Question

what is the answer to this question:
y= integral of (e^x^2) dx from 0 to infinity

Answer 1

6?

Answer 2

explain how please....

Answer 3

oh never mind it says from 0 to infinity right?

Answer 4

yes sir

Answer 5

\[\infty\]

Answer 6

please show your steps....thank you

Answer 7

you do the integration by parts, but there is no reason to because this thing gets huge

Answer 8

what zarkon said. think of what the graph of
\[y=e^x\] looks like

Answer 9

no what that this integral converges

Answer 10

yes i know the graph of y=e^x but i need the full solution to this problem

Answer 11

\[e^{x^2}\ge1\] for all \[x\ge 0\]

Answer 12

@zarkon
got a better explanation?

Answer 13

\[\int\limits_{0}^{\infty}e^{x^2}dx\ge\int\limits_{0}^{\infty}1dx=\infty\]

Answer 14

i already know the answer to this problem from wolfram but need to know the steps to solve it

Answer 15

oh i even read the problem wrong. i thought it said
\[\int_0^{\infty} x^2e^x dx\]
i am going to make a bet. my bet is that the question is
\[\int\limits_{0}^{\infty}e^{-x^2}dx\]

Answer 16

that would make for a more interesting problem

Answer 17

yes sir thats the problem

Answer 18

this question is from advanced statistics class

Answer 19

write as a double integral and integrate in polar coordinates

Answer 20

zarkon...please show your method

Answer 21

\[I=\int\limits_{0}^{\infty}e^{-x^2}dx\]

\[I^2=\int\limits_{0}^{\infty}e^{-x^2}dx\int\limits_{0}^{\infty}e^{-y^2}dy\]

\[I^2=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}e^{-(x^2+y^2)}dxdy\]


now switch to polar coordinates

Answer 22

but that would give me r^2 instead of x^2 and put me back to same problem

Answer 23

\[I^2=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}e^{-(x^2+y^2)}dxdy\]

\[I^2=\int\limits_{0}^{\pi/2}\int\limits_{0}^{\infty}e^{-r^2}r\,dr\,d\theta\]

\[I^2=\int\limits_{0}^{\pi/2}\left[-\frac{1}{2}e^{-r^2}\right]_{0}^{\infty}\,d\theta\]

\[I^2=\int\limits_{0}^{\pi/2}\frac{1}{2}\,d\theta=\frac{\pi}{4}\]

\[I=\frac{\sqrt{\pi}}{2}\]