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 an antiderivative of \( v(x) = \sec^2(8x) + 3x^3 \). Assume that the value of the arbitrary constant is 0.
**Solution Box:**
The answer section includes a boxed space for \( V(x) = \), where the antiderivative will be written.
**Explanation:**
The function given is composed of two parts: \( \sec^2(8x) \) and \( 3x^3 \).
1. **Antiderivative of \( \sec^2(8x) \):**
The antiderivative of \( \sec^2(kx) \) is \( \frac{1}{k} \tan(kx) \). Thus, for \( \sec^2(8x) \), the antiderivative is:
\[
\frac{1}{8} \tan(8x)
\]
2. **Antiderivative of \( 3x^3 \):**
The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Therefore, for \( 3x^3 \), the antiderivative is:
\[
\frac{3x^4}{4}
\]
By adding these two results together, we find the antiderivative \( V(x) \):
\[
V(x) = \frac{1}{8} \tan(8x) + \frac{3x^4}{4}
\]
Given that the arbitrary constant is assumed to be 0, this is the complete expression for \( V(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bb6e943-00f0-4d56-9172-c2b4ab2d26a2%2F751d5281-b0a2-45d4-8821-20dedead4a8a%2Fxvqssha_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an antiderivative of \( v(x) = \sec^2(8x) + 3x^3 \). Assume that the value of the arbitrary constant is 0.
**Solution Box:**
The answer section includes a boxed space for \( V(x) = \), where the antiderivative will be written.
**Explanation:**
The function given is composed of two parts: \( \sec^2(8x) \) and \( 3x^3 \).
1. **Antiderivative of \( \sec^2(8x) \):**
The antiderivative of \( \sec^2(kx) \) is \( \frac{1}{k} \tan(kx) \). Thus, for \( \sec^2(8x) \), the antiderivative is:
\[
\frac{1}{8} \tan(8x)
\]
2. **Antiderivative of \( 3x^3 \):**
The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Therefore, for \( 3x^3 \), the antiderivative is:
\[
\frac{3x^4}{4}
\]
By adding these two results together, we find the antiderivative \( V(x) \):
\[
V(x) = \frac{1}{8} \tan(8x) + \frac{3x^4}{4}
\]
Given that the arbitrary constant is assumed to be 0, this is the complete expression for \( V(x) \).
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