**Problem Statement:**Use a power series to compute the integral:\[\int_0^1 e^{-x^2} \, dx =\]**Explanation:**This problem asks you to evaluate the integral of the function $ e^{-x^2} $ from 0 to 1 using a power series expansion. The function $ e^{-x^2} $ does not have an elementary antiderivative, so power series expansion is one of the methods to approach this integral. The power series expansion for $ e^{-x^2} $ can be derived from the Taylor series for $ e^x $ by substituting $ -x^2 $ into the series. Once the series is established, you integrate term by term within the limits from 0 to 1.

Given that, the integral is ∫01e-x2dx. It is known that, the power series representation of…

Answered: Use a power series to compute: dx

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Problem Statement:**

Use a power series to compute the integral:

\[
\int_0^1 e^{-x^2} \, dx =
\]

**Explanation:**

This problem asks you to evaluate the integral of the function $ e^{-x^2} $ from 0 to 1 using a power series expansion. The function $ e^{-x^2} $ does not have an elementary antiderivative, so power series expansion is one of the methods to approach this integral. The power series expansion for $ e^{-x^2} $ can be derived from the Taylor series for $ e^x $ by substituting $ -x^2 $ into the series. Once the series is established, you integrate term by term within the limits from 0 to 1.

Expert Solution

Step 1

Given that, the integral is $\int_{0}^{1} e^{- x^{2}} d x$ .

It is known that, the power series representation of $e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$ .

Substitute $x = - x^{2} i n e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$ .

$\begin{array}{rcl} e^{- x^{2}} & = & \sum_{n = 0}^{\infty} \frac{{(- x^{2})}^{n}}{n!} \\ = & \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{{(x)}^{2 n}}{n!} \end{array}$

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