Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

plz help me top solve-integrate e^-2(y)^2..limit is 0 to infinity

OpenStudy (anonymous):

please write it with help of equation button

OpenStudy (anonymous):

i just joined the group,donno where it is

OpenStudy (anonymous):

there is a button named 'equation' just below the space to type reply

OpenStudy (anonymous):

\[\int\limits_{0}^{\infty} (e ^{-2y ^{2}})dy\]

OpenStudy (anonymous):

this is the question

OpenStudy (anonymous):

is it some type of error function?

OpenStudy (dumbcow):

yes it is

OpenStudy (anonymous):

limit x approaches infinity?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

HINT: substitue with x = y^2, replace dy, then use 'integration by parts' the replacement for dy involves an x to the power -1/2, and the squared y terms is eliminated from the e, being replaced by -2x.

OpenStudy (anonymous):

NOTE: the limits will stay the same

OpenStudy (anonymous):

the answer has \[\Pi\] in the term,integration by parts wont simply solve the problem

OpenStudy (dumbcow):

you can solve it numerically since it is definite integral --> equals approx 0.63

OpenStudy (anonymous):

i want to know is it representing any function for which there is direct answer,for example we know the results of beta function without solving it analytically

OpenStudy (dumbcow):

i don't think there is for this integral

OpenStudy (anonymous):

this is differential calculus

OpenStudy (anonymous):

answer is \[(1\div2)\sqrt{\pi/2}...anybody who knows what function properties \it used??\]

OpenStudy (anonymous):

anybody who knows what function properties it used??

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!