Question

anyone knows integration here?

Answer 1

Of course

Answer 2

and

Answer 3

\[\huge \mathbb{\color{navy}{Welcome~To~Openstudy}}\]

Answer 4

What exactly is your question?

Answer 5

i have to integrate the expression in the attached file

Answer 6

\[\Huge I(x)=\frac{1}{\sigma\sqrt{2\pi}}\int e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrm dx\]

Answer 7

what are you limits of integration?

Answer 8

from a to b

Answer 9

well its a bit like the error function, can you make a subsitution?

\[\operatorname{erf}(x)=\frac2{\sqrt \pi}\int\limits_0^x{e^{-u^2}}{\mathrm du}\]

Answer 10

can anyone provide me all the steps for this integration

Answer 11

@UnkleRhaukus can you provide me all the steps for this integration

Answer 12

let \[\dfrac{x-\mu}{2\sigma^2}=z\\\
\frac{\mathrm dx}{2\sigma^2}=\mathrm dz\\\mathrm dx=2\sigma^2\mathrm dz \]
\[x=a\to z=\dfrac{a-\mu}{2\sigma^2}\\
x=b\to z=\dfrac{b-\mu}{2\sigma^2}\]

Answer 13

\[\huge I(x)=\frac{2\sigma^2}{\sigma\sqrt{2\pi}}\int\limits_{\frac{a-\mu}{\sigma^2}}^{\frac{b-\mu}{\sigma^2}}e^{-z^2}\mathrm dz\]

Answer 14

\[\Large\qquad =\frac{\sqrt2\sigma}{\sqrt{\pi}}\left[\tfrac{2}{\sqrt \pi}\operatorname{erf}(\tfrac{b-\mu}{\sigma^2})-\tfrac{2}{\sqrt \pi}\operatorname{erf}(\tfrac{a-\mu}{\sigma^2})\right]\\
\Large=
\frac{2\sqrt2\sigma}{\pi}\left[\operatorname{erf}(\tfrac{b-\mu}{\sigma^2})-\operatorname{erf}(\tfrac{a-\mu}{\sigma^2})\right]\]

Answer 15

if we take limits from -infinity to plus infinity .Can we prove that area under the curve is equal to one

Answer 16

** Whoops\[\Large\qquad =\frac{\sqrt2\sigma}{\sqrt{\pi}}\left[\tfrac{\sqrt \pi}2\operatorname{erf}(\tfrac{b-\mu}{\sigma^2})-\tfrac{\sqrt \pi}2\operatorname{erf}(\tfrac{a-\mu}{\sigma^2})\right]\\
\Large=
\frac{\sqrt2\sigma}2\left[\operatorname{erf}(\tfrac{b-\mu}{\sigma^2})-\operatorname{erf}(\tfrac{a-\mu}{\sigma^2})\right]\]

Answer 17

now you can use
\[\operatorname{\text{erf}}(∞)=1\] and
\[\operatorname{\text{erf}}(-x)=-\operatorname{\text{erf}}(x)\]

Answer 18

i think i must have made an error with the constants somewhere else as well