Asked by Hayden
Evaluate the improper integral
ā« [4,0] 1/(š„ā4)^3 šš„
I am stumped please help
ā« [4,0] 1/(š„ā4)^3 šš„
I am stumped please help
Answers
Answered by
mathhelper
write ā« [4,0] 1/(š„ā4)^3 šš„ as
ā« [4,0] (š„ā4)^-3 šš„
the indefinite integral would be 1/(2(x-4)^2) + c
when we try to evaluate
= (x-4)^-2 / -2 | from 0 to 4
= 1/(2(x-4)^2) | from 0 to 4
= 1/(2(0)..... ahhh, we are dividing by zero
time to look at the graph:
there is a vertical asymptote at x = 4, and x=4 is your upper boundary
You don't have a closed region, so the "area" is infinitely large.
ā« [4,0] (š„ā4)^-3 šš„
the indefinite integral would be 1/(2(x-4)^2) + c
when we try to evaluate
= (x-4)^-2 / -2 | from 0 to 4
= 1/(2(x-4)^2) | from 0 to 4
= 1/(2(0)..... ahhh, we are dividing by zero
time to look at the graph:
there is a vertical asymptote at x = 4, and x=4 is your upper boundary
You don't have a closed region, so the "area" is infinitely large.
Answered by
Chris
I am having problem with a similar one can you explain it further