find-a-better-approximation-for-the-integrals-1-0-1-e-x-2-dx-2-1-e-x-2-dx-

Question Number 37281 by abdo.msup.com last updated on 11/Jun/18

f i n d a b e t t e r a p p r o x i m a t i o n f o r t h e

i n t e g r a l s

1) \int_{0}^{1} e^{- x^{2}} d x

2) \int_{1}^{+ \infty} e^{- x^{2}} d x .

Commented by prof Abdo imad last updated on 18/Jun/18

w e h a v e e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!} \Rightarrow

\int_{0}^{1} e^{- x^{2}} d x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} \int_{0}^{1} x^{2 n} d x

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n! (2 n + 1)} l e t t a k e 8 t e r m s w e g e t

\int_{0}^{1} e^{- x^{2}} d x \sim 1 - \frac{1}{3} + \frac{1}{5.2!} - \frac{1}{7.3!} + \frac{1}{9.4!}

- \frac{1}{11.5!} + \frac{1}{13.6!} - \frac{1}{15.7!} + \frac{1}{17.8!} \Rightarrow \dots

2) \int_{0}^{+ \infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} = \int_{0}^{1} e^{- x^{2}} d x + \int_{1}^{+ \infty} e^{- x^{2}} d x

\Rightarrow \int_{1}^{+ \infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} - \int_{0}^{1} e^{- x^{2}} d x a n d w e u s e

a p p r o x i m a t e v a l u e f r o m Q 1 \dots