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

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∻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.⚹◊How can the Maclaurin series expansion be used to approximate the integral $\int e^{x^2} \, dx$⍭ Generate me a similar question◊

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