Evaluate the integral as a power series. 1. f(x) 3. f(x) S J) = freds f(2) 1 4. f(x) 2. f(x) = Σ n=0 5. f(x) = = = Σ η = 3 = Σ n=0 n=0 Σ n=0 z³n 3η + 2 (−1)n3n+2 3η + 2 x3n+2 3η + 2 23η 3η η -1) 3η 3η .
Evaluate the integral as a power series. 1. f(x) 3. f(x) S J) = freds f(2) 1 4. f(x) 2. f(x) = Σ n=0 5. f(x) = = = Σ η = 3 = Σ n=0 n=0 Σ n=0 z³n 3η + 2 (−1)n3n+2 3η + 2 x3n+2 3η + 2 23η 3η η -1) 3η 3η .
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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![**Evaluate the integral**
\[ f(z) = \int_{0}^{z} \frac{s}{1 - s^3} \, ds. \]
as a power series.
1. \( f(z) = \sum_{n=3}^{\infty} \frac{z^{3n}}{3n + 2} \)
2. \( f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{3n+2}}{3n + 2} \)
3. \( f(z) = \sum_{n=0}^{\infty} \frac{z^{3n+2}}{3n + 2} \)
4. \( f(z) = \sum_{n=0}^{\infty} \frac{z^{3n}}{3n} \)
5. \( f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{3n}}{3n} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4a40cee-b6e7-4175-94ff-7eded3f3eb47%2Feb9b7d9f-7692-4963-8f96-555fb074f7fb%2Fpuv53gm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Evaluate the integral**
\[ f(z) = \int_{0}^{z} \frac{s}{1 - s^3} \, ds. \]
as a power series.
1. \( f(z) = \sum_{n=3}^{\infty} \frac{z^{3n}}{3n + 2} \)
2. \( f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{3n+2}}{3n + 2} \)
3. \( f(z) = \sum_{n=0}^{\infty} \frac{z^{3n+2}}{3n + 2} \)
4. \( f(z) = \sum_{n=0}^{\infty} \frac{z^{3n}}{3n} \)
5. \( f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{3n}}{3n} \)
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