Question

Can you help me to evaluate integral?

Answer 1

\[\int\limits_{-1}^{1} (x+e^{-\frac{ 1 }{ 2 }x^2}) dx\]

Answer 2

I know how to evaluate \[\int\limits_{-\infty}^{\infty} e^{-\frac{1}{2}x^2} dx\] but I don't know how to do it over interval [-1,1]

Answer 3

you know what the integral of an x is going to be in this case.

Answer 4

\(\large\color{black}{ e^{\frac{1}{2}x^2}}\) is not really an odd function, because the exponent wouldn't make it negative.

Answer 5

wait, are you sure it is: \(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{\frac{1}{2}x^2}~dx}\) ? (without an x next to it) ?

Answer 6

yeah, please verify whether x is added to e term or multiplied...

Answer 7

yes. only there is minus sign in exponent

Answer 8

\(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{-\frac{1}{2}x^2}~dx}\) ?

Answer 9

\[\int\limits_{-1}^{1} (x+e^{-\frac{1}{2}x^2}) dx\]

Answer 10

are you expecting answer in terms of error function erf ?
http://www.wolframalpha.com/input/?i=integral+-1+to+1+++%28++e%5E%7B-x%5E2%2F2%7D%29

Answer 11

I took a look at wolfram before. In fact, I need only the value of that integral but I would like to know how to get it. I tried doing it using polar coordinates, but I got another value (1,572336)

Answer 12

I was thinking of:

\(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{-\frac{1}{2}x^2}~dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{-\frac{1}{2}x^2}~dx=xe^{-\frac{1}{2}x^2}+\displaystyle\int\limits_{~}^{~} x^2e^{-\frac{1}{2}x^2}~dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{-\frac{1}{2}x^2}~dx=xe^{-\frac{1}{2}x^2}+\displaystyle\int\limits_{~}^{~} \frac{x^2\times x}{x}e^{-\frac{1}{2}x^2}~dx}\)

u=x^2, -4du=x dx, (again x^2=u, and) x=sqrt{-2u}

\(\large\color{black}{\displaystyle\int\limits_{~}^{~} e^{-\frac{1}{2}x^2}~dx=xe^{-\frac{1}{2}x^2}-4\displaystyle\int\limits_{~}^{~} \frac{u}{\sqrt{-2u}}e^{u}~dx}\)

Answer 13

is that the original question ? or a subpart of it...may we have a screenshot of the original Q ?

Answer 14

I tried wolfram too, to see if it is that I am dump that can't do it, or is the problem un-do-able...

Answer 15

in last line should be du, but it doesn't matter I either way can't do it...

Answer 16

or if you have the answer, it may help

Answer 17

the exponent is negative, so the graph is this: https://www.desmos.com/calculator/rf8bpeho1i

Answer 18

even function, so you can as well try to evaluate from 0 to 1

Answer 19

Original question is to generate random numbers from distribution\[f(x) \approx x+e^{-\frac{x^2}{2}} , x\in[-1,1]\]. So I have to find a coefficient C such that \[\int\limits_{-1}^{1} Cf(x)=1\] so that f(x) is probability density function.

Answer 20

I think you can forget about \(x\) as it is odd :
\[\int\limits_{-1}^{1} (x+e^{-\frac{1}{2}x^2}) dx = \int\limits_{-1}^{1} x~dx+\int\limits_{-1}^{1}e^{-\frac{1}{2}x^2}dx = 0 + \int\limits_{-1}^{1}e^{-\frac{1}{2}x^2}dx \]

Answer 21

evaluating that integral is trivial if you use error function

Answer 22

\[ \int\limits_0^x e^{-t^2} dt = \frac{\sqrt{\pi}}{2}\mathrm{erf x} \]

Answer 23

As a start substitute \(\frac{x}{\sqrt{2}} = u\)

Answer 24

erf?

Answer 25

Ok, I have this:
\[\int\limits_{-1}^{1} e^{-\frac{x^2}{2}} dx=\int\limits_{-1}^{1} e^{-u^2} \sqrt{2} du=\sqrt{2}\int\limits_{-1}^{1} e^{-u^2}du\]
but now I stuck here :/ Can you help me about error function. How to use it?

Answer 26

so far so good!

