Y=1/sqrt 2π^e−x2/5 y = 0 x = 0 x = 1

[From: ] [author: ] [Date: 11-06-19] [Hit: ]

y = [ 1/√(2π) ] * e^(−x^2/5)-can u post the question again so i can answer it for you?Report Abuse -i got this question but can you answer this question for me please.http://answers.yahoo.com/question/index;_ylt=Auwlcm73.n96qA4imZjGj03ty6IX;_ylv=3?......

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis. (Round your answer to three decimal places.)

-

do u mean this function:

y = [ 1/√(2π) ] * e^(−x^2/5)

-

can u post the question again so i can answer it for you?

Report Abuse

-

i got this question but can you answer this question for me please.

http://answers.yahoo.com/question/index;_ylt=Auwlcm73.n96qA4imZjGj03ty6IX;_ylv=3?qid=20110619052326AAFZlu7

Report Abuse

1

keywords: minus,sqrt,pi,Y=1/sqrt 2π^e−x2/5 y = 0 x = 0 x = 1

New

Hot