^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2}x^{\frac{2}{3}} \).
- For \( x^{\frac{1}{3}} \), the integral is \( \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4}x^{\frac{4}{3}} \).
The general anti-derivative will include both integrals and a constant of integration \( C \):
\[ F(x) = \frac{3}{2}x^{\frac{2}{3}} + \frac{3}{4}x^{\frac{4}{3}} + C \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30c87a43-7b2c-40fe-85f1-990c225cb51e%2Fbdc602eb-15cf-4b1c-a985-b02fb2dda3e1%2Fir340nq_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the general anti-derivative of \( f(x) = \frac{1}{\sqrt[3]{x}} + \sqrt[3]{x} \) on a continuous interval.
**Explanation:**
To solve this problem, you will need to integrate the function \( f(x) \). The function is composed of two terms:
1. \( \frac{1}{\sqrt[3]{x}} \) can be rewritten as \( x^{-\frac{1}{3}} \).
2. \( \sqrt[3]{x} \) can be rewritten as \( x^{\frac{1}{3}} \).
**Integration Steps:**
- For \( x^{-\frac{1}{3}} \), the integral is \( \frac{x^{-\frac{1}{3} + 1}}{-\frac{1}{3} + 1} = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2}x^{\frac{2}{3}} \).
- For \( x^{\frac{1}{3}} \), the integral is \( \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4}x^{\frac{4}{3}} \).
The general anti-derivative will include both integrals and a constant of integration \( C \):
\[ F(x) = \frac{3}{2}x^{\frac{2}{3}} + \frac{3}{4}x^{\frac{4}{3}} + C \]
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