Q. 60 Use Theorem 8.12 and the results... [FREE SOLUTION]

Chapter 8: Q. 60 (page 702)

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.
${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$

Short Answer

Expert verified

${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} [\frac{{(0.3)}^{6 k + 8}}{6 k + 8}]$

Step by step solution

Step 1. Given information is:

${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$

Step 2. Definite integral

$From Q 50 . Maclaurin series for f (x) = x^{4} \tan^{- 1} (3 x^{3}) is x^{4} \tan^{- 1} (3 x^{3}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} x^{6 k + 7} Also, F = 鈭� f F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} (\frac{x^{6 k + 8}}{6 k + 8}) Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} {[\frac{x^{6 k + 8}}{6 k + 8}]}_{0}^{0.3} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} [\frac{{(0.3)}^{6 k + 8} - {(0)}^{6 k + 8}}{6 k + 8}] = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(3)}^{2 k + 1} [\frac{{(0.3)}^{6 k + 8}}{6 k + 8}]$

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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.
${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$

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鈥�50 to find series equal to the definite integrals in Exercises 51鈥�60.鈭�00.3x4tan-1(3x3)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

Applied Mathematics

Mechanics Maths

Logic and Functions

Decision Maths

Geometry

Statistics

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.
${鈭�}_{0}^{0.3} x^{4} \tan^{- 1} (3 x^{3}) dx$