Chapter 8: Q. 53 (page 701)

Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

Short Answer

Expert verified

$\int_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx = \sum_{k = 0}^{\infty} {(- \frac{1}{3})}^{k} \frac{1}{k!} [\frac{2}{2 k + 1}]$

Step by step solution

Step 1. Given information is:

$\int_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

Step 2. Definite integral

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

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Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$

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 41–50 to find series equal to the definite integrals in Exercises 51–60.∫-11e-x23dx

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 41–50 to find series equal to the definite integrals in Exercises 51–60.
$\int_{- 1}^{1} e^{\frac{- x^{2}}{3}} dx$