`Solve the integral $int e^{x^2} , dx$ using a series expansion method.`

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

First, we need the series expansion of $e^{x^2}$. The Taylor series expansion of $e^{x^2}$ around $x=0$ is given by:
$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \frac{x^8}{24} + O(x^9)$

step 2

Next, we integrate the series term by term from 0 to 1:
$\int_0^1 \left(1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \frac{x^8}{24}\right) dx$

step 3

Integrate each term separately:
$\int_0^1 1 \, dx = x \Big|_0^1 = 1$
$\int_0^1 x^2 \, dx = \frac{x^3}{3} \Big|_0^1 = \frac{1}{3}$
$\int_0^1 \frac{x^4}{2} \, dx = \frac{x^5}{10} \Big|_0^1 = \frac{1}{10}$
$\int_0^1 \frac{x^6}{6} \, dx = \frac{x^7}{42} \Big|_0^1 = \frac{1}{42}$
$\int_0^1 \frac{x^8}{24} \, dx = \frac{x^9}{216} \Big|_0^1 = \frac{1}{216}$

step 4

Sum the results of the integrals:
$1 + \frac{1}{3} + \frac{1}{10} + \frac{1}{42} + \frac{1}{216} = \frac{11051}{7560} \approx 1.4618$

Answer

The integral of $e^{x^2}$ from 0 to 1 using the series expansion method is approximately 1.4618.

Key Concept

Series Expansion

Explanation

The series expansion method involves expanding the function into a Taylor series and then integrating term by term. This approach is useful for functions that do not have elementary antiderivatives.

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