How would I find the Maclaurian series for f(x)? (attached) (2n)!2"n!(2n-1)Find a Maclaurin series for f(x). (Usefor 1: 3· 5 ... (2n – 3).)f(x) =I V1+5 dtf(x) = x +8Σn = 2+

given fx=∫0x1+t3dt to find the maclaurin series

Answered: Find a Maclaurin series for f(x). (Use…

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Find a Maclaurin series for f(x). (Use for 1·3· 5·.. (2n – 3).) 2"n!(2n-1) f(x) V1 + t3 dt f(x) = x + 8 Σ n = 2

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 49E

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Question

How would I find the Maclaurian series for f(x)? (attached)

(2n)!
2"n!(2n-1)
Find a Maclaurin series for f(x). (Use
for 1: 3· 5 ... (2n – 3).)
f(x) =
I V1+5 dt
f(x) = x +
8
Σ
n = 2
+

Expert Solution

Step 1

given

$f (x) = \int_{0}^{x} \sqrt{1 + t^{3}} d t$

to find the maclaurin series

Step 2

$\sqrt{1 + t^{3}} = 1 + \sum_{k = 1}^{\infty} \frac{\frac{1}{2} (\frac{1}{2} - 1) . . . . . . (\frac{1}{2} - k + 1)}{k!} t^{3 k} = 1 + \sum_{k = 1}^{\infty} \frac{1 (- 1) (- 3) . . . . . . (- 2 k + 3)}{2^{k} k!} t^{3 k} = 1 + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k} (2 k)!}{2^{k} k! (2 k - 1) k!} t^{3 k} = 1 + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k + 1} (2 k)!}{2^{2 k} k! (2 k - 1) k!} t^{3 k} = 1 + \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k + 1} (2 k)!}{4^{k} {(k!)}^{2} (2 k - 1)} t^{3 k}$

substituting the value of $\sqrt{1 + t^{3}}$ into $f (x) = \int_{0}^{x} \sqrt{1 + t^{3}} d t$

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