Q. 58 Use Theorem 8.12 and the results... [FREE SOLUTION]

Chapter 8: Q. 58 (page 701)

Use Theorem 8.12 and the results from Exercises 41鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.
${鈭�}_{0.5}^{2} x^{3} \cos (\frac{x}{2}) dx$

Short Answer

Expert verified

${鈭�}_{0.5}^{2} x^{3} \cos (\frac{x}{2}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {(\frac{1}{2})}^{2 k + 3} [\frac{2^{2 k + 4} - {(0.5)}^{2 k + 4}}{2 k + 4}]$

Step by step solution

Step 1. Given information is:

${鈭�}_{0.5}^{2} x^{3} \cos (\frac{x}{2}) dx$

Step 2. Definite integral

$From Q 48 . Maclaurin series for f (x) = x^{3} \cos (\frac{x}{2}) is x^{3} \cos (\frac{x}{2}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {(\frac{x}{2})}^{2 k + 3} Also, F = 鈭� f F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {(\frac{1}{2})}^{2 k + 3} (\frac{x^{2 k + 4}}{2 k + 4}) Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {(\frac{1}{2})}^{2 k + 3} {[\frac{x^{2 k + 4}}{2 k + 4}]}_{0.5}^{2} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k)!} {(\frac{1}{2})}^{2 k + 3} [\frac{2^{2 k + 4} - {(0.5)}^{2 k + 4}}{2 k + 4}]$

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Use Theorem 8.12 and the results from Exercises 41鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.
${鈭�}_{0.5}^{2} x^{3} \cos (\frac{x}{2}) dx$

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite integral

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91影视

Use Theorem 8.12 and the results from Exercises 41鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.鈭�0.52x3cosx2dx

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

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Pure Maths

Theoretical and Mathematical Physics

Applied Mathematics

Logic and Functions

Statistics

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Use Theorem 8.12 and the results from Exercises 41鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.
${鈭�}_{0.5}^{2} x^{3} \cos (\frac{x}{2}) dx$