Answer 27

http://en.wikipedia.org/wiki/Error_function
just for the reference

Answer 28

erf(x) is just a new function defined based on integrals like log function :
\[\int e^{-t^2} dt = \frac{\sqrt{\pi}}{2}\mathrm{erf(x)} + C\]

Answer 29

\(\int \limits_0^xe^{-t^2} dt = \frac{\sqrt{\pi}}{2}\mathrm{erf(x)} \)

Answer 30

I think that:
\[\int\limits_{-1}^{1}e ^{-x ^{2}/2}dx=\sqrt{2 \pi}*68.27\]
since the integral is proportional to the function erf(t)

Answer 31

\[\sqrt{2}\int\limits_{x=-1}^{x=1} e^{-u^2}du = \sqrt{2}\cdot \frac{\sqrt{\pi}}{2} \mathrm{erf(u)} \Bigg|_{x=-1}^{x=1}\]

Answer 32

evaluate ^

Answer 33

or since \(e^{-u^2}\)is an even function, you can as well evaluate it from 0 to 1 and them multiply it by 2

Answer 34

\[\int\limits_{-1}^{1} e^{-\frac{u^2}{2}} du=-\int\limits_{0}^{-1}e^{-u^2} du+\int\limits_{0}^{1}e^{-u^2}du\]

Can I do it like this? And then to use error function ?

Answer 35

you can do that but the indefinite integral form is more easy to use here. dont get so stuck up with the bounds

Answer 36

if you want to split the integral, use the fact that the integrand is even function :

\[\int\limits_{-1}^{1} e^{-u^2} du=2\int\limits_{0}^{1}e^{-u^2}du \]

Answer 37

watch out - the bounds are referring to x

Answer 38

Let me summarize:
1. I write \[\int\limits_{1}^{1} e^{-\frac{x^2}{2}} dx = \int\limits_{-1}^{1} e^{-u^2} du\]
2. I use \[\int\limits_{0}^{x} e^{-u^2} du=\frac{ \sqrt{\pi} }{ 2 }erf(x)\]
3. I use \[\int\limits_{-1}^{1}e^{-u^2} du=2\int\limits_{0}^{1}e^{-u^2} du\]

From 2 . I know that \[\int\limits_{0}^{1} e^{-u^2} du=\frac{ \sqrt{\pi} }{ 2 } erf(1)\] so\[\int\limits_{-1}^{1} e^{-u^2} du=\sqrt{\pi} \ erf(1)\]

Is this ok? What is the value of erf(1)?

Answer 39

your steps are fine but you forgot to convert the bounds when you did u substitution

Answer 40

yes, right :) thanks

Answer 41

\[\int\limits_{-1}^{1} e^{-\frac{x^2}{2}}
dx ~~\stackrel{u = x/\sqrt{2}}{===}~~\int\limits_{-1/\sqrt{2}}^{1/\sqrt{2}} e^{-u^2}
\sqrt{2} ~du\]

rest should be easy.. keep messing with it till you get the wolfram answer :)

Answer 42

\[\int\limits_{-1}^{1}e^{-\frac{x^2}{2}}dx=\int\limits_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{2}\ e^{-u^2}\ du\]\[=2\int\limits_{0}^{\frac{1}{\sqrt{2}}}\sqrt{2}e^{-u^2}du=2\sqrt{2}\frac{ \sqrt{\pi} }{ 2 }\ erf(\frac{1}{\sqrt{2}})\]
\[=\sqrt{2\pi}\ erf(\frac{1}{\sqrt{2}})\] which is the result from wolfam, but I still don't know how how the value is 1.71125, i.e. I don't know what is the value of\[erf(\frac{1}{\sqrt{2}})\]

Answer 43

you'll need to use calculator
or a table like this :
http://www.geophysik.uni-muenchen.de/~malservisi/GlobaleGeophysik2/erf_tables.pdf

since , 1/sqrt 2 = 0.707
search for 0.7 row and column 0

Answer 44

Actually, I can use this since I have to use it in Matlab.
A big Thank you to all of you. I haven't heard about error function for 5 years of studying and I think that's a shame. Thank you once again!!!

Answer 45

its ok, thats why error function is not call as 'standard' function :P
most of us are not familiar/comfortable using it.
good luck!

Answer 46

Thank you very much :)

Answer 47

from my table I got:
erf(0.71)= 26.11

Answer 48

If so, result wouldn't be the same as in wolfram

Answer 49

so you have to multiply 26.11 by 2 =52.22

Answer 50

and now?

Answer 51

erf(0.70) = 0.6778
multiply that with \(\sqrt{2\pi}\)
and you will have your wolf result :)

Answer 52

my result is multiplied by 100, so I have
0.5222

Answer 53

Thank you :) I think I can't express my thankfulness to all of you. You spent more than one hour helping me. I appreciate. I'm sorry I can give only one medal...

Answer 54

Thank you! @gorica

Answer 55

we're not here for medals :)
you understand, thats our medal,
we're glad to help :)

Answer 56

Then I am sorry I can't send you a chocolate now :D

Answer 57

lol
i'll eat it myself, assuming you sent it ;)

Answer 58

bdw, we won't mind pizza if not chocolate ;P