Q. 53 Use Theorem 8.12 and the results... [FREE SOLUTION]

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Chapter 8: Q. 53 (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.
${鈭�}_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information is:

${鈭�}_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

Step 2. Definite integral

$From Q 43 . Maclaurin series for f (x) = e^{\frac{- x^{2}}{3}} dx is e^{\frac{- x^{2}}{3}} = {鈭�}_{k = 0}^{鈭�} (- \frac{1}{3}) \frac{1}{k!} x^{2 k} Also, F = 鈭� f F (x) = = {鈭�}_{k = 0}^{鈭�} {(- \frac{1}{3})}^{k} \frac{1}{k!} (\frac{x^{2 k + 1}}{2 k + 1}) Adding the limits, F (x) = {鈭�}_{k = 0}^{鈭�} {(- \frac{1}{3})}^{k} \frac{1}{k!} {[\frac{x^{2 k + 1}}{2 k + 1}]}_{- 1}^{1} = {鈭�}_{k = 0}^{鈭�} {(- \frac{1}{3})}^{k} \frac{1}{k!} [\frac{1^{2 k + 1} - {(- 1)}^{2 k + 1}}{2 k + 1}] = {鈭�}_{k = 0}^{鈭�} {(- \frac{1}{3})}^{k} \frac{1}{k!} [\frac{2}{2 k + 1}]$

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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.
${鈭�}_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

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�-11e-x23dx

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Definite integral

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Discrete Mathematics

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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.
${鈭�}_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$