Q. 54 Use Theorem 8.12 and the results... [FREE SOLUTION]

Chapter 8: Q. 54 (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.
${鈭�}_{0.1}^{0.2} \tan^{- 1} (x^{2}) dx$

Short Answer

Expert verified

${鈭�}_{0.1}^{0.2} \tan^{- 1} (x^{2}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} [\frac{0 . 2^{4 k + 3} - 0 . 1^{4 k + 3}}{4 k + 3}]$

Step by step solution

Step 1. Given information is:

${鈭�}_{0.1}^{0.2} \tan^{- 1} (x^{2}) dx$

Step 2. Definite integral

$From Q 44 . Maclaurin series for f (x) = \tan^{- 1} (x^{2}) is \tan^{- 1} (x^{2}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} x^{4 k + 2} Also, F = 鈭� f F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} [\frac{x^{4 k + 3}}{4 k + 3}] Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {[\frac{x^{4 k + 3}}{4 k + 3}]}_{0.1}^{0.2} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} [\frac{0 . 2^{4 k + 3} - 0 . 1^{4 k + 3}}{4 k + 3}]$

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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.
${鈭�}_{0.1}^{0.2} \tan^{- 1} (x^{2}) dx$

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鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.鈭�0.10.2tan-1(x2)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

Decision Maths

Statistics

Calculus

Geometry

Logic and Functions

Study anywhere. Anytime. Across all devices.

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.
${鈭�}_{0.1}^{0.2} \tan^{- 1} (x^{2}) dx$