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 3
**Objective**: Evaluate the integral \(\int x^4 \cos (3x^2) \, dx\) as an infinite series. (You do not need to find the radius of convergence.)
---
**Solution Approach**:
1. **Express the Integrand Using a Series**: Begin by expanding \(\cos(3x^2)\) into its Taylor series. The general Taylor series expansion for \(\cos(u)\) is given by:
\[
\cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!}
\]
Here, \(u = 3x^2\).
2. **Substitute into the Integral**: Replace \(\cos(3x^2)\) in the integral with its series expansion:
\[
\cos(3x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (3x^2)^{2n}}{(2n)!}
\]
Therefore, our integral becomes:
\[
\int x^4 \cos (3x^2) \, dx = \int x^4 \left( \sum_{n=0}^{\infty} \frac{(-1)^n (3x^2)^{2n}}{(2n)!} \right) \, dx
\]
3. **Combine and Simplify**: Simplify the expression inside the integral:
\[
\int x^4 \sum_{n=0}^{\infty} \frac{(-1)^n (3^{2n} x^{4n})}{(2n)!} \, dx = \sum_{n=0}^{\infty} \int x^4 \frac{(-1)^n 3^{2n} x^{4n}}{(2n)!} \, dx
\]
This equation follows by interchanging the summation and integration (justified if the series converges uniformly).
4. **Simplify Further**: Combine \(x^4\) and \(x^{4n}\):
\[
\sum_{n=0}^{\infty} \frac{(-1)^n 3^{2n}}{(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa216c9ee-ecac-40ec-b583-a361985d4bbd%2F80101ab7-abc8-4551-9490-15eabafb8209%2Fk0lljki_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 3
**Objective**: Evaluate the integral \(\int x^4 \cos (3x^2) \, dx\) as an infinite series. (You do not need to find the radius of convergence.)
---
**Solution Approach**:
1. **Express the Integrand Using a Series**: Begin by expanding \(\cos(3x^2)\) into its Taylor series. The general Taylor series expansion for \(\cos(u)\) is given by:
\[
\cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!}
\]
Here, \(u = 3x^2\).
2. **Substitute into the Integral**: Replace \(\cos(3x^2)\) in the integral with its series expansion:
\[
\cos(3x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (3x^2)^{2n}}{(2n)!}
\]
Therefore, our integral becomes:
\[
\int x^4 \cos (3x^2) \, dx = \int x^4 \left( \sum_{n=0}^{\infty} \frac{(-1)^n (3x^2)^{2n}}{(2n)!} \right) \, dx
\]
3. **Combine and Simplify**: Simplify the expression inside the integral:
\[
\int x^4 \sum_{n=0}^{\infty} \frac{(-1)^n (3^{2n} x^{4n})}{(2n)!} \, dx = \sum_{n=0}^{\infty} \int x^4 \frac{(-1)^n 3^{2n} x^{4n}}{(2n)!} \, dx
\]
This equation follows by interchanging the summation and integration (justified if the series converges uniformly).
4. **Simplify Further**: Combine \(x^4\) and \(x^{4n}\):
\[
\sum_{n=0}^{\infty} \frac{(-1)^n 3^{2n}}{(
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