what is integral of

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- Nov 29th 2011, 11:18 AMmoonnightingaleintegral of exp(-x^2)
what is integral of

- Nov 29th 2011, 01:20 PMTheEmptySetRe: integral of exp(-x^2)
$\displaystyle e^{-x^2}$ does not have an elementry antiderivative. The integral can be calulated using Taylor series.

$\displaystyle e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} $

So we can integrate this term by term to get

$\displaystyle \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!} $

This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating. - Nov 30th 2011, 04:02 AMmoonnightingaleRe: integral of exp(-x^2)
- Nov 30th 2011, 04:16 AMHallsofIvyRe: integral of exp(-x^2)
What do

**you**mean by "manual method"? - Nov 30th 2011, 05:34 AMmoonnightingaleRe: integral of exp(-x^2)
i mean how to solve this by hand

by standard integration method - Nov 30th 2011, 05:35 AMmoonnightingaleRe: integral of exp(-x^2)
Kindly tell me how u will solve this integration question on paper

- Nov 30th 2011, 06:20 AMCaptainBlackRe: integral of exp(-x^2)
You have already been told that this is not expressible a closed form in terms of elementary functions. You have been given a series approximation which will give the numerical value to any precision you require.

$\displaystyle \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}\\ \\ \phantom{SSSSSSS}=\sum_{n=0}^{N-1}\frac{(-1)^n}{(2n+1)n!} +R_N$

Where $\displaystyle |R_N|<\frac{1}{(2N+1)N!} $

CB - Nov 30th 2011, 08:28 AMmoonnightingaleRe: integral of exp(-x^2)
can u tell me answer of my question

with simpson rule it comes to be

0.7455 - Nov 30th 2011, 08:49 AMHallsofIvyRe: integral of exp(-x^2)
You have been told twice now, by TheEmptySet and Captain Black, that [itex]e^{-x^2}[/tex] does not have an "elementary" anti-derivative.

If you insist upon such an answer then it is $\displaystyle \frac{\sqrt{\pi}}{2} erf(1)$ where "erf" is the Gauss error function. It is**defined**as $\displaystyle \frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx$. Does that help? - Nov 30th 2011, 08:53 AMOpalgRe: integral of exp(-x^2)
If you multiply it by $\displaystyle 2/\sqrt\pi$, this integral becomes the error function, for which detailed tables of values are available (for example on the Wikipedia page that I just linked to). This gives erf(1) = 0.8427008. So your integral is that number times $\displaystyle \sqrt\pi/2$, or approximately 0.7468241.

- Nov 30th 2011, 09:16 AMAmerRe: integral of exp(-x^2)
cant we use that method suppose

$\displaystyle I = \int_{0}^{1} e^{-x^2} \cdot dx $

$\displaystyle I^2 = \int_{0}^{1} \int_{0}^{1} e^{-x^2}\cdot e^{-y^2} \; dx \;dy $

then transfer to polar ?? - Nov 30th 2011, 09:19 AMTheEmptySetRe: integral of exp(-x^2)