Q. 57 Use Theorem 8.12 and the results... [FREE SOLUTION]

Chapter 8: Q. 57 (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.
${鈭�}_{1}^{2} x^{2} \sin (5 x^{2}) dx$

Short Answer

Expert verified

${鈭�}_{1}^{2} x^{2} \sin (5 x^{2}) dx = = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} [\frac{2^{4 k + 5} - 1}{4 k + 5}]$

Step by step solution

Step 1. Given information is:

${鈭�}_{1}^{2} x^{2} \sin (5 x^{2}) dx$

Step 2. Definite integral

$From Q 47 . Maclaurin series for f (x) = x^{2} \sin (5 x^{2}) is x^{2} \sin (5 x^{2}) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} x^{4 k + 4} Also, F = 鈭� f F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} \frac{x^{4 k + 5}}{4 k + 5} Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} {[\frac{x^{4 k + 5}}{4 k + 5}]}_{1}^{2} = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} [\frac{2^{4 k + 5} - 1^{4 k + 5}}{4 k + 5}] = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} 5^{2 k + 1} [\frac{2^{4 k + 5} - 1}{4 k + 5}]$

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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.
${鈭�}_{1}^{2} x^{2} \sin (5 x^{2}) dx$

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�12x2sin(5x2)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

Geometry

Decision Maths

Mechanics Maths

Theoretical and Mathematical Physics

Applied Mathematics

Discrete Mathematics

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.
${鈭�}_{1}^{2} x^{2} \sin (5 x^{2}) dx$