Chapter 8: Q. 51 (page 701)

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0}^{2} \sin (x^{3}) dx$

Short Answer

Expert verified

$\int_{0}^{2} \sin (x^{3}) dx = = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} [\frac{2^{6 k + 3}}{3 k + 2}]$

Step by step solution

Step 1. Given information is:

$\int_{0}^{2} \sin (x^{3}) dx$

Step 2. Definite Integral

$From Q 41 . Maclaurin series for f (x) = \sin (x^{3}) is \sin (x^{3}) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{6 k + 3} Also, F = \int f F (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} (\frac{x^{6 k + 4}}{6 k + 4}) Adding the limits, F (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} {[\frac{x^{6 k + 4}}{6 k + 4}]}_{0}^{2} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} [\frac{2^{6 k + 4} - 0}{6 k + 4}] = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} [\frac{2^{6 k + 4}}{6 k + 4}] = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} [\frac{2^{6 k + 3}}{3 k + 2}]$

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Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0}^{2} \sin (x^{3}) dx$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite Integral

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Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.∫02sin(x3)dx

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite Integral

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Discrete Mathematics

Geometry

Statistics

Applied Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{0}^{2} \sin (x^{3}) dx